maths chapter wise important questions

1

QUESTION BANK II PUC SCIENCE

I. Very Short answer questions. (1x19=19)

1. Define Symmetric relation.

Ans: A relation „R‟ on the set „A‟ is said to be symmetric if for all a, b, A, aRb Implies

bRa. i.e. (a, b) R (b, a) R

2. Let A = {4, 6, 8, 20} R = {(a, b); a + b = 25, a, b A} Show that the relation „R‟ Is

empty.

Ans: This is an empty set, as no pair (a, b) satisfies the condition a + b = 25.

3. Give an example of a relation defined on a suitable set which is

i. reflexive, symmetric and transitive.

ii. reflexive, symmetric but not transitive.

iii. reflexive, transitive but not symmetric.

iv. symmetric, transitive but not reflexive.

vi. symmetric, but not reflexive and not transitive.

vii. not reflexive, not symmetric, and not transitive.

Solution: Consider a Set A = {a, b, c}

i. Define a relation R, on „A‟ as.

R1 = {(a, a), (b, b), (c, c)}. Clearly R1 is reflexive, symmetric and transitive

ii. Consider the relation R2 on A as.

R2 = {(a, a), (b, b), (c, c), (a, b), (b, a), (a, c), (c, a)}.

R2 is symmetric, reflexive

But R2 is not transitive – for (b, a) R2 and (a, c) R2 but (b, c) R2.

iii. Consider the relation R3 on A as

R3 = {(a, a), (b, b), (c, c), (a, b)}

R3 is reflexive and transitive but not symmetric for (a, b) R3 but (b, a) R3

iv. Consider the relation R4 on A as

R4 = {(a, a), (b, b), (c, c), (b, a)} is symmetric and transitive but not reflexive.

Since (c, c) R4.

v. Consider the relation R5 on A as

R5 = {(a, a), (b, b), (c, c), (a, b), (c, a)}

R5 is reflexive. R5 is not symmetric because (a, b) R5 but (b, a) R5 Also R5 is

not transitive because (c, a) R5, (a, b) R5 but (c, b) R5.

vi. Consider the relation R6 on A by

R6 = {(a, a), (c, c), (a, b), (b, a)}

R6 is Symmetric but R6 is not reflexive because (b, b) R6 . Also R6 is not transitive

for (b, a) R6 , (a, b) R6 but (b, b) R6.

vii. The relation R7 defined on A as

R7 = {(a, b), (b, c) } is not reflexive, not symmetric and not transitive.

* FUNCTION * (One mark question and answers)

4. Let A = {1, 2, 3} B = {4, 5, 6, 7,} and let f = {(1, 4), (2, 5), (3, 6)} be a function from

„A‟ to „B‟. show that „f‟ is not onto.

Ans: 7 B has no pre image in A. so „f‟ is not onto.

5. If, f : R R is defined by f(x) = 4x-1 x R prove that „f‟ is one-one.

Solution: For any two elements x1, x2 R such that f(x1) = f (x2)

we have 4x1–1 = 4x2–1 4x1 = 4x2

x1 = x2 Hence „f‟ is one – one.

6. Define transitive relation.

2

Ans: A relation R on the set “S” is said to be transitive relation if aRb and bRc aRc.

i.e. if (a, b) R and (b, c) R (a, c) R.

7. Let f : R R be the function defined by f (x) = 2x-3 x R . write 1

f

Ans; 1

= f(x) = 2x–3 x = (y + 3)2

y

–1 –11 1

( ) = (y+3) i.e. ( ) = (x+3)2 2

f y f y

8. Define a binary operation on a set.

Ans: Let S be a non-empty set. The function *: S X S S which associates each ordered

pair (a, b) of the elements of S to a unique element of S denotes by a*b is called a

binary operation or a binary composition on S.

9. Determine whether or not each of the definition defined below is a binary operation

justify.

a) on Z+ defined by a*b = | a-b |

b) on Z+ defined by a*b = a

Solution: a) We have for a, b, Z+ a*b = | a – b |. We know the

| a – b | is always positive. + a, b Z , a*b = | a–b | is a positive integer.

Hence * is a binary operation on Z+

b) Cleary a*b = a Z for all ,a b Z

Thus * is a binary operation on Z+

* SHORT ANSINER QUESTIONS* (Answers to two marks questions)

10. Define an equivalence relation. Give on example of a relation which is transitive but

not reflexive.

Solution: A relation “R” on a Set S is called on equivalence relations. if „R‟ is reflexive,

symmetric, and transitive.

Ex. The relation “<” (less than) defined on the set R of all real numbers is transitive,

but not reflexive.

11. Show that the signum function f : R R

defined by

1 x > 0

( ) 0 x = 0

–1 x < 0

if

f x if

if

is neither one-one nor onto.

Solution: From the graph of the function we have

f(2) = 1 and f(3) = 1, i.e. f(2) = f(3) = 1 but 2 3

„f‟ is not one-one

Again for 4 R (co domain) there exists no x R (domain)

Such that f(x) = 4 because f(x) = 1 or –1 for x 0.

„f‟ is not onto.

3

12. State whether the function f : R defined by f(x) = 1 + x2 x R is one-one, onto

or objective justify your answer.

Solution: We have f(x) = 1 + x2 for all x R .

clearly we observe that f(x) = 1 + 22 = 5 and f(–2) = 1 + (–2)

2 = 5 i.e, f(2) = f(–2)

but 2 –2 „f‟ is not one-one.

Again for y –2 there exist no real number x, such that f(x) = –2 because if f(x) = –2

1 + x2 = –2 x

2 = –3

x = –3 R


13. Show that the greatest integer function f : R R defined by f(x) = [x] is neither one-

one nor onto.

Solution: We have f(x) = [x] = greatest integer less than or equal to x.

i.e. [2.3] = 2, [2] = 2 but 2.3 2.

Now we have [x] = 2 for all x [2, 3]

„f‟ is not one-one.

Again 3

2R (codomain), but there existing no x R (domain) such that f(x) = 3

2

because [x] is always an integer x R


14. Cheek the injectivity and surjectivity of the function f : Z Z defined by f(x) = x3

x Z

Solution: We have f(x1) = f(x2) 3 3

1 2 1 2x x x x

„f‟ is injective.

Now 7 Z (codomain) and it has no pre-image in Z because if f(x) = 7 then 3 37 7x x Z

„f‟ is not surjective.

15. If f(x) = | x | and g(x) = | 5x – 2 | then find (i) got and (ii) fog.

Solution: We have f(x) = | x | and g(x) = | 5x–2 |

consider go f(x) = g( | x | ) = 5 | x | – 2

| 5 – 2 | 0

= | –5 – 2 | 0

x x

x x

Now fog(x) = f [g(x)] = f( |5x–2 | )

= | 5x–2 |

16. If f ; R R be given by 3( ) 3 –f x x1

3

find fo f(x).

Solution: We have 1

33( ) 3 –f x x

Now to 1

33( ) ( ) 3 – = f(y)fof x f f x f x

where 1

33 = (3 –y x

1

33( ) ( ) 3 –fof x f y y

1

3 31 13 33 3= 3 – 3 – = (x ) .x x

Hence fo f(x) = x.

4

17. Consider f : {1, 2, 3} {a, b, c} given by f(1) = a, f(2) = b f(3) = C find –1

f and

show that

–1–1

f f

Solution: Consider f : {1, 2, 3} {a, b, c} given by

f(1) = a, f(2) = b and f(3) = C

(1) , (2) and f(3)=4f a f b

1 1

( ) 1, ( ) 2f a f b

and 1

( ) 3f c

–1

,1 , ,2 , ,3 ( )f a b c g say

we observe that ‟g‟ is also objective.

–1

1, , 2, , 3,g a b c f

–1–1 –1

g f f f

18. If f(x) = 4 3 2,

36 – 4

xx

x

show that

(fof) (x) = x for all 23

x . what is the inverse f ?

Solution:

4 34 3

4 ( ) 3 6 – 4( ) = 4 36 ( ) – 4

6. – 46 – 4

x

f x xfof x f f xxf x

x

4 (4 3) 3(6 – 4) 34

6 (4 3) – 4( – 4) 34

x x xx

x x

–1

fof I f f

19. Let f : R R be defined by f(x) = 10x+7, Find the function g : R R such that

gof = fog = Ig

Solution: We have f(x) = 10x+7

By data, gof(x) = Ig(x) g [f(x)] = x.

g [10x+7] = x.

Let y = 10x+7 x – 7

10

y

Then g(10x+7) = – 7

10 710

yg y

20. Consider a function f : : 0,2

f R

given by f(x) = Sinx. and

g : : 0,2

f R

given by g(x) = cosx. Show that „f‟ and „g‟ are one-one but

(f + g) is not one-one.

Solution: Since for any two distinct elements x1 and x2 in 0,2

, sin x1 sin x2

and cos x1 cos x2

Both „f‟ and g are one-one

But (f + g) (o) = f(o) + g(o) = 1 and

5

= f sin cos 12 2 2 2 2

f g g

Therefore (f + g) is not one – one

21. Examine whether the binary operation * defined below are commutative, associative

a. On Q defined loy a * b = ab + 1

b. On Z+ defined loy a * b = a

b

c. On Q defined loy a * b = ab/2

Solution: a) For every a, b, Q we have

i) a * b = ab + 1 Q

Now a * b = ab + 1 = ba + 1 (Usual multiplication is commutative)

= b * a

* is commutative in Q.

ii) consider 4, 5, 6 Q.

4 * ( 5 * 6) = 4 * (30 + 1) = 4 * 31)

= 4 (31) + 1 = 124 + 1 = 125

and (4*5) * 6 = 21 * 6 = (21 x 6) + 1 = 157

a * (b * c) = (a * b) * c is not true for all a, b, c, Q. Hence * is not associative.

b) For all a, b, * Z+, clearly a * b = a

b Z

+

i.e. * is a binary operation in Z+

i) If a then ab b

a

i.e. a * b b * a for a b.

Thus * is not commutative in Z+

ii) Consider 2, 3, 4, Z+

(2 * 3) * 4 = 23 * 4 = 8 * 4 = 8

4 = 2

12

and 2 * (3 * 4) = 2 * 34 = (2)

34 = 2

81

( 2 * 3) * 4 2 * ( 3 * 4)

Thus (a*b)*c = a*(b*c) is not true , , ,a b c Z

Hence * is not associative in Z+.

C. We have a * b = a,b Q2

ab

i) Now a * b = 2

ab = 2

ba = b * a

(Usual multiplication is commutative)

* is commutative in Q.

ii. Now consider a * (b * c) = a * 2

bc

2

42

bca

abc

( * )* *2 4

ab abca b c c

(a * b) * c = a * (b * c) for all a, b, c, Q.

Hence * is associative in Q.

22. Write the multiplicative modulo 12 table for the set A = {1, 5, 7, 11}. Find the

identity element w.r .t X12 X12 1 5 7 11

Solution: The elements in the row corresponding 1 1 5 7 11

to the element 1, coincides with the elements 5 5 1 11 7

of a above the horizontal line in the same order. 7 7 11 1 5

Thus 1 acts as an identify element. 11 11 7 5 1

6

* ANSIEWERS TO THREE MARKS QUESTIONS*

23. The relation R defined on the Set A = {1, 2, 3, 4, 5} by R = {(a, b) ; | a–b | is even}.

Show that the relation „R‟ is on equivalence relation. (consider „O‟ as an even num

ber)

Proof: For any a A we have a–a = o considered as even i.e. (a, a,) R for all a A

Thus R is reflexive relation.

Let (a, b) R |a–b | is even

|b–a | is also even (b, a ) R.

Thus the relation R is symmetric.

Let (a, b) R and (b, c) R.

| a–b | is even and | b–c | even 1 1 2a b k and | b – c | =2l for k, l Z.

| a–b | + | b–c | = 2k + 2 l = 2 (k + l)

a–b + b–c = 2 (n) where n = (k+l)

a–c = 2n

| a–c | = even (a, c) R

„R‟ Is transitive.

Hence „R‟ Is an equivalence relation.

24. Let A = R–{3} and B = R–{1}. Consider the function f : A B defined by f(x) =

23

xx

Is „f‟ is one-one and onto ? Justify your answer.

Solution: We have f(x) = 2

3

x

x

for all x R

Now let f(x1) = f(x2) 1 2

2 2

2 2

3 3

x x

x x

1 2 2 1 1 2 1 22 3 6 2 3 6x x x x x x x x

1 2x x

Thus 1 2 1 2( ) ( )f x f x x x i.e. „f‟ is one-one.

Let y B thus .y l

Now 2

( ) 2 ( 3)3

xf x y y x y x

x

3 23 2 ( 1) ( 1).

1

yy x y x y

y

Also 2 3 2 3

3 for if 31 1

y y

y y

2 3 3 3 2 3y y which is not true.

Thus for every y B there exists

2 3

1

yx A

y

such that ( )f x y

i.e. „f‟ is onto.

Hence „f‟ is bijective function.

25. If „f‟ and g are two functions defined by f(x) = 2x + 5 and g(x) = x2 + 1 find 1 gof(2),

ii) fog(2) and iii) gog(2).

Solution: We have 2( ) 2 5, ( ) 1.f x x g x x

Now ( ) [ ( )] (2 5)gof x g f x g x

7

2( ) (2 5) 1gof x x

24 25 20 1x x

24 20 26x x

2(2) 4(2) 20(2) 26gof

16 40 26 82

ii) 2 2( ) ( 1) 2( 1) 5fog x f x x

2( ) 2 7fog x x

2(2) 2(2) 7 8 7 15fog

iii) 2( ) ( 1)gog x g x

2 2 4 2( 1) 1 2 2x x x

(2) 16 8 2 26gog

26. Let * be the binary operation on N given by a * b = l.c.m of „a‟ and „b‟ Is *

commutative? Is * associative? Find the identify in N w.r.t *.

Solution: We have a * b = l.c.m. of „a‟ and „b‟ ,a b N clearly l.c.m of two positive

integers is a positive integer. Thus N is closed under the operation *.

i) Clearly l.c.m of „a‟ and b = l.c.m. of „b‟ and „a‟

a*b = b*a for all a, b, N

Thus * is commutative.

ii) Let a, b, c, N be arbitrary.

Now a*(b * c) = a*(l.c.m. of b and c)

= l.c.m. of [„a‟ and (l.c.m of b and c)]

= l.c.m of (a,b,c) ................ (1)

Again, (a * b) * c

= (l.c.m of „a‟ and b) * c

= l.c.m of [(l.c.m. of „a‟ and b) and c]

= l.c.m of (a,b,c)................(2)

Thus (a * b) *c = a* (b * c) , ,a b c N

Hence * is associative.

iii) Let e be the identify element in N. Then for all a N, we have

a * e = e * a = a

* = l.c.m of a and e = a

* = l.c.m. of e and a = a

a e

e a

l.c.m of a and e = l.c.m. of e and a = a

for all a N. implies e = 1.

Thus 1 acts as identify element in N.

* FIVE MARKS QUESTIONS AND ANSWERS*

27. Consider : [ 5, )f R defined by 2( ) 9 6 5f x x x show that „f‟ is invertible

with –1 6 1

( )3

yf y

Solution: We have : [ 5, )f R by 2( ) 9 6 5f x x x

Let 1,a a R Such that f(a1) = f(a2)

2 2

1 1 2 29 6 5 9 6 5a a a a

8

2 2

1 2 1 29 9 6 6 0a a a a

1 2 1 2 1 29 6 0a a a a a a

1 2

2 2

1 2 9 9 6 0a a a a

1 2 1 2 1 20 9 9 6 as ,a a a a o a a R

1 2a a Thus „f‟ is one-one.

Let 2( ) 9 6 5y f x x x

29 6 5 0x x y

6 36 36( 5)

18

yx

1 1 5 1 6

3 3

y yx

1 1 6( )

3

yf y

For every element ,y B there exists a pre-Image x in [ 5, . So „f‟ is onto Thus „f‟

is one-one and onto and therefore invertible Hence inverse function of „f‟ is given by

1 1 6( )

3

yf y

28. If :f A A defined by 4 3

( )6 4

xf x

x

where 2 .

3A R show that „f‟ is

invertible and 1

f f

Solution: We have 4 3

( )6 4

xf x

x

with 2

3x

Now 1 21 2

1 2

4 3 4 3( ) ( )

6 4 6 4

x xf x f x

x x

1 2 1 2 1 2 1 224 16 18 12 24 18 16 12x x x x x x x x

2 1 1 234 34x x x x

Thus „f‟ is one-one.

Now let y A and f(x) ( )f x y

y A and 4 3

6 4

xy

x

y A and 4 3

6 3

yx

x

Thus for every y A , there exists

4 3 236 3

yx A y

y

such that

( ) .f x y That is „f‟ is onto.

Hence „f‟ is bijective.

Now 1

( ) ( )f x y f y x

9

4 3 4 3

6 4 6 4

y xx y

x x

Thus 1

:f A A

is defined by 1 4 3( )

6 4

xf x

x

Clearly 1

.f f

29. Let f; w :f w w be defined by

1 n is odd

( )1 if 'n' iseven

nf n

n

Show that „f‟ is invertible and find

The inverse of „f‟. Hence w is the set of all whole numbers.

Solution: Let x1, x2 w be arbitrary

Let x1 and x2 are even.

1 2 1 2 1 2( ) ( ) 1 1f x f x x x x x

Let 1x and 2x are odd

1 2 1 2 1 2( ) ( ) 1 1f x f x x x x x

Let x1 and x2 are odd 1 2 1 2( ) ( ) 1 1f x f x x x

1 2x x

In both the cases, 1 2 1 2( ) ( )f x f x x x

Supposing 1x is odd and 2x is even

then 1 2x x . Now 1 1( ) 1f x x and 2 2( ) 1f x x

Also 1 21 1x x

i.e. 1 2 1 2 ( ) ( )x x f x f x

Hence „f‟ is one-one

Let 0m w (co domain) and f(n) = m

Now 1 if 'n' is odd

( ) ( )1 if 'n' is even.

n mf n m f n

n m

1 if 'n' is odd

( )1 if 'n' is even.

n mf n

n m

Thus for every m w (co domain)

there exists n w (= m+1 or m–1)

Such that f(n) = m.

Also (considering O as even)

( ) 0 1 1f o i.e. 1

(1) 0f

Thus „f‟ is onto

Hence „f‟ is a bijection.

Let 1

( ) ( ) .f n m f m n

1 is odd

( )1 is even

m n mf m

m n m

1 is even

( )1 is odd

m n nf m

m n n

10

Thus 1 1 is even( )

1 is odd

n nf n

n n

Is inverse function.

INVERSE TRIGONOMETRIC FUNCTIONS

1 MARK QUESTIONS WITH SOLUTIONS

1. Find the principal value of (

√ )

Solution let (

√ )

√

(

√ )

2. Find the principal value of √

( √ )

√

(

)

( √ ) (

)

3. Find the principal value of (

)

Solution (

)

it does not lie b/w –

–

(

) * ( —

)+ * (

)+

(

)

solution

( (

))

, it does not lie b/w 0 and

( (

)) ( (

))

= (

)

=

(

)

5. ( √ )

√

√

( √ )

6. Evaluate ( ) [– ]

[– ]

[ ( )]

7. Find the value of

Solution we have

8. Find the value of

(

)

2 MARKS QUESTIONS WITH SOLUTIONS

(

) (

)

(

)

(

)

(

)

(

)

*

+

–

(

)

(

)

(

)

(

) (

)

*

–

+ (

)

*

+

(

)

*

+

(

)

* (

)+ (

)

(

)

(

)

(

)

(

)

(

)

(

)

√

√

√ –

(√

)

(√

) (

√

)

(

)

(

) (

) (

)

(

)

.

(

) (

)

(

) (

)

(

) *

(

)

(

)+

[ (

)]

(

)

(

)

(

)

(

)

3 marks Question with solutions

(

) (

)

(

) (

)

*

(

) (

)+

*

(

) (

)+

(

+

–

√

√ √

√

.

√ √

√

(

)

(

)

Topic: Matrices

Question bank with solutions

One mark question ( V S A)

1. Define matrix

2. Define a diagonal matrix

3. Define scalar matrix

4. Define symmetric matrix

5. Define skew-symmetric matrix

6.In a matrix [

2 5 19 −17

35 −25

2 12

√3 1 −5 17

]

find 1) order of the matrix

2) Write the elements of 𝑎13 , 𝑎21 , 𝑎33 , 𝑎24 , 𝑎23

7. If a matrix 8 elements what is the possible order it can have ?

8. If a matrix 18 elements what is the possible order it can have?

9. construct 2 × 2 matrix [𝑎𝑖𝑗] whose elements are given by

1) 𝑎𝑖𝑗 = (𝑖 + 𝑗) 2 2) 𝑎𝑖𝑗 = (𝑖+𝑗) 2

2

10. construct the 2 × 3 matrix whose elements are given by 𝑎𝑖𝑗= |𝑖 − 𝑗|

11. Construct the 3× 2 matrix whose elements are given by 𝑎𝑖𝑗= 𝑖

𝑗

12. Find x, y, z if [4 3𝑥 5

]= [𝑦 𝑧1 5

]

13. Find x, y, z if [𝑥 + 𝑦 25 + 𝑧 𝑥𝑦

]= [6 25 8

]

14. Find the matrix x such that 2A + B + X =0 where A = [−1 23 4

] and B = [3 −21 5

]

15. If A = [1 2 32 3 1

] B = [3 −1 3

−1 0 2] Find 2A – B

16. Find X if Y =[3 21 4

] and 2X+Y = [1 0

−3 2]

17. Find X If X+Y = [7 02 5

] and X-Y = [3 00 3

]

18. Simplify cos 𝜃 [cos 𝜃 sin 𝜃−sin 𝜃 cos 𝜃

] + sin 𝜃 [sin 𝜃 −cos 𝜃cos 𝜃 sin 𝜃

]

19. Find X If 2[1 30 𝑥

] + [𝑦 01 2

] = [5 61 8

]

20. If A = [1 23 4

] Find A + 𝐴1

21. A = [1 −2 30 1 4

] and B = [0 2 56 −3 1

] Find 3A + 2B

22. if A = [sin 𝜃 cos 𝜃

−cos 𝜃 sin 𝜃] Verify A A1 = I

23. if B = [cos 𝜃 sin 𝜃−sin 𝜃 cos 𝜃

] verify B B1= I

24. If A = [012] B = [1 5 7] Find AB

25. Compute 1) [1 −22 3

] [1 2 32 3 1

]

2) [3 −1 3

−1 0 2] [

2 −31 03 1

]

26. Find X and Y [2𝑥 + 𝑦 3𝑦

6 4] = [

6 06 4

]

27. What is the number of possible square matrix order 3 with each entries 0 or 1

28. Find X and Y if [5 − 𝑥 2𝑦 − 8

0 3] is a scalar matrix

29. Find X [4 𝑥 + 2

2𝑥 − 3 𝑥 + 1] is a symmetric matrix

II. Two mark and Three marks questions (SA)

1.Radha , fauzia, simran are the student of 12th class Radha has 15 note book and 6 pens ,

Fauzia has 10 books 2 pens and Simran has 13 books and 5 pens express this in to matrix

forms.

2. Construct 3× 2 matrix whose elements are given by 𝑎𝑖𝑗 =1

2 |𝑖 − 3𝑗|

3. Find X,Y,Z from the equation [

𝑥 + 𝑦 + 𝑧𝑥 + 𝑧𝑦 + 𝑧

] = [957]

4. Find a,b,c, d From the equation [𝑎 − 𝑏 2𝑎 + 𝑐2𝑎 − 𝑏 3𝑐 + 𝑑

] = = [−1 50 13

]

5. If A = [8 04 −23 6

] B = [2 −24 2

−5 1] Find X such that 2A + 3X = 5B

6. Find X and Y 2[𝑥 57 𝑦 − 3

] + [3 −41 2

] = [7 615 14

]

7. Find X and Y if x [23] + 𝑦 [

−11

]= [105

]

8. Given 3 [𝑥 𝑦𝑧 𝑤

] = [𝑥 6

−1 2𝑤] + [

4 𝑥 + 𝑦𝑧 + 𝑤 3

] Fine the values of X,Y,Z and W

9. If 𝐴𝑋 = [cos 𝑥 sin 𝑥−sin 𝑥 cos 𝑥

] and 𝐴𝑌 = [cos 𝑦 sin 𝑦−sin 𝑦 cos 𝑦

] Show that 𝐴𝑋𝐴𝑌 = 𝐴𝑋+𝑌

10. If A = [3 −24 −2

] and I = [1 00 1

] Find K If A2 = KA – 2I

11. If A = [3 √3 24 2 0

] and B =[2 −1 21 2 4

] verify (A + B )1 = A1 +B1

12. For any matrix A with real number entries , A+ A1 is symmetric matrix and A – A1

Skew-symmetric matrix

13. For any matrix A = [1 56 7

] verify that A+ A1 is symmetric matrix

14. For any matrix A = [1 56 7

] verify that A – A1 Skew-symmetric matrix

15. If A and B be the invertible matrices of same order then (AB)-1 = B-1A-1

16. By using elementary operation Find the inverse of the matrix [1 22 −1

]

17. By using elementary operation Find the inverse of the matrix [1 32 7

]

18. By using elementary operation Find the inverse of the matrix [1 −22 1

]

19. Find P-1 if it exists and P = [10 −2−5 1

]

20. If A = [3 1

−1 2] Show that A2 -5A +7I = 0

21. If A = [2 30 −4

] and B = [1 52 0

] Show that (AB)1 = B1A1

III. Five mark questions ( LA)

1.If A = [1 1 −12 0 33 −1 2

] B = [1 30 2

−1 4] and C =[

1 2 3 −42 0 −2 1

]

Find A B , BC and show that (AB )C = A(BC)

2. If A = [0 6 7

−6 0 87 −8 0

] B = [0 1 11 0 21 2 0

] C = [2

−23

] calculate AC, BC and (A+B) C

Deduce that (A+B) C = AC + BC

3. If A = [1 2 33 −2 14 2 1

] Show that A3 – 23A - 40 I = 0

4. If A = [1 2 −35 0 21 −1 1

] B = [3 −1 24 2 52 0 3

] and C = [4 1 20 3 21 −2 3

]

verify A+ (B-C) = (A+B ) –C

5. If A =

[ 2

31

5

31

3

2

3

4

37

32

2

3]

and B =

[ 2

5

3

51

1

5

2

5

4

57

5

6

5

2

5]

find 3A – 5B

6. If A = [2 0 12 1 31 −1 0

] find A2 – 5 A + 6 I ?

7. If A = [1 0 20 2 12 0 3

] prove that A3 – 6A2 + 7A + 2I = 0

8. Express the matrix B = [2 −2 −4

−1 3 41 −2 −3

] Find the sum of symmetric and skew-

symmetric matrix

9. Express the matrix B = [6 −2 2

−2 3 −12 −1 3

] Find the sum of symmetric and skew-

symmetric matrix

10. If A = [1 22 1

] B = [2 01 3

] C = [1 12 3

] calculate AB , BC, A(B+C)

Verify that AB + AC = A(B+C)

11. If F(x) = [cos 𝑥 − sin 𝑥 0sin 𝑥 cos 𝑥 0

0 0 1] show that F(x) F(y) = F(x+y)

12. If A =[−245

] and B = [1 3 −6] verify (AB)1 = B1A1

13. If A = [cos 𝜃 sin 𝜃−sin 𝜃 cos 𝜃

] Prove that An = [cos 𝑛𝜃 sin 𝑛𝜃−sin 𝑛𝜃 cos 𝑛𝜃

]

********

Solutions

One mark questions (VSA)

1. The numbers arranged in rectangular array of rows and columns by the brockets is

called matrix

2. A square matrix is said to be diagonal matrix if all non diagonals elements are zeros

3. A diagonal matrix is said to be scalar marics if it’s diagonal elements are equal

4. If a square matrix A = [𝑎𝑖𝑗]𝑚×𝑚is said to be symmetric if and only if A1 = A

5. If a square matrix A = [𝑎𝑖𝑗]𝑚×𝑚is said to be skew-symmetric if and only if A1 = -A

6. 1) order of the matrix is 3 × 4 2) 19, -2, -5 , 12, 5

2

7. Possible orders are (1,8) (8,1) (2,4) (4,2) is 1X8 , 8X 1 , 2X 4 , 4X 2

8. Possible orders are (1,18) (18,1) (3,6) (6,3) (2,9) (9,2) is 1 X18 , 18X1 , 3X6 , 6X3 ,

2X9, 9X2 ,

9. 1) [𝑎𝑖𝑗] = [4 99 16

] 2) [𝑎𝑖𝑗] = [2

9

29

28]

10. [0 1 21 0 2

]

11. [

11

2

2 1

33

2

]

12. X= 1 Y = 4 Z= 3

13. X = 2 Y = 4 Z = 0

14. X = -2A - B = [2 −4

−6 −8] - [

3 −21 5

] = [−1 −2−7 −13

]

15. 2A –B = [−1 5 35 6 0

]

16. By solving X = [−1 −1−2 −1

]

17. By solving above matrix X = [5 01 4

] and Y = [2 01 1

]

18. By multiplying we get the answer [1 00 1

] = I

19. 2+Y = 5 implies Y = 3 and 2x+2 = 8 implies x =3

20. A + A1 = [2 55 8

]

21. 3 A +2 B = [3 −2 1912 −3 14

]

22. A A1 = [sin 𝜃 cos 𝜃

−cos 𝜃 sin 𝜃] [

sin 𝜃 −cos 𝜃cos 𝜃 sin 𝜃

] = [1 00 1

] = I after multiplying

23. BB 1 = [cos𝜃 sin 𝜃−sin 𝜃 cos 𝜃

] [cos 𝜃 −sin 𝜃sin 𝜃 cos𝜃

] = [1 00 1

]= I after multiplying

24. (AB)1 = [0 1 20 5 100 7 14

]

25. 1) [−3 −4 18 13 3

] 2) [14 −64 5

] after multiplying

26. 3 |𝐴| = K |𝐴| implies K = 3

27. Y = 0, X = 3 by solving

28. The square matrix of order 3X3 = 9 and 2 entries

Then possible entries is 29 = 512

29. [5 − 𝑥 2𝑦 − 8

0 3]= [

3 00 3

] then X = 2 Y = 4

30. [4 𝑥 + 2

2𝑥 − 3 𝑥 + 1] = [

4 2𝑥 − 3𝑥 + 2 𝑥 + 1

] implies X = 5

Solutions : Two mark and Three marks questions (SA)

1. books pens

Radha : 15 6 this can be expressed as [15 610 213 5

] or [15 10 136 2 5

]

Fauzia : 10 2

Simran: 13 5

2. 𝑎𝑖𝑗 =

[ 1

5

21

22

03

2]

3. X+ Y + Z = 9 X +Z = 5 Y + Z = 7

7 + Z = 9 X + 2 = 5 Y + 2 = 7

Z= 2 X = 3 Y = 5

4. By solving equality a =1 , b= 2, c =3 and d = 4

5. X =

[ −2 −

10

3

414

3

−31

3

−7

3 ]

6. compare two matrices X = 2, Y = 9

7. by solving we get X = 3 , Y = -4

8. by solving and compare we get X = 2 , Y = 4, Z = 1, w = 3

9. 𝐴𝑋 𝐴𝑌 = [cos 𝑥 sin 𝑥−sin 𝑥 cos 𝑥

] [cos 𝑦 sin 𝑦−sin 𝑦 cos 𝑦

] = [cos(𝑥 + 𝑦) sin(x +𝑦)−sin(𝑥 + 𝑦) cos(𝑥 + 𝑦)

] = 𝐴𝑋+𝑌

10. A2 = KA – 2I

[1 −24 −4

] = [3𝑘 − 2 −2𝑘

4𝑘 −2𝑘 − 2]

Then 4K = 4

K = 1

11. ( A+B)1 = [5 5

√3 − 1 44 4

] and A1 + B 1 = [5 5

√3 − 1 44 4

]

Hence ( A+B)1 = A1 + B 1

12. B = A + A1 , B1 = (A + A1 )1 = A1+A = B ∴ B = A +A1 is symmetric

C = A –A 1 , C 1 = (A -A1 )1 = A1-A = -(A- A1) = - C ∴ C = A – A1 is skew- symmetric

13. Z = A + A1 = [1 56 7

] + [1 65 7

] = [2 1111 14

] = Z 1 ∴ Z = Z1 = A + A1 is symmetric

14. Z1 = (A - A1)1 = ([1 56 7

] − [1 65 7

])1

= ([0 −11 0

] )1

= -Z ∴ Z1 = - Z, A –A1

skew- symmetric

15. (AB) (AB)-1 = I

A-1(AB) (AB)-1 = A-1I I A = A

B(AB)-1 = A-1 IA-1 = A-1

B-1B(AB)-1 = B-1A-1 AA-1 = I

(AB)-1 = B-1A-1 BB-1 I

16. A = [1 22 −1

]

A = IA

[1 22 −1

] = [1 00 1

] A 𝑅2 = 𝑅2 − 2𝑅1

[1 20 −5

] = [1 0

−2 1]A 𝑅2 = −

1

5𝑅2

[1 20 1

] = [1 02

5−

1

5

] A 𝑅1 = 𝑅1 − 2𝑅2

[1 00 1

] = [

1

5

2

52

5−

1

5

] A

∴ A -1 = [

1

5

2

52

5−

1

5

]

17. By above process ∴A -1 = [7 −3

−2 1]

18. By above prose’s ∴A -1 = [

1

5

2

5−2

5

1

5

]

19. P = [10 −2−5 1

]

P = IP

[10 −2−5 1

] = [1 00 1

] P

By elementary operation

[1 −

1

5

0 0] = [

1

100

1

21] p

p-1 does not exists

20. A2 -5A +7I = [8 5

−5 3] - [

15 5−5 10

] + [7 00 7

] = [0 00 0

] = 0

21. By mathematical induction we get the solution

22. If A=A1 , B = B1 , (AB)1 = AB

(AB)1 = B1 A1 = BA ∴ AB = BA AB Is symmetric

23. A2 = A A By product of two matrix get the solution

24. (AB)1 =[8 −810 0

]

B1A1 = [8 −810 0

]

∴ (AB)1 = B1A1

25. By solving x = 2 , y = 4 , z = 3

Solutions : Five mark questions (LA)

1.AB =[2 1

−1 181 5

] (AB) C = [4 4 4 −735 −2 −39 2231 2 −27 11

]

BC = [7 2 −3 −14 0 −4 27 −2 −11 8

] A(BC) = [4 4 4 −735 −2 −39 2231 2 −27 11

]

Hence (AB) C = A(BC)

2. (A +B) C = [102028

] AC = [91230

] BC = [18

−2] AC + BC = [

102028

]

Hence (A +B) C = AC + BC

3. A = [1 2 33 −2 14 2 1

] A2 = [19 4 81 12 814 6 15

] A3 = [63 46 6960 −6 2392 46 63

]

LHS = A3 – 23A – 40 I = 0 By simplification

4. A + (B – C) = [0 0 −39 −1 52 1 1

] and (A+B) –C = [0 0 −39 −1 52 1 1

]

Hence A + (B – C) = (A+B) –C

5. 3A -5B = [2 3 51 2 47 6 2

] - [2 3 51 2 47 6 2

] = [0 0 00 0 00 0 0

] = 0

6. A2 – 5A + 6I = [1 −1 −3

−1 −1 −10−5 4 4

] by simplification

7. If A = [1 0 20 2 12 0 3

] by calculating A2 , A3 take LHS = RHS

8. B = [2 −2 −4

−1 3 41 −2 −3

] by theorem number 2

B = 1

2 (B +B 1) +

1

2 (B -B 1) hence they are equal

9. B = [6 −2 2

−2 3 −12 −1 3

] by theorem number 2

B = 1

2 (B +B 1) +

1

2 (B -B 1) hence they are equal

11. If AB = [4 65 3

] AC = [5 74 5

] A(B+C) = [9 139 8

] = AB + AC

12. F(x).F(y) = [cos 𝑥 − sin 𝑥 0sin 𝑥 cos 𝑥 0

0 0 1] [

cos 𝑦 − sin 𝑦 0sin 𝑦 cos 𝑦 0

0 0 1]

= [cos(𝑥 + 𝑦) − sin(𝑥 + 𝑦) 0sin(𝑥 + 𝑦) cos(𝑥 + 𝑦) 0

0 0 1

] = F(x+y)

13. LHS = (AB)1 = [−2 4 5−6 12 1512 −24 −30

] = B1A1 = RHS

14. By mathematical induction we get the solution

1

Question Bank

5. CONTINUITY AND DIFFERRENTIABILITY

ONE MARK QUESTIONS

1. Find the derivative of 2cos( )x with respect to x.

Sol: let 2cos( )y x

2 2 2sin( ) ( ) 2 sin( )dy d

x x x xdx dx

2. Find the derivative of 2 3 4 5x x x x xe e e e e with respect to x.

Sol : let 2 3 4 5x x x x xy e e e e e

2 3 4 52 3 42 3 4 5x x x x xdy

e xe x e x e x edx

3. Find the derivative of log(logx) with respect to x.

Sol : log(log )y x

1 1

(log )log log

dy dx

dx x dx x x

4. Find the derivative of cos(sinx) with respect to x.

Sol : cos(sin )y x

sin(sin ) (sin ) cos sin(sin )dy d

x x x xdx dx

5. Find the derivative of sec(tan )x with respect to x

Sol : sec(tan )y x

2sec tan tan tan sec ( )

sec tan tan tan (tan )2

x x xdy dx x x

dx dx x

6. Find the derivative of the function cos( )x with respect to x.

Sol : cosy x

sin

sin2

xdy dx x

dx dx x

2

7. If 2 33 2x xy e e find dy

dx

Sol : 2 33 2x xy e e

2 3 2 3 2 33 2 6 6 6( )x x x x x xdy d de e e e e e

dx dx dx

8. Find the derivative of 5cosx – 3sinx with respect to x.

Sol : 5cos 3siny x x

5sin 3cosdy

x xdx

9. The function 1

( )5

f xx

is not continuous at x = 5. Justify the statement.

Sol : 1

( )5

f xx

is a quotient function. The function f(x) is not defined at x = 5 because

1 1(5)

5 5 0f

is not defined. Therefore f(x) is continuous for all values of x except x = 5.

10. Find dy

dx if x y

Sol : x y

d d

x ydx dx

( ) ( ) 0d d

x ydx dx

1dy

dx

11. If tan(2 3)y x find dy

dx

Sol : tan(2 3)y x

2 2sec (2 3) 2 3 2sec (2 3)dy d

x x xdx dx

12. Find the derivative of f given by 1( ) tanf x x assuming it exists.

Sol : 1tany x 2

1

1

dy

dx x

3

13. Prove that the function ( ) nf x x is continuous at x = n, where n is a positive integer.

Sol : ( ) nf x x , n N . Here, f(x) is a polynomial function and fD R

lim ( ) lim ( ).n n

x n x nf x x n f n

Therefore f(x) is continuous at n N .

14. Find the derivative of 1sin xe

with respect to x.

Sol : 1sin xy e

1

1sin

sin 1

2sin

1

xxdy d e

e xdx dx x

15. Find dy

dx, if 2x at , 2y at .

Sol : 2x at , 2y at

2dx

atdt

, 2dy

adt

2 1

2

dy

dy adtdxdx at t

dt

TWO MARK QUESTIONS:

1. Examine whether the function f given by 2( )f x x is continuous at x = 0.

Sol : 2( )f x x at x = 0; (0) 0f .

Then 2

0 0lim ( ) lim 0x x

f x x

0

lim ( ) 0 (0)x

f x f

.

f is continuous at x = 0.

2. Discuss the continuity of the function f defined by 1

( )f xx

, 0x .

Sol : Fix any non zero real number c, we have 1 1

lim ( ) limx c x c

f xx c

4

Also, since for 0c , 1

( )f cc

, we have lim ( ) ( )x c

f x f c

and hence, f is continuous at every

point in the domain of f. Thus f is continuous function.

3. Find the derivative of the function sin

xey

x with respect to x.

Sol : sin

xey

x

2

sin sin

sin

x xd dx e e x

dy dx dx

dx x

2 2

sin cossin cos

sin sin

xx x e x xdy e x e x

dx x x

4. Discuss the continuity of the function f given by 3 2( ) 1f x x x

Sol : Clearly f is defined at every real number c and its value at c is 3 2 1c c . We also know that

3 2 3 2lim ( ) lim 1 1x c x c

f x x x c c

Thus lim ( ) ( )x c

f x f c

, and hence f is continuous at every real number. This means f is a

continuous function.

5. Verify Rolle’s theorem for the function 2 2y x , a = -2 and b = 2

Sol : The function 2 2y x is continuous in [- 2 , 2] and differentiable in (-2, 2). Also

( 2) (2) 6f f and hence the value of ( )f x at -2 and 2 coincide. Rolle’s theorem states that

there is a point ( 2,2)c , where |( ) 0f c . Since |( ) 2f x x , we get c = 0. Thus at c = 0, we

have |( ) 0f c and 0 ( 2,2)c .

6. If f and g be two real functions continuous at real number c. Then show that f + g is continuous at

x = c.

Sol : The continuity of f + g at x = c, clearly it is defined at x = c we have

lim lim ( ) ( )x c x c

f g x f x g x

lim ( ) lim ( ) ( ) ( )x c x c

f x g x f c g c

( )f g c

Hence f + g is continuous at x = c.

5

7. Find dy

dx if, cosx a , siny a .

Sol : cosx a , siny a

sindx

ad

cos

dya

d

cos

cotsin

dy

dy addxdx a

d

8. Discuss the continuity of the function f given by ( ) | |f x x at x = 0.

Sol : By definition , if x<0

( ), if x 0

xf x

x

Clearly the function is defined at x = 0 and f(0) = 0.

Let hand limit of f at x = 0 is 0 0

lim ( ) lim( ) 0x x

f x x

Right hand limit of f at x = 0 is 0 0

lim ( ) lim 0x x

f x x

.

Thus the left hand limit, right hand limit and the value of the function coincide at x = 0.Hence, f is

continuous at x = 0.

9. Find the derivative of the function sin( )

cos( )

ax b

cx d

with respect to x.

Sol : sin( )

cos( )

ax by

cx d

2

cos( ) sin( ) sin( ) cos( )

cos ( )

d dcx d ax b ax b cx d

dy dx dx

dx cx d

2

cos( )cos( ) sin( )sin( )

cos ( )

dy a cx d ax b c ax b cx d

dx cx d

10. Discuss the continuity of sine function.

Sol : ( ) sinf x x is defined for every real number. Let c be a real number. Put x = c + h.

If x c we know that 0h . Therefore

lim ( ) limsinx c x c

f x x

6

0 0

limsin( ) lim sin .cosh cos .sinhh h

c h c c

0 0

limsin .cosh limcos .sinh sin 0 sin ( )h h

c c c c f c

Thus lim ( ) ( )x c

f x f c

Therefore f is a continuous function.

11. Differentiate sin xx 0x with respect to x.

Sol : sin xy x

Taking log on both sides

sinlog log xy x

log sin logy x x

1

sin (log ) (sin ) logdy d d

x x x xy dx dx dx

1 sin

cos logdy x

x xy dx x

sin

cos logdy x

y x xdx x

sin sincos logxdy x

x x xdx x

12. Differentiate the function 1sin tan xe with respect to x.

Sol : 1sin tan xy e

1 1cos tan tanx xdy de e

dx dx

1

2

cos tan

1

x x

x

e edy

dx e

13. If 22x at , 4y at find dy

dx

Sol : 22x at , 4y at

4dx

atdt

, 34dy

atdt

7

3

24

4

dy

dy atdt tdxdx at

dt

14. If x yxy e , prove that ( 1)

( 1)

dy y x

dx x y

Sol : x yxy e

Differentiating both sides with respect to x.

( ) x yd dxy e

dx dx

1x ydy dyx y e

dx dx

x y x ydy dyx y e e

dx dx

x y x ydy dyx e e y

dx dx

x y x ydyx e e y

dx

( 1)

( 1)

x y

x y

dy e y xy y y x

dx x e x xy x y

15. If cos cos2 cos3y x x x find dy

dx

Sol : cos cos2 cos3y x x x

Taking log on both sides, we get

log log(cos cos2 cos3 )y x x x

log logcos logcos2 logcos3y x x x

1 sin 2sin 2 3sin3

cos cos 2 cos3

dy x x x

y dx x x x

cos cos 2 cos3 tan 2 tan 2 3tan3dy

x x x x x xdx

16. If x y a prove that dy y

dx x

8

Sol : x y a

Differentiate w.r.t x we get

1 1

02 2

dy

dxx y

1 1

2 2

dy

dxy x

dy y

dx x

17. Find the derivative of sin cos

sin cosx x

x x

with respect to x.

Sol : (sin cos )

sin cosx x

y x x


(sin cos )log log(sin cos ) x xy x x

log (sin cos )log(sin cos )y x x x x

Differentiate w.r.t x

1

sin cos log(sin cos ) (sin cos ) log(sin cos )dy d d

x x x x x x x xy dx dx dx

1 cos sin

(sin cos ) (cos sin ) log(sin cos )sin cos

dy x xx x x x x x

y dx x x

(sin cos )

sin cos cos sin 1 log(sin cos )x xdy

x x x x x xdx

18. If 1sinx

y x find dy

dx

Sol : 1sinx

y x

Taking logarithm on both sides

1log log sinx

y x

1log log siny x x

1 11log sin ( ) log sin

dy d dx x x x

y dx dx dx

1

1 2

1log sin

sin 1

dy xx

y dx x x

9

1

1 2log sin

sin 1

dy xy x

dx x x

= 1 1

1 2sin log sin

sin 1

x xx x

x x

19. If sin logey x prove that 21 ydy

dx x

Sol : sin logey x

cos log

cos log loge

e e

xdy dx x

dx dx x

221 sin(log 1e x ydy

dx x x

THREE MARK QUESTIONS:

1. If 2

1

2

1cos

1

xy

x

0 < x < 1 Find

dy

dx

Sol : 2

1

2

1cos

1

xy

x

Put tanx

2

1 1

2

1 tancos cos cos 2

1 tany

12 2tany x

Differentiate w.r. t x

2

1

1

dy

dx x

2. If 3 2 2 3 81x x y xy y . Find dy

dx

Sol : 3 2 2 3 81x x y xy y


2 2 2 23 2 2 3 0dy dy dy

x x xy x y y ydx dx dx

2 2 2 22 3 3 2dy

x xy y x xy ydx

10

2 2

2 2

3 2

2 3

x xy ydy

dx x xy y

3. Diffentiate 2

2

3 4

3 4 5

x xy

x x

w.r. t x.

Sol : 2

2

3 4

3 4 5

x xy

x x

Taking logarithm on both sides, we have

2 21log log 3 log 4 log 3 4 5

2y x x x x

Differentiating on both sides w.r.t x, we get

2 2

1 1 1 2 6 4

2 3 4 3 4 5

dy x x

y dx x x x x

2 2

1 2 6 4

2 3 4 3 4 5

dy y x x

dx x x x x

2

2 2 2

3 41 1 2 6 4

2 3 4 5 3 4 3 4 5

x xdy x x

dx x x x x x x

4. Find dy

dx if 1

2

2sin

1

xy

x

Sol : 1

2

2sin

1

xy

x

Put tanx

1 1

2

2 tansin sin sin 2

1 tany

12 2tany x

Differentiating w.r.t x, we get

2

2

1

dy

dx x

5. Differentiate the function cos

logx


11

Sol : cos

logx

y x


cos

log log logx

y x

log cos log logy x x

Differentiating w.r. t. x on both sides, we get

1

cos log log cos log logdy d d

x x x xy dx dx dx

1 cos

sin log loglog

dy xx x

y dx x x

coscos cos

sin log log log sin log loglog log

xdy x xy x x x x x

dx x x x x

6. Find dy

dx if x yy x

Sol : x yy x


log logx yy x

log logx y y x

Differentiating with respect to x, on both sides, we get

(log ) log ( ) (log ) log ( )d d d d

x y y x y x x ydx dx dx dx

log logx dy y dy

y xy dx x dx

log logx y x dy y x y

y dx x

log

log

y y x ydy

dx x x y x

7. Differentiate 2sin x with respect to cos xe

Sol : let 2sinu x and cos xv e


12

2sin cosdu

x xdx

and cossin xdvxe

dx

cos cos

2sin cos 2cos

sin x x

du

du x x xdxdvdv xe e

dx

8. Differentiate 1 sintan

1 cos

x

x

with respect to x.

Sol : 1 sintan

1 cos

xy

x

1 1 1

2

2sin cos sin2 2 2tan tan tan tan

22cos cos

2 2

x x xx

yx x

2

xy

Differentiating w.r.t x on both sides

1

2

dy

dx

9. Verify mean value theorem if 3 2( ) 5 3f x x x x in the interval [a, b] where a = 1 and b = 3. Find

(1,3)c for which |( ) 0f c .

Sol: Given 3 2( ) 5 3f x x x x [1,3]x which is a polynomial function.

Since a polynomial function is continuous and derivable at all x R

(1) f(x) is continuous on [1,3] (2) f(x) is derivable on (1, 3)

Therefore condition of mean value theorem satisfied on [1,3]. Hence, at least one real (1,3)c

3 2

|3 5(3) 3(3) 1 5(1) 3(1)(3) (1)

( )3 1 2

f ff c

| 20( ) 10

2f c

| 2( ) 3 10 3f x x x ; | 2( ) 3 10 3 10f c c c

23 10 7 0c c

23 7 3 7 0c c c ;

13

(3 7) (3 7) 0c c c c = 1 (1,3) 7

(1,3)3

c .

Hence the mean theorem satisfied for given function in the given interval.

10. If 1cosy x find 2

2

d y

dx in terms of y alone.

Sol : 1cosy x

cosx y

Differentiating w.r.t y, we get

sindx

ydy

cosdy

ecydx

Again differentiating w.r. t x , we get

2

2cos cot

d y dyecy y

dx dx

22

2cos cot

d yec y y

dx

11. Find the derivative of log

logx


Sol : log

logx

y x


log

log log logx

y x

log log log logy x x

Differentiating w.r. t x on both sides

1

log log log log log logdy d d

x x x xy dx dx dx

log log1 log

log

xdy x

y dx x x x

loglog log log log1 1log

xx xdyy x

dx x x x x

14

12. Find dy

dx if 3 2 5cos .siny x x

Sol : 3 2 5cos siny x x


3 2 5 2 5 3cos sin sin cosdy d d

x x x xdx dx dx

3 5 5 4 5 3 2cos 2sin cos 5 sin sin 3dy

x x x x x x xdx

4 5 5 3 2 5 310 sin cos cos 3 sin sindy

x x x x x x xdx

13. Verify Rolle’s theorem for the function 2( ) 2 8f x x x , [ 4,2]x .

Sol : Given 2( ) 2 8f x x x , [ 4,2]x . Since a polynomial function is continuous and derivable on

R. (1) f(x) is continuous on [-4,2] (2) f(x) is derivable on [-4,2].

Also 2( 4) ( 4) 2( 4) 8 0f and 2(2) 2 2 2 8 0f ( 4) (2)f f .

This means that all the conditions of Rolle’s theorem are satisfied by f(x) in [-4,2].

Therefore there exist at least one real number ( 4,2)c such that |( ) 0f c .

2( ) 2 8f x x x |( ) 2 2f x x

|( ) 0f c 2 2 0c 1 ( 4,2)c

Rolle’s theorem is verified with c = - 1

14. If 1sin tx a

and 1cos ty a

then prove that dy y

dx x

Sol : 1sin tx a

1cos ty a

11

sin2

t

x a

11

cos2

t

y a

Differentiating w.r.t “t” we get

11

sin12

1log sin

2

tdx da a t

dt dx

11cos

121

log cos2

tdy da a t

dt dx

11sin

2

2

log

2 1

t

dx a a

dt t

11cos

2

2

log

2 1

t

dy a a

dt t

15

1

1

11

1cos

2

sin2

1cossin

2

2

log

2 1

log

2 1

t

t

tt

ady ady atdt

dxdx a aadt t

dy y

dx x

15. Find the derivative of sin2x xx with respect to x.

Sol : let sin2x xy x = u – v

Where xu x and sin2 xv


log log xu x and sinlog log 2 xv

log logu x x and log sin log2v x

Differentiate with respect to x we get

1

log logdu d d

x x x xu dx dx dx

and 1

cos log 2dv

xv dx

1.log 1 logxdu xu x x x

dx x

and sincos log 2 2 cos log 2xdv

v x xdx

y u v

sin1 log 2 cos log 2x xdy du dvx x x

dx dx dx

16. If ( sin )x a and (1 cos )y a prove that tan2

dy

dx

Sol : ( sin )x a (1 cos )y a


(1 cos )dx

ad

(0 sin ) sin

dya a

d

sin

(1 cos )

dy

dy addxdx a

d

2

2 sin cos sin2 2 2 tan

22 cos cos

2 2

ady

dxa

17. If a function f(x) is differentiable at x = c , prove that it is continuous at x = c.

16

Sol : Since f is differentiable at c, we have |( ) ( )lim ( )x c

f x f cf c

x c

But for x c , we have ( ) ( )

( ) ( ) .f x f c

f x f c x cx c

Therefore ( ) ( )

lim[ ( ) ( )] lim .x c x c

f x f cf x f c x c

x c

Or ( ) ( )

lim ( ) lim ( ) lim limx c x c x c x c

f x f cf x f c x c

x c

|( ) . 0f c = 0

lim ( ) ( )x c

f x f c

. Hence f is continuous at x = c.

FIVE MARK QUESTIONS

1. If 3cos log 4sin logy x x prove that 2

2 1 0x y xy y .

Sol : 3cos log 4sin logy x x


1

3sin log 4cos logx xy

x x

1 3sin log 4cos logxy x x

Again differentiating on both sides we get

2 1

3cos log 4sin log1

x xxy y

x x

2

2 1 3cos log 4sin logx y xy x x

2

2 1x y xy y 2

2 1 0x y xy y

2. If 2 33 2x xy e e prove that 2

25 6 0

d y dyy

dx dx

Sol : 2 33 2x xy e e

2 3 2 36 6 6x x x xdye e e e

dx

2

2 3 2 3

212 18 6 2 3x x x xd y

e e e edx

17

Hence 2

3 3 2 3 2 3

25 6 6(2 3 ) 30 6 3 2x x x x xd y dy

y e e x e e e edx dx

2

3 3 2 3 2 3

25 6 12 18 30 30 18 12 0x x x x xd y dy

y e e x e e e edx dx

3. If 2

1tany x prove that 2

2 2

2 11 2 1 2x y x x y .

Sol : 2

1tany x

Differentiating w.r.t on both sides

1

1 2

2 tan

1

xy

x

2 1

11 2tanx y x

Again differentiating w.r.t x on both sides

2

2 1 2

21 2

1x y xy

x

On cross multiplication, we get

2

2 2

2 11 2 1 2x y x x y

4. If mx nxy Ae Be , Show that 2

20

d y dym n mny

dx dx .

Sol : mx nxy Ae Be


mx nxdyAme Bne

dx Again differentiate w.r.t x on both sides

2

2 2

2

mx nxd yAm e Bn e

dx

Hence 2

2 2

2

mx nx mx nxd y dym n mny m Ae n Be m n Ame Bne mny

dx dx

2 2 2 2mx nx mx nx mx nxm Ae n Be Am e Bmne Amne n Be mny

nx mx mx nxBmne Amne mny mn Ae Be mny

0mny mny mx nxy Ae Be

18

5. If 1siny x prove that 2

2

21 0

d y dyx x

dx dx

Sol : 1siny x

Differentiate w.r.t x, we get

2

1

1

dy

dx x

On cross multiplication

21 1dy

xdx

Again Differentiate w.r.t x, we get

2

2

2 2

21 0

2 1

d y x dyx

dx dxx

Taking Lcm and simplifying, we get

2

2

21 0

d y dyx x

dx dx

6. If 1cosy x prove that 2

2 11 0x y xy

Sol : 1siny x

Differentiate w.r.t x, we get

2

1

1

dy

dx x

On cross multiplication

2

11 1x y

Again Differentiate w.r.t x, we get

2

2 12

21 0

2 1

xx y y

x

Taking Lcm and simplifying, we get 2

2 11 0x y xy

7. If 5cos 3siny x x , prove that 2

20

d yy

dx

Sol : 5cos 3siny x x

Differentiating w.r.t x , on both sides

19

5sin 3cosdy

x xdx

Again differentiating w.r.t x we get

2

25cos 3sin 5cos 3sin

d yx x x x

dx

2

2

d yy

dx

2

20

d yy

dx

8. If 1 1ye x , prove that

22

2

d y dy

dx dx

Sol : 1 1ye x

Differentiate w.r.t x on both sides

1 1 0y yd de x x e

dx dx

1 0y y dye x e

dx

1

1

dy

dx x

Again differentiate w.r.t x , we get

2

22

1

1

d y

dx x

22

2

d y dy

dx dx

9. If 7 7500 600x xy e e then show that 2

249

d yy

dx

Sol : 7 7500 600x xy e e


7 7500 7 600 7x xdye e

dx

Again differentiate w.r.t x

2

7 7

2500 49 600 49x xd y

e edx

2

7 7

249 500 600x xd y

e edx

2

249

d yy

dx

APPLICATION OF DERIVATIVES

TWO MARK QUESTIONS:

1) Find the rate of change of the area of a circle w.r.t to its radius ‘r’

when r = 4 cm?

Ans: Area of circle A = r2, dA/dr = ? when r = 4 cm

Differentiate w.r.t. ‘r’

dA/dr = (2r)

= (2)(4)

= 8 sq. cms

Therefore area of the circle is increasing at the rate of 8 sq. cms.

2) An edge of a variable cube is increasing at the rate of 3cm/s. How fast is

the volume of the cube increasing when the edge is 10cm long?

Ans: Volume of a cube V = x3., Given: dx/dt = 3cm/s. dV/dt = ?

when x = 10cm

Didifferentiate w.r.t ‘t’

dV/dt = 3x2(dx/dt)

= 3(10)2 . (3)

= 900 c.c/s

Therefore volume of the cube increasing at the rate of 900 c.c/s.

3) Show that the function f(x) = x3 -3x2 + 4x, xR is strictly increasing on R.

Ans: f(x) = x3 – 3x2 + 4x


f(x) = 3x2 – 6x + 4

= 3(x2 -2x+1) + 1

= 3(x-1) 2 + 1 0, xR

Therefore f is strictly increasing on R.

4) Show that the function f(x) = e2x is strictly increasing on R.

Ans: f(x) = e2x,


f(x) = 2. e2x

clearly f(x) 0 xR (since exponential function is always

positive)

Therefore f is strictly increasing on R.

5) Find the intervals in which f(x) = x2 + 2x – 5 is strictly increasing or

decreasing.

Ans: f(x) = x2 + 2x – 5


f(x) = 2x + 2

= 2(x+1)

Now f(x) = 0, 2(x+1) = 0

x = -1.

Now x = -1 divides the real line into 2 disjoint intervals namely

(-, -1) and (-1, ).

In (-, -1), f(x) 0

In (-1,), f(x) > 0.

f is strictly decreasing in (-, -1) and f is strictly increasing in (-1,).

6) Find the slope of the tangent to the curve y = (x-1)/(x-2), x≠2 at x = 10.

Ans: y =


dy/dx = (x-1) [(-1)/(x-2)2] + [1/(x-2)](1)

slope of tangent = dy/dx x = 10

= (10-1)[(-1)/(10-2)2] + [1/(10-2)](1)

= -9/64 + 1/8 = -1/64

7) Find the points at which the tangent to the curve y = x3 – 3x2 – 9x + 7

is parallel to x axis.

Ans: y = x3 – 3x2 – 9x + 7.

Differentiate w. r. t. x

dy/dx = 3x2 – 6x -9 = slope of the tangent.

Given tangent is parallel to x axis.

Slope of the tangent = slope of x axis.

3x2 – 6x – 9 = 0

X2 -2x -3 = 0

(x-3)(x+1) = 0

x = 3, x = -1

When x = 3, y = (3)3 – 3(3)2 – 9(3) + 7 = -20

When x = -1, y = (-1)3 -3(-1)2 -9(-1) + 7 = 12

Therefore the points are (3,-20) , ( -1,12) .

8) Using differentials, find approximate value of 25.3 up to 3 decimal

places.

Ans: y = x, Let x = 25 and x = 0.3

Then y = x + x - x

= 25.3 -25

25.3 = y + 5

Now dy = (dy/dx) x = (1/2x) (0.3)

= (1/225) (0.3) = 0.3/10 = 0.03.

Therefore approximate value of 25.3 is 5 + 0.03 = 5.03

9) If the radius of a sphere is measured as 7 m with an error of 0.02m,

then find the approximate error in calculating its volume.

Ans: Given radius of the sphere r = 7m and r = 0.02 m.

Volume of sphere V = (4/3) r3.

Differentiate w.r.t ‘r’

dV/dr = (4/3) (3r2)

Therefore dV = (dV/dr) r

= (4r2)(r)

= (4) (49) (0.02) = 3.92 m3.

Therefore the approximate error in calculating the volume is 3.92 m3.

10) If the radius of sphere is measured as 9m with an error of 0.03m,

then find the approximate error in calculating its surface area.

Ans: Radius of the sphere r = 9m, r = 0.03m.

Surface area of sphere S = 4r2.

Differentiate w.r.t. ‘r’

dS/dr = 4(2r)

Now dS = (dS/dr) (r)

= (4)(2)(r)r

=(8) (9) (0.03)

=2.16m3


1) Find the local maxima and local minima if any, of the function f(x) = x2

and also find the local maximum and local minimum values.

Ans: f(x) = x2 Differentiate w.r.t. x

f’(x) = 2x

f (x) = 0 2x = 0 x=0

f(x) =2x

f(x) = 2 > 0

By second derivative test x= 0 is a point of local minima .

local minimum m value = f(0) = 02 = 0

2) Find the local maxima and local minima if any, of the function

f(x) =x3 – 6x2 + 9x + 15 and also find the local maximum and local

minimum values.

Ans: f(x) = x3- 6x2+9x+15

Differentiate w.r.t. x

f(x) = 3x2-12x+9

f(x) = 6x-12

Now f(x) = 0 3x2-12x+9=0

X2-4x+3=0

(x-3)(x-1) = 0

x=3, x=1

Now f(3) = 6(3)- 12 = 6>0

f(1) = 6(1) – 12 = -6<0

By second derivative test, x= 3 is a point of local minima and

x = 1 is a point of local maxima

local maximum value = f(1) = (1)3-6(1)2+9(1)+15 =19

Local minimum value = f(3) = (3)3-6(3)2+9(3)+15=15.

3) Prove that the function f(x) = logx do not have maxima or minima.

Ans: f(x) = logx


f(x) = 1/x

f(x) = -1/x2

Now f(x) = 0 1/x = 0

x=

The function do not have maxima or minima.

4) Prove that the function f(x) = x3 + x2 + x+ 1 do not have maxima or

minima.

Ans: f(x) = x3+x2+x+1


f(X) = 3X2+2X+1

f(x) = 6x+2

Now f(x) = 0 3x2+2x+1=0

X = [-2±4-4(3)(1)]/2(3)

X = [-2 ± -8] /6 which is imaginary

The given function do not have maxima or minima for all reals.

5) Find the absolute maximum value and the absolute minimum value of

the function f(x) = sinx + cosx, x[0,].

Ans: f(x) = sinx+cosx,


f(x) = cosx - sinx

Now f (x) = 0 Cosx - sinx =0

sinx = cosx tanx = 1

x = /4 and 5/4

Now the value of the function f(x) at x= /4, 5/4 and end points

of intervals that is 0 and is f(0) = sin0+cos0 = 0+1 = 1

f(/4) = sin(/4) +cos (/4) = 1/2 + 1/2 =2/2 = 2

f(5/4) = sin(5/4) +cos (5/4) =(-1/2)+( - 1/2) =-2/2 = -2

f() = sin + cos = 0+(-1) = -1

Absolute maximum value of f on [ 0,] is 2 occurring at x=/4.

Absolute minimum value of f on [ 0,] is -2 occurring at

x=5/4.

6) Find two numbers whose sum is 24 and whose product is as large as

possible.

Ans: Let the numbers be ‘x’ & ‘y’

Given S = x+y = 24

y = 24-x

Product of numbers, P= x y is large

P = x(24-x) = 24x-x2

Differentiate w.r.t. x dP/dx = 24-2x

Differentiate w.r.t. x d2P/dx2 = -2<0 Product is maximum

For the product to be maximum dP/dx = 0

24-2x = 0 x = 12

The numbers are x & 24-x, 12 & 24-12 12 & 12

The numbers are 12 & 12

7) Find two positive numbers whose sum is 16 and the sum of whose cubes

is minimum.

Ans: Let numbers be x and y

Sum = x+y = 16 y = 16 –x

Given S = x3 +y3 is minimum

= x3 + (16-x)3

Differentiate w.r.t. x dS/dx = 3x2 + 3(16-x)2(-1)

d2S/dx2 = 6x - 3(2) (16-x) (-1)

=6x+6(16-x)

For S to be minimum dS/dx = 0

3x2-3(16-x)2 = 0

x2-(16-x)2 = 0

32x – 256 = 0

x = 8 y = 16-x = 16-8 = 8

Hence the numbers are 8 and 8.

8) Show that of all rectangles inscribed in a given fixed circle, the squares

has the maximum area.

Ans : Let ‘r’ be the radius of circle ABCD is a rectangle.

OA = r , OE = x , AE = y ,In le OAE ,

OA2 = OE2 + AE2

r2 = x2 +y2

y2 = r2 –x2 y = r2 - x2

Area of rectangle A = x. y

= xr2-x2

Squaring both sides A2 = x2(r2 –x2)

Let A2 = B B = x2(r2 –x2)


dB/dx = x2(-2x)+(r2-x2)(2x)

= 2x(r2 – 2x2)

d2B/dx2 = 2x(-4x) + (r2-2x2)(2)

= 2r2 -12x2

For the area to be maximum dB/dx = 0

2x(r2-2x2) = 0 x = 0 & x2 = r2/2 x = r/2

d2B/dx2x=r/2 = 2r2 – 12(r2/2) = -4r2< 0

Area is maximum

Y2 = r2 - x2 = r2- r2/2 = r2/2

Since x= y = r/2 , ABCD is a square.

9) Find the equation of the normal at the a point (am2, am3) for the curve

ay2 = x3.

Ans: ay2 = x3

Y2 = x3/a


2y dy/dx = [1/a] 3x2

dy/dx = 3x2/2ay

Slope of tangent = dy/dx(am2

,am3

) = 3(am2)2/ 2a(am3)

= 3a2m4/2a2m3 = 3m/2

slope of normal = -2/3m

Equation of normal at (am2, am3) having slope -2/3m is

Y - am3 = (-2/3m) (x - am2).

B

C D

O E

A

10) Find the equation of the normals to the curve y = x3 + 2x + 6 which are

parallel to the line x + 14y + 4 = 0.

Ans: y = x3+2x+6


dy/dx = 3x2+2 = slope of tangent

slope of normal = -1/(3x2+2)

Normal is parallel to x +14y+4 = 0

Slope of normal = slope of x + 14y +4 = 0

-1/(3x2+2) = -1/14

3x2+2 = 14

3x2 = 12 x = ± 2

When x =2, y = (2)3 + 2(2) + 6 = 18, (2,18)

When x = -2, y =(-2)3 +2(-2)+6 = -6, (-2,-6)

Slope of normal = -1/14

equation of normal at (2,18) is y – 18 = (-1/14) (x – 2)

x+14y – 254 = 0

Also equation of normal at (-2,-6) is y+6 = (-1/14) (x+2)

x + 14y+86 = 0.

11) Find the points on the curve x2/9 +y2/16 = 1 at which the tangents are

parallel to y axis.

Ans: x2/9 +y2/16 = 1


(1/9) 2x + (1/16) 2y (dy/dx) = 0

dy/dx = (-2x/9)/(y/8) = -16x/9y

Tangent parallel to y axis.

Slope of tangent = slope of y axis

-16x/9y = 1/0

y = 0

When y = 0 , x2/9 +0/16 = 1 x2= 9, x = ± 3

The points are (±3,0).

12) Find the equation of all lines having slope two which are tangents to the

curve y = 1/(x-3), x≠ 3.

Ans: y = 1/(x-3)


dy/dx =

Given dy/dx = 2

-1/(x-3)2 = 2

2(x-3)2 =-1

2(x2-6x+9)=-1

2x2-12x+19 = 0

X = (12 ± 144-152) / 2(2) which is complex

No tangent to the curve which has slope two.

13) Prove that the function ‘f’ given by f(x) = log(sinx) is strictly increasing

on (0,/2) and strictly decreasing on (/2,).

Ans: f(x) = log(sinx)


f(x) = (1/sinx) (cosx)

= cotx

Since for each x (0,/2), cotx > 0 f(x) > 0

So f is strictly increasing in (0,/2)

Since for each x (/2,), cotx < 0 f(x)< 0

So f is strictly decreasing in (/2,).

FIVE MARKS QUESTIONS:

1) The volume of a cube is increasing at the rate of 8 c.c/s. How fast is the

surface area increasing when the length of an edge is 12cm?

Ans: Let x , V, S be the length of side , volume and surface area of the

cube respectively.

Given dV/dt = 8c.c/s, dS/dt = ? when x = 12cm

Volume of cube = V = x3

Differentiate w.r.t. t

dV/dt = 3x2 . dx/dt

8 = 3(12)2 dx/dt

dx/dt = 8/3(144) = 1/54

Surface area of a cube S = 6x2


dS/dt = 6(2x) ( dx/dt)

= 12(12) (1/54) = 24/9 = 2.6 sq.cm/s

surface area of a cube is increasing at the rate of 2.6 sq.cm/s.

2) A stone is dropped into a quiet lake and waves in circles at the speed of

5cm/s. At the instant when the radius of circular wave is 8 cm, how fast

is the enclosed area is increasing?

Ans: Let r , A be the radius and Area of a circle respectively

Given dr/dt = 5cm/s dA/dt = ? when r = 8cm

Area of a circle A = r2


dA/dt = 2r dr/dt

= 2(8).(5)

= 80 cm2 /s

The enclosed area is increasing at the rate of 80 cm2/s

when r = 8cm.

3) The length ‘x’ of a rectangle is decreasing at the rate of 5cm/m and the

width ‘y’ increasing at the rate of 4cm/m. When x = 8cm and y = 6cm,

find the rates of changes of

(a) the perimeter and (b) the area of the rectangle.

Ans: Since the length ‘x’ is decreasing and width ‘Y ‘is increasing with

respect to time,

we have dx/dt = -5 cm / m , dy/dt = 4 cm /m

(a) The perimeter P of a rectangle is given by

P = 2(x+y)


dP/dt = 2dx/dt + 2 dY/dt

= 2(-5) + 2 (4)

= -2 cm/min

(b) The area ‘A’ of the rectangle is given by A = x.y

dA/dt = dx/dt y + x dy/dt

= (-5) (6) + (8) (4)

= 2cm2/m

The perimeter and area of a rectangle is decreasing and

increasing at the rate of 2cm/m and 2 cm2/m respectively .

4) A balloon, which always remains spherical, has a variable diameter

(3/2)(2x+1). Find the rate of change of its volume w.r.t. x

Ans: Volume of a sphere = V = (4/3)r3

Given 2r = (3/2) (2x+1) r =( 3/4) (2x+1)

V = (4/3)[(3/4)(2x+1)]3

= (4/3) (27/64)(2x+1)3

V = (9/16)(2x+1)3


dV/dx = (9/16) 3(2x+1)2(2)

= (27/8) (2x+1)2

volume of a sphere increases at the rate of (27/8) (2x+1)2

5) A balloon, which always remains spherical has a variable radius. Find

the rate at which its volume is increasing with the radius when the

radius is 10 cm.

Ans: Let ‘r’ and ‘ V’ be the radius and volume of a sphere.

To find dV/dr = ? when r = 10cm

Volume of a sphere V =( 4/3)r3

Differentiate w.r.t. r

dV/dr = (4/3) 3r2

= 4(10)2

= 400 cm3/cm

The volume of the spherical balloon is increasing with radius is

400 cm3/cm.

6) A water tank has the slope of an inverted right circular cone with its

axis vertical and lower most. Its semi-vertical angle is tan-1(0.5). Water

is poured into it at a constant rate of 5 cubicmeter per hour. Find the

rate at which the level of water is rising at the instant when the depth

of water in the tank is 4m.

Ans: Let r, h and be the radius , height and semi-vertical angle of

cone.

Tan = r/h = tan-1(r/h)

Given = tan -1(0.5)

r/h= 0.5 = ½ r = h/2 Given dV/dt = 5 c.m/h

volume of a cone V = (1/3)r2h : dh/dt = ? when h = 4 m

= (1/3) (h2/4) h

= (/12).h3 OA = h

Differentiate w.r.t. t AB = r

dV/dt = (/12).3h2(dh/dt)

5 = (/4)(4)2(dh/dt)

(dh/dt) = 5/4 = 35/88 m/h ( = 22/7)

Rate of change of water level = (35/88) m/h.

7) A ladder 5m long is leaning against a wall. The bottom of the

ladder is pulled along the ground, away from the wall at the rate

of 2cm/s. How fast is its height of the ladder decreasing when the

foot of the ladder is 4cm away from the wall?

Ans: Let AB be the ladder, AC wall, BC ground.

Let BC = x, AC = y

Given: AB = 5m, dx/dt = 2cm/s, dy/dt = ? when x = 4m.

From the fig, x2 + y2 = 52

(4)2 + y2 = 25

y2 =9, y =3.

Consider x2 + y2 = 52

Differentiate w.r.t ‘t’

2x(dx/dt) + 2y(dy/dt) = 0

2 (4)(2) + 2(3)(dy/dt) = 0

6(dy/dt) = -16

dy/dt = -8/3.

Height of the ladder is decreasing at the rate

of 8/3 cm/s.

O

B A

A

C B

8) A man 6ft tall moves away from a source of light 20ft above the ground

level, his rate of walking being 4 m.p.h. At what rate is the length of his

shadow changing? At what rate is the tip of his shadow moving?

Ans: At any time t, let AB = 6ft be the position of the man. Let C be the

source of light. OC = 20 ft. Then AD is the shadow and D is the tip

of the shadow.

Let OA = x and AD = y (be measured in miles)

Given: dx/dt = 4 m.p.h; dy/dt = ?; d(x+y)/dt = ?

From the figure,

,

20y = 6x+6y

14y = 6x ; y =

Differentiate w.r.t ‘t’

=

=

The shadow is changing at the rate of

m.p.h.

Now

Therefore tip of the shadow is changing at the rate of

m.p.h.

9) A stone is dropped into a pond, waves in the form of circles are

generated and the radius of the outer most ripple increases at the rate

of 2 inches/sec. How fast is the area increasing when the radius is 5

inches?

Ans: Let ‘r’ and ‘A’ be the radius and area of the circle respectively.

Given:

,

Area of circle, Differentiate w.r.t. t

sqinches/sec.

Therefore area of the circle increases at the rate of 20 sq. inches/sec.

C

O

B

A D

KHV’S PAGE 1 http://pue.kar.nic.in

MATHEMATICS: QUESTION BANK

CHAPTER 7: INTEGRALS(INDEFINITE)

Standard forms 1mark questions:

Write an antiderivative for each of the

following functions using differentiation

Question 1: i)sin 2x

Soln: The anti derivative of sin 2x is a

function of x whose derivative is

sin 2x.

It is known that,

Therefore, the anti derivative of

Question 2: Cos 3x

The anti derivative of cos 3x is a function of x

whose derivative is cos 3x.

It is known that,


.

Question 3: e2x

The anti derivative of e2x

is the function of x

whose derivative is e2x

.

It is known that,

Therefore, the anti derivative of .

Evaluate the following integrals:

Question 4:

Question 5:

Question 6:

Question 7: Find an anti-derivative

of with respect to x.

Ans: cot2x=cosec2x-1; antiderivative

of cot2 x is -cotx-x+c


of √ with respect to x.

√

√

√( ) = sinx+cosx

Antiderivative of sinx+cosx is

cosx-sinx+c

Question 9: Evaluate∫ (

) .

Ans: e5x+c


TWO MARK QUESTIONS: Evaluate the following integrals:


following functions using

differentiation :

Question 1:

The anti derivative of is the function

of x whose derivative is .

It is known that,


.

Question 2:

The anti derivative of is the


.

It is known that,


is .


Question 3:

Question 4:

Question 5:

Question 6:

Question 7:

On dividing, we obtain


Question 8:

Question 9:

Question 10:

Question 11:

Question 12:

Question 13:

Question 14:

Question 15:Find the anti derivative of

Solution:


3 mark questions:

If such that f(2) = 0, find f(x)

Solution: It is given that

∴Anti derivative of

∴

Also,

INTEGRATION BY SUBSTITUTION:

ONE MARK QUESTIONS:

1`. Evaluate 2tan (2 ).x dx

Solution:

2 2tan (2 ). (sec 2 1)

1tan 2

2

x dx x dx

x x c

2. Evaluate 2 xcosec dx

2

.

= -2cot

+c

TWO MARK QUESTIONS:

Integrate the following w.r.t x

1.

Hint: = t Ans: log(1+x2) +c

2.

Hint: log |x| = t ∴

Question 3:

Let 1 + log x = t

∴

Question 4:sin x ⋅ sin (cos x)

sin x ⋅ sin (cos x)

Let cos x = t

∴ −sin x dx = dt


Question 5:

Let ax + b = t ⇒ adx = dt

Question 6:

Let 1 + 2x2 = t ∴ 4xdx = dt

Question 7:

Let ∴ (2x + 1)dx = dt

Question 8:

Let

∴

Question 9:

Let

∴ 2dx = dt

Question 10:

Let

∴ 9x2 dx = dt

Question 11:

Let log x = t ∴

Question 12:

Let

∴ −8x dx = dt


Question 13:

Let ∴ 2xdx = dt

Question 14:

Let ∴ 2xdx = dt

Question 15:

Let ∴

Question 16:

Let ∴

Question 17:

Let sin 2x = t ∴

Question 18:

Let

∴ cos x dx = dt

Question 19: cot x log sin x

Let log sin x = t

Question 20:

Let 1 + cos x = t ∴ −sin x dx = dt


Question 21:


Question 22:

Let 1 + log x = t ∴

Question 23: equals

Let Also, let

⇒

Question 24: Find

Let ∴

Question 25: =

THREE MARKS QUESTIONS

Integrate the following :

Question 1:

Let ∴ 2adx = dt

Question 2:

Let ∴ dx = dt


Question 3:

Let ∴ dx = dt

Question 4:

Let ∴

Question 5:

Dividing numerator and denominator by ex,

we obtain

Let ∴

Question 6:

Let ∴

Question 7:

Let 2x − 3 = t ∴ 2dx = dt

Question 8:

Let 7 − 4x = t ∴ −4dx = dt

Question 9:

Let ∴


Question 10:

Let

∴

Question 11:

Let ∴

Question 12:

Let ∴

Question 13:

Let sin x + cos x = t ⇒ (cos x − sin x) dx = dt

Question 14:

Put cos x − sin x = t ⇒ (−sin x − cos x) dx = dt


Question 15:

Question 16:

Let

∴

Question 17:

Let x4 = t

∴ 4x3 dx = dt

Let

∴ From (1), we obtain

******

INTEGRATION USING TRIGONOMETRIC IDENTITIES:

THREE MARKS QUESTIONS:

Integrate the following functions:

Question 1:

Question 2: It is known that,


Question 3: cos 2x cos 4x cos 6x

It is known that,

Question 4: sin

3 (2x + 1)

Let

Question 5: sin

3 x cos

3 x

Question 6: sin x sin 2x sin 3x

It is known that

Question 7:sin 4x sin 8x

It is known that

Question 8:


Question 9:

Question 10: sin

4 x

Question 11: cos

4 2x

Question 12:

Question 13:

Question 14:


Question 15:

Question 16: tan

4x

From equation (1), we obtain

Question 17:

Question 18:

Question 19:

Question 20:


Question 21: sin−1

(cos x)

It is known that,

Substituting in equation (1), we obtain

Question 22:

Question 23:

Question 24:

Let exx = t


INTEGRALS OF SOME PARTICULAR FUNCTIONS

TWO MARK QUESTIONS: Integrate the following w.r.t x

Question 1:

Let x3 = t ∴ 3x

2 dx = dt

Question 2:

Let 2x = t ∴ 2dx = dt

Question 3:

Let 2 − x = t ⇒ −dx = dt

Question 4:


Question 5:

Question 6:

Let x3 = t ∴ 3x

2 dx = dt

Question 7:

From (1), we obtain

Question 8:

Let x3 = t ⇒ 3x

2 dx = dt


Question 9:

Let tan x = t ∴ sec2x dx = dt

Question 10:

Question 11:

Question 12:

Question 13:

Question 14:


THREE MAKRS QUESTIONS:

Question 1:

Question 2:

Question 3:

Question 4:

Equating the coefficients of x and

constant term on both sides, we obtain

4A = 4 ⇒ A = 1

A + B = 1 ⇒ B = 0

Let 2x2 + x − 3 = t

∴ (4x + 1) dx = dt

Question 5:

Equating the coefficients of x and constant

term on both sides, we obtain


From (1), we obtain


Question 6: Integrate 2

x 2

x 2x 3

with respect to x.

2 2

12x 2 1

x 2 2I dx dxx 2x 3 x 2x 3

2 2

1 2x 2 1dx dx

2 x 2x 3 x 2x 3

2

2

1 12 x 2x 3 dx

2 x 1 2

2 2x 2x 3 log x 1 x 2x 3 c

ADDITIONAL 4 TO 5 MARK QUESTIONS: Integrate the following:

Question 1:

Equating the coefficient of x and constant term

on both sides, we obtain


Substituting equations (2) and (3) in equation

(1), we obtain

Question 2:


term, we obtain 2A = 6 ⇒ A = 3

−9A + B = 7 ⇒ B = 34

∴ 6x + 7 = 3 (2x − 9) + 34

Substituting equations (2) and (3) in (1), we

obtain

Question 3:




Using equations (2) and (3) in (1), we obtain

Question 4:

Let x

2 + 2x +3 = t

⇒ (2x + 2) dx =dt


Question 5:



Substituting (2) and (3) in (1), we obtain

Question 6:


term, we obtain



INTEGRATION BY PARTIAL FRACTIONS

TWO MARK QUESTIONS:

Question 1:

Let


term, we obtain

A + B = 1 ; 2A + B = 0

On solving, we obtain A = −1 and B = 2

Question 2:

Let


term, we obtain

A + B = 0 ; −3A + 3B = 1

On solving, we obtain


Question 1:

Let

Substituting x = 1, 2, and 3 respectively in

equation (1), we obtain

A = 1, B = −5, and C = 4

Question 2:

Let




Question 3:

Let

Substituting x = −1 and −2 in equation (1), we

obtain A = −2 and B = 4

Question 4:

It can be seen that the given integrand is not a

proper fraction. Therefore, on dividing (1 −

x2) by x(1 − 2x), we obtain

Let

Substituting x = 0 and in equation (1), we

obtain A = 2 and B = 3


Question 4:

Let

Equating the coefficients of x

2, x, and constant

term, we obtain

A + C = 0 ;−A + B = 1 ; −B + C = 0

On solving these equations, we obtain


Question 5:

Let

Substituting x = 1, we obtain

Equating the coefficients of x2 and constant

term, we obtain

A + C = 0 ;−2A + 2B + C = 0


Question 6:


Let

Substituting x = 1 in equation (1), we obtain

B = 4

Equating the coefficients of x2 and x, we

obtain A + C = 0; B − 2C = 3


Question 7:

Let


2 and x, we

obtain

Question 8:

Let

Substituting x = −1, −2, and 2 respectively in


Question 9:


proper fraction.

Therefore, on dividing (x3 + x + 1) by x

2 − 1,

we obtain

Let

Substituting x = 1 and −1 in equation (1), we

obtain

Question 10:

Equating the coefficient of x and constant

term, we obtain A = 3

2A + B = −1 ⇒ B = −7


Question 11:

[Hint: multiply numerator and denominator by

xn − 1

and put xn = t]

Multiplying numerator and

denominator by xn − 1

, we obtain

Substituting t = 0, −1 in equation (1), we

obtain

A = 1 and B = −1

Question 12: [Hint: Put

sin x = t]

Substituting t = 2 and then t = 1 in equation

(1), we obtain A = 1 and B = −1

Question 13:

Let x2 = t ⇒ 2x dx = dt

Substituting t = −3 and t = −1 in equation (1),

we obtain

Question 14:

Multiplying numerator and denominator by x

3,

we obtain

Let x

4 = t ⇒ 4x

3dx = dt

Substituting t = 0 and 1 in (1), we obtain

A = −1 and B = 1


Question 15:

[Hint: Put ex = t]

Let ex = t ⇒ e

x dx = dt

Substituting t = 1 and t = 0 in equation (1), we


Question 16:

Substituting x = 1 and 2 in (1), we obtain

A = −1 and B = 2

Question 17:


2, x, and constant

term, we obtain

A + B = 0; C = 0 ; A = 1


A = 1, B = −1, and C = 0

ADDITIONAL QUESTIONS: 4 TO 5 MARKS:

Question 1:

Equating the coefficient of x

2, x, and constant

term, we obtain

A − B = 0; B − C = 0; A + C = 2


A = 1, B = 1, and C = 1


Question 2:

Equating the coefficient of x

3, x

2, x, and

constant term, we obtain


Question 18:


3, x

2, x, and

constant term, we obtain

A + C = 0; B + D = 4 ;4A + 3C = 0

4B + 3D = 10


A = 0, B = −2, C = 0, and D = 6

INTEGRATION BY PARTS TWO MARKS QUESTIONS:

Question 1: x sin x

Let I =

Taking x as first function and sin x as second

function and integrating by parts, we obtain

Question 2:

Let I =

Taking x as first function and sin 3x as second


Question 3: ∫ . Given Integral

= ∫ ∫

.

=


Question 4: x logx

Let

Taking log x as first function and x as second


Question 5: x log 2x

Let

Taking log 2x as first function and x as second


Question 6: x

2 log x

Let

Taking log x as first function and x2 as second


Question 7:

Let

Taking x as first function and sec2x as second



Integrate the following w.r.t. x

Question 1:

Let

Taking x2 as first function and e

x as second


Again integrating by parts, we obtain

Question 2:

Let

Taking as first function and x as

second function and integrating by parts, we

obtain


Question 3:

Let



obtain

Question 4: Evaluate: ∫ .

∫ ∫

∫

∫ .

= ∫

∫

=

| |

Question 5:

Let

Taking cos−1

x as first function and x as second


Question 6:

Let

Taking as first function and 1 as


obtain


Question 7:

Let

Taking as first function and

as second function and integrating by parts, we

obtain

*******

INTEGRAL OF THE FORM ∫ [ ( ) ( )]

TWO MARK QUESTIONS

Question 1:

Let

Let

⇒

∴ It is known that,

Question 2:

Also, let ⇒

It is known that,

2. Evaluate: x 1 sin xe dx

1 cos x

.

x

2

x

2

x 2 x

x

x x1 2sin cos

2 2I e dxx

2cos2

1 xe tan dx

x 22cos

2

1 x xe sec tan dx e f '(x) f (x) dx

2 2 2

xe tan

2

THREE MARK QUESTIONS


Question 1:

Let


Let ⇒

It is known that,

Question 2:

Let ⇒

It is known that,


Question 3:

Let ⇒

It is known that,

Question 3:

Let

Integrating by parts, we obtain


Question 4:

Let ⇒

= 2θ

⇒



Question 5:

equals

Let

Also, let ⇒

It is known that,

INDEFINITE INTEGRALS

(FIVE MARK QUESTIONS)

1. Find the integral of

√ with

respect to x and hence evaluate

∫

√ .

Solution: Let then

∴ ∫

√ ∫

√

∫ (

)

Consider ∫

√

∫

√ ( )

(

)


√ with


∫

√ .

Solution: Let

√

Let then

∴ ∫

√ ∫

| |

|

√

|

| √ | | |

| √ | ,

| |

Consider ∫

√

∫

√( )

|( ) √( ) | .


with respect

to x and evaluate ∫

.

Solution: ∫

Consider

( )( )

[( ) ( )

( )( )]

[

( )]

[

( )]

∴ ∫

*

( )+ ∫

*

( )+

[ | | | |]

|

|

Consider∫

∫

( ) Let

then

∴

∫

( )

|

| .



with respect

to x and hence evaluate ∫

( ) .

Solution: ∫

Consider

( )( )

[( ) ( )

( )( )]

[

( )]

[

( )]

∴ ∫

*

( )+ ∫

*

( )+

[ | | | |]

|

|

Consider ∫

( )

Let ( ) then

∴

∫

|

| .

=

|

( )

( )|

5.Find the integral of 2 2

1

x awith

respect to and hence evaluate

2

1dx

5x 2x

Substituting

∫

√ =∫

√

=∫

√ ( ) ∫

√

=∫

=log|sec +tan |+C1

=log|

√

|+C1

= log| √ |-log|a|+C1

=log| √ |+C

where C=C1 –log|a|

Now ∫

√

=x2+2x+1=(x2+2x+1)+1

=(x+1)2+(1)2

∫

√ =∫

√( )

= | √ |


1

x awith

respect to and hence evaluate

∫

Let I= 2 2

1dx;

x aPut x=atan ,

then dx=a sec2 ;

x2+a2 =a2 tan2 +a2

=a2(tan2 +1)=a2sec2

2

2 2 2 2

1

1 asec d 1I dx d

x a a sec a

1 1 xtan c

a a a

Consider ∫

7.Find the integral of √ with

respect to x and evaluate ∫ √ .

Solution: Let ∫√ applying

integration by parts

We get √

∫

√

√ ∫

√

√ ∫

√


√ | √ |

√

| √ |

Consider ∫ √

√

| √ |

8. Find the integral of √ with

respect to x and evaluate

∫ √ .



We get √ ∫

√

√ ∫

√

√ ∫

√

√ (

)

√

(

)

Consider ∫ √

∫ √ ( )

( )

√ ( )

(

√ )

Find the integral of √ with


∫



We get √

∫

√

√ ∫( )

√

= √ ∫ *√

√ +

√ ∫

√

√ | √ |

√

| √ |

Now consider ∫

x2+2x+5=(x2+2x+1)+4=(x+1)2+4

I=∫[( ) ]

Put x+1=t dx=dt

√

√ +

| √ |

=

√

+ 2log| x+1+√ |

Note: The above questions is for 5 mark

questions in part D of the question paper for

second PUC.

Note: In this chapter “Indefinite Integrals”

Some of the solved examples given in the text

book are not included in the question bank.

The students are advised to go through the

those questions also.

*******


Assignments (i) Integration by substitution

LEVEL I

1. 2. 3.

LEVEL II

1. 2. 3.

LEVEL III

1. 2. 3.

(ii) Application of trigonometric function in integrals LEVEL I

1. 2. 3.

LEVEL II

1. 2.

LEVEL III

1. 2.

(iii) Integration using standard results LEVEL I

1. 2. 3.

LEVEL II

1. 2. 3.

LEVEL III

1. 2. 3.

4. 5. [CBSE 2011]

(iv) Integration using Partial Fraction

LEVEL I

dxx

)x(logsec2

dxx1

e2

x1tanm

dx

x1

e

2

x1sin

dxxx

1

dx

1xx

1

6

dx

1e

1

x

dxxcos.xsin

xtan

dxxcosxsec

xtan dx

xcos.xsin

13

dx.xsin3 dx.x3cos2

dx.x3cos.x2cos.xcos

dx.xtan.xsec4 dx

xsin

x4sin

dx.xcos5 dx.xcos.xsin 32

9x4

dx

2

dx

10x2x

12

13x12x9

dx2

dx1xx

x24

dx

5xsin4xsin

xcos2

2xx67

dx

dxxx1

x2

42

dx

1xx

1xx2

2

dx

6x5x

2x

2

dx

x1

x1

4x5x

7x6


1. 2. 3.

LEVEL II

1. 2. 3.

LEVEL III

1. 2. 3.

(v) Integration by Parts LEVEL I

1. 2. 3.

LEVEL II

1. 2. 3.

4. 5.

LEVEL III

1. 2. 3.

4. 5.

(vi) Some Special Integrals LEVEL I

1. 2.

LEVEL II

1. 2.

LEVEL III

1. 2.

(vii) Miscellaneous Questions LEVEL II

1. (Hint: Divide the Numerator and Denominator by cos2x and use the relation

sec2x=1+tan

2x; and put tanx=t

LEVEL III

1.∫

dx

)1x)(1x(

1x2

dx)3x)(2x)(1x(

x2

dx

)3x()1x(

2x32

dx)2x)(1x(

8x2x2

dx

)2x(x

1xx2

2

dx

)3x()1x(

1x2

2

dx)4x)(2x(

82 x2sinxsin

dx

dx

x1

13

dx.xsec.x 2

dx.xlog

dx)xseclogx(tanex

dx.xsin 1

dx.xsin.x 12

dx

x1

xsin.x

2

1

dx.x1

x1cos

2

21

dx.xsec3

dxxlogcos

dx)x2(

)x1(e2

x

dx)xlog1(

xlog2

dx.e

x2cos1

xsin2 x

dx.x3cos.e x2

dx.x4 2

dx.x41 2

dx.6x4x2

dx.xx41 2

dx.xx1)1x( 2 dxxx)5x( 2

xcos5xsin4

dx22


SOLUTIONS: ASSIGNMENTS: INDEFINITE INTEGRALS

(i) Integration by substitution

LEVEL I 1. tan(logex) + C 2. 3.

LEVEL II 1. 2. 3.

LEVEL III 1. 2. 3.

(ii) ) Application of trigonometric function in integrals

LEVEL I 1. 2.

3.

LEVEL II 1. 2.

LEVEL III 1. 2.

(iii) Integration using Standard results

LEVEL I 1. 2. + C 3. + C

LEVEL II 1. + C 2. 3.

LEVEL III 1. 2.

3. C

4. [Hint: Put x=cos2 ]

5.

(iv) Integration using Partial Fraction

LEVEL I 1. 2.

3.

LEVEL II 1. x – 11log(x – 1) + 16log(x – 2) + C 2.

3.

Cem

1 xtanm 1

Ce xsin 1

Cx1log2 e Cxsec3

1 31 Ce1log x

e

Cxtan2 Cxcostan 1 Cxtanlog2

xtane

2

Cx3cos12

1xcos

4

3 C

6

x6sinx

2

1

Cx2sin8

1x4sin

16

1x6sin

4

1

4

x

C4

xtan

2

xtanORCxsec

4

1 424 Cxsin2x3sin

3

2

Cxsin5

1xsin

3

2xsin 53 C

5

xsin

3

xsin 53

C9x42

1xlog

2

1 2e

3

1xtan

3

1 1

3

2x3tan

9

1 1

3

1x2tan

3

1 21 C2xsintan 1 C

5

1x2sin 1

C5

1x2sin

21

C

3

1x2log

3

21xxlogx 2

6x5x

2

5xlog

2

16x5x 22

Cx1xsin 21

C20x9x2

9x2log3420x9x6 22

C)2xlog(3

5)1xlog(

3

1 C3xlog

2

32xlog21xlog

2

1

C

1x2

5

3x

1xlog

4

11

C2xlog4

3

x2

1xlog

4

1

C3xlog8

5

1x2

11xlog

8

3


LEVEL III 1. log(x +2) 2.

3. + C [Hint: Partial fractions]

(v) Integration by Parts

LEVEL I 1.x.tanx + logcosx + C 2.xlogx – x + C 3.ex.logsecx + C

LEVEL II 1. 2.

3. 4.

5.

LEVEL III 1. 2. [Hint: =

3. 4. ex.tanx + C 5.

(vi) Some Special Integrals

LEVEL I 1. 2.

LEVEL II 1.

2.

LEVEL III 1.

2.

(vii) Miscellaneous Questions

LEVEL II 1.

LEVEL III 1.

+

log|cosx+sinx|+C

2

xtan4xlog

2

1 12

C

3

xcos21log2

2

xcos1log

6

xcos1log

3

1x2tan

3

1xx1log

6

1x1log

3

1 12

Cx1xsinx 21 C

9

x12xxsin

3

x 221

3

Cxxsinx1 12 Cx1logxtanx2 21

Cxtanxseclogxtan.xsec2

1

Cxlogsinxlogcos2

x C

x2

ex

Cxlog1

x

Cx3cos2x3sin3

13

e x2

Cx4xlog22

x4x 22

Cx2sin4

1

2

x41x 12

C6x4x2xlog

2

6x4x2x 22

C

5

2xsin

2

5

2

xx412x 12

C5

1x2sin

16

5xx11x2

8

1xx1

3

1 122/32

Cxx21x2log16

11xx1x2

8

11xx

3

1 222/32

C5

xtan2tan

52

1 1


Detail of the concepts to be mastered by every student of class second PUC

with exercises and examples of NCERT Text Book.

Indefinit

e

Integrals

(i) Integration by substitution * Text book , Vol. II Examples 5&6

Page 300, 302,301,303

(ii) ) Application of trigonometric

function in integrals

** Text book , Vol. II Ex 7 Page 306,

Exercise 7.3 Q13&Q24

(iii) Integration of some particular

function

22 ax

dx, ,

, ,

, ,

*** Text book , Vol. II Exp 8, 9, 10

Page 311,312,313, Exercise 7.4 Q

3,4,8,9,13&23

(iv) Integration using Partial Fraction *** Text book , Vol. II Exp 11&12

Page 318 Exp 13 319,Exp 14 & 15

Page320

(v) Integration by Parts ** Text book , Vol. II Exp 18,19&20

Page 325 Exs 7.6 QNO ,10,11,

17,18,20

(vi)Some Special Integrals

,

*** Text book , Vol. II Exp 23 &24

Page 329

(vii) Miscellaneous Questions ** Text book , Vol. II Solved Ex. 40,

41

viii)Some special integrals Text book Supplimentary material

Page 614,615

SYMBOLS USED :

* : Important Questions, ** :Very Important Questions, *** : Very-Very Important Questions

22 ax

dx

dx

xa

1

22 cbxax

dx2

cbxax

dx

2

cbxax

dx)qpx(2

cbxax

dx)qpx(

2

dxxa 22

dxax 22


MATHEMATICS: QUESTION BANK

CHAPTER 7: INTEGRALS(INDEFINITE)

Standard forms 1mark questions:


following functions using differentiation

Question 1: i)sin 2x

Ans: The anti derivative of sin 2x is a


sin 2x.

It is known that,


Question 2: Cos 3x

The anti derivative of cos 3x is a function of x

whose derivative is cos 3x.

It is known that,


.

Question 3: e2x

The anti derivative of e2x is the function of x

whose derivative is e2x.

It is known that,

Therefore, the anti derivative of .


Question 4:

Question 5:

Question 6:


of cot2 𝑥 with respect to x.

Ans: cot2x=cosec2x-1;

antiderivative of cot2 x is

-cotx-x+c


of √1 + sin 2𝑥 with respect to x.

√1 + sin 2𝑥

= √sin2 𝑥 + cos2 𝑥 + 2𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥

= √(𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥)2= sinx+cosx

Antiderivative of sinx+cosx is

cosx-sinx+c


TWO MARK QUESTIONS: Evaluate the following integrals:


following functions using differentiation :

Question 1:

The anti derivative of is the function

of x whose derivative is .

It is known that,


.

Question 2:

The anti derivative of is the


.

It is known that,


is .


(Question 3 to 14)

Question 3:

Question 4:

Question 5:

Question 6:

Question 7:

Simplifying and dividing by x-1, we obtain


Question 8:

Question 9:

Question 10:

Question 11:

Question 12:

Question 13:

Question 14:

Question 15:Find the anti derivative of

Solution:


3 mark questions:

If such that f(2) = 0, find f(x)

Solution: It is given that

∴Anti derivative of

∴

Also,

INTEGRATION BY SUBSTITUTION:

ONE MARK QUESTIONS:

1`. Evaluate 2tan (2 ).x dx

Solution:

2 2tan (2 ). (sec 2 1)

1tan 2

2

x dx x dx

x x c

2. Evaluate2 x

cosec dx2

.

= -2cot𝑥

2+c

TWO MARK QUESTIONS:

Integrate the following functions w.r.t x

Question 1.

Hint: = t Ans: log(1+x2) +c

Question 2.

Hint: log |x| = t ∴

Question 3:

Let 1 + log x = t

∴

Question 4:sin x ⋅ sin (cos x)

sin x ⋅ sin (cos x)

Let cos x = t

∴ −sin x dx = dt

Question 5:

Let ax + b = t ⇒ adx = dt


Question 6:

Let 1 + 2x2 = t ∴ 4xdx = dt

Question 7:

Let ∴ (2x + 1)dx = dt

Question 8:

Let

∴

Question 9:

Let

∴ 2dx = dt

Question 10:

Let

∴ 9x2 dx = dt

Question 11:

Let log x = t ∴

Question 12:

Let

∴ −8x dx = dt

Question 13:

Let ∴ 2xdx = dt


Question 14:

Let ∴ 2xdx = dt

Question 15:

Let ∴

Question 16:

Let ∴

Question 17:

Let sin 2x = t ∴

Question 18:

Let

∴ cos x dx = dt

Question 19: cot x log sin x

Let log sin x = t

Question 20:


Question 21:



Question 22:

Let 1 + log x = t ∴

Question 23: equals

Let Also, let

⇒

Question 24: Find

Let ∴

Question 25: =

THREE MARKS QUESTIONS

Integrate the following :

Question 1:

Let ∴ 2adx = dt

Question 2:

Let ∴ dx = dt


Question 3:

Let ∴ dx = dt

Question 4:

Let ∴

Question 5:

Dividing numerator and denominator by ex,

we obtain

Let ∴

Question 6:

Let ∴

Question 7:

Let 2x − 3 = t ∴ 2dx = dt

Question 8:

Let 7 − 4x = t ∴ −4dx = dt

Question 9:

Let ∴


Question 10:

Let

∴

Question 11:

Let ∴

Question 12:

Let ∴

Question 13:

Let sin x + cos x = t ⇒ (cos x − sin x) dx = dt

Question 14:

Put cos x − sin x = t ⇒ (−sin x − cos x) dx = dt


Question 15:

Question 16:

Let

∴

Question 17:

Let x4 = t

∴ 4x3 dx = dt

Let

∴ From (1), we obtain

******

INTEGRATION USING TRIGONOMETRIC IDENTITIES:


Integrate the following functions:

Question 1:

Question 2: It is known that,

Question 3: cos 2x cos 4x cos 6x

It is known that,


Question 4: sin3 (2x + 1)

Let

Question 5: sin3 x cos3 x

Question 6: sin x sin 2x sin 3x

It is known that

Question 7:sin 4x sin 8x

It is known that

Question 8:


Question 9:

Question 10: sin4 x

Question 11: cos4 2x

Question 12:

Question 13:

Question 14:


Question 15:

Question 16: tan4x


Question 17:

Question 18:

Question 19:

Question 20:


Question 21: sin−1 (cos x)

Sin-1(cosx)=sin-1[sin(2

-x)]= x

2

1sin (cos x) ( x)dx2

=

2xx

2 2

+C

Question 22:

Question 23:

Question 24:

Let exx = t

INTEGRALS OF SOME PARTICULAR FUNCTIONS

TWO MARK QUESTIONS: Integrate the following w.r.t x

Question 1:

Let x3 = t ∴ 3x2 dx = dt

Question 2:


Question 3:

Let 2 − x = t ⇒ −dx = dt

Question 4:



Question 5:

Question 6:

Let x3 = t ∴ 3x2 dx = dt

Question 7:

From (1), we obtain

Question 8:

Let x3 = t ⇒ 3x2 dx = dt

Question 9:

Let tan x = t ∴ sec2x dx = dt

Question 10:

Question 11: 1

9𝑥2+6𝑥+5


Question 12:

Question 13:

Question 14:

THREE MAKRS QUESTIONS:

Question 1:

Question 2:


Question 3:

Question 4:

Equating the coefficients of x and

constant term on both sides, we obtain

4A = 4 ⇒ A = 1

A + B = 1 ⇒ B = 0

Let 2x2 + x − 3 = t

∴ (4x + 1) dx = dt

Question 5:



From (1), we obtain



Question 6: Integrate 2

x 2

x 2x 3

with respect to x.

2 2

12x 2 1

x 2 2I dx dxx 2x 3 x 2x 3

2 2

1 2x 2 1dx dx

2 x 2x 3 x 2x 3

2

2

1 12 x 2x 3 dx

2 x 1 2

2 2x 2x 3 log x 1 x 2x 3 c

INTEGRATION BY PARTIAL FRACTIONS

TWO MARK QUESTIONS:

Question 1:

Let


term, we obtain

A + B = 1 ; 2A + B = 0

On solving, we obtain A = −1 and B = 2

Question 2:

Let


term, we obtain

A + B = 0 ; −3A + 3B = 1



Question 1:

Let



A = 1, B = −5, and C = 4

Question 2:

Let




Question 3:

Let

Substituting x = −1 and −2 in equation (1), we


Question 4:


proper fraction. Therefore, on dividing (1 −

x2) by x(1 − 2x), we obtain

Let

Substituting x = 0 and in equation (1), we

obtain A = 2 and B = 3


Question 4:

Let

Equating the coefficients of x2, x, and constant

term, we obtain

A + C = 0 ;−A + B = 1 ; −B + C = 0



Question 5:

Let

Substituting x = 1, we obtain

Equating the coefficients of x2 and constant

term, we obtain

A + C = 0 ;−2A + 2B + C = 0


Question 6:


Let

Substituting x = 1 in equation (1), we obtain

B = 4


obtain A + C = 0; B − 2C = 3


Question 7:

Let


obtain

Question 8:

Let

Substituting x = −1, −2, and 2 respectively in


Question 9:


proper fraction.

Therefore, on dividing (x3 + x + 1) by x2 − 1,

we obtain

Let

Substituting x = 1 and −1 in equation (1), we

obtain

Question 10:

Equating the coefficient of x and constant

term, we obtain A = 3

2A + B = −1 ⇒ B = −7

Question 11:

[Hint: multiply numerator and denominator by

xn − 1 and put xn = t]


Multiplying numerator and

denominator by xn − 1, we obtain

Substituting t = 0, −1 in equation (1), we

obtain

A = 1 and B = −1

Question 12: [Hint: Put

sin x = t]

Substituting t = 2 and then t = 1 in equation

(1), we obtain A = 1 and B = −1

Question 13:

Let x2 = t ⇒ 2x dx = dt

Substituting t = −3 and t = −1 in equation (1),

we obtain

Question 14:

Multiplying numerator and denominator by x3,

we obtain

Let x4 = t ⇒ 4x3dx = dt

Substituting t = 0 and 1 in (1), we obtain

A = −1 and B = 1


Question 15:

[Hint: Put ex = t]

Let ex = t ⇒ ex dx = dt

Substituting t = 1 and t = 0 in equation (1), we


Question 16:

Substituting x = 1 and 2 in (1), we obtain

A = −1 and B = 2

Question 17:

Equating the coefficients of x2, x, and constant

term, we obtain

A + B = 0; C = 0 ; A = 1


A = 1, B = −1, and C = 0

INTEGRATION BY PARTS TWO MARKS QUESTIONS:

Question 1: x sin x

Let I =

Taking x as first function and sin x as second


Question 2:

Let I =

Taking x as first function and sin 3x as second


Question 3: ∫ 𝑙𝑜𝑔𝑥 𝑑𝑥. Given Integral

=log 𝑥 ∙ ∫ 1 𝑑𝑥 − ∫ 1.𝑑

𝑑𝑥log 𝑥 𝑑𝑥.

== 𝑥 log 𝑥 − 𝑥 + 𝑐


Question 4: x logx

Let

Taking log x as first function and x as second


Question 5: x log 2x

Let

Taking log 2x as first function and x as second


Question 6: x2 log x

Let

Taking log x as first function and x2 as second


Question 7:

Let

Taking x as first function and sec2x as second



Integrate the following w.r.t. x

Question 1:

Let

Taking x2 as first function and ex as second



Question 2:

Let



obtain


Question 3:

Let



obtain

Question 4: Evaluate: ∫ tan−1 𝑥 𝑑𝑥.

∫ tan−1 𝑥 𝑑𝑥 = 𝑡𝑎𝑛−1 𝑥 . ∫ 1 𝑑𝑥

− ∫𝑑

𝑑𝑥𝑡𝑎𝑛− 1𝑥. ∫ 1 𝑑𝑥 . 𝑑𝑥.

=𝑥 tan−1 𝑥 − ∫𝑥

1+𝑥2 𝑑𝑥

= 𝑥 tan−1 𝑥 −1

2∫

2𝑥

1 + 𝑥2

= 𝑥 tan−1 𝑥 −1

2log|1 + 𝑥2| + 𝑐

Question 5:

Let

Taking cos−1 x as first function and x as second


Question 6:

Let


Taking as first function and 1 as


obtain

Question 7:

Let

Taking as first function and

as second function and integrating by parts, we

obtain

*******

INTEGRAL OF THE FORM ∫ 𝒆𝒙[𝒇(𝒙) + 𝒇′(𝒙)]𝒅𝒙

TWO MARK QUESTIONS

Question 1:

Let

Let

⇒

∴ It is known that,

Question 2:

Also, let ⇒

It is known that,

2. Evaluate: x 1 sin x

e dx1 cos x

.

x

2

x

2

x 2 x

x

x x1 2sin cos

2 2I e dxx

2cos2

1 xe tan dx

x 22cos

2

1 x xe sec tan dx e f '(x) f (x) dx

2 2 2

xe tan

2

THREE MARK QUESTIONS


Question 1:

Let


Let ⇒

It is known that,

Question 2:

Let ⇒

It is known that,


Question 3:

Let ⇒

It is known that,

Question 3:

Let



Question 4:

Let ⇒

= 2θ

⇒



Question 5:

equals

Let

Also, let ⇒

It is known that,

INDEFINITE INTEGRALS

(FIVE MARK QUESTIONS)

1. Find the integral of 𝟏

√𝒂𝟐− 𝒙𝟐 with


∫𝟏

√𝟕−𝟔𝒙−𝒙𝟐 𝒅𝒙.

Solution: Let 𝑥 = 𝑎 sin 𝜃 then 𝑑𝑥 =

𝑎 cos 𝜃 𝑑𝜃

∴ ∫𝑑𝑥

√𝑎2−𝑥2= ∫

𝑎 cos 𝜃 𝑑𝜃

√𝑎2−𝑎2 sin2 𝜃

∫ 1 𝑑𝜃 = 𝜃 + 𝑐 = sin−1 (𝑥

𝑎) + 𝑐

Consider 𝐼 = ∫1

√7−6𝑥−𝑥2 𝑑𝑥

= ∫1

√16 − (𝑥 + 3)2 𝑑𝑥

= sin−1 (𝑥+3

4) + 𝑐


√𝒙𝟐+ 𝒂𝟐 with


∫𝟏

√𝒙𝟐+𝟐𝒙+𝟐 𝒅𝒙.

Solution: Let 𝐼 =1

√𝑥2+ 𝑎2 𝑑𝑥

Let 𝑥 = 𝑎 tan 𝜃 then

𝑑𝑥 = 𝑎 sec2 𝜃 𝑑𝜃

∴ 𝐼 = ∫𝑎 sec2 𝜃 𝑑𝜃

√𝑎2 tan2 𝜃+ 𝑎2= ∫ sec 𝜃 𝑑𝜃

= log|sec 𝜃 + tan 𝜃| + 𝑐1

= log | 𝑥

𝑎+ √

𝑥2

𝑎2+ 1 | + 𝑐1

= log|𝑥 + √𝑥2 + 𝑎2| − log|𝑎| + 𝑐1

= log|𝑥 + √𝑥2 + 𝑎2| + 𝑐 ,

𝑐 = 𝑐1 − log|𝑎|

Consider ∫1

√𝑥2+2𝑥+2 𝑑𝑥

= ∫1

√(𝑥 + 1)2 + 1 𝑑𝑥

= log |(𝑥 + 1) + √(𝑥 + 1)2 + 1| + 𝑐.


𝒙𝟐− 𝒂𝟐 with respect

to x and evaluate ∫𝒙

𝒙𝟒− 𝟏𝟔 𝒅𝒙.

Solution: Let 𝐼 = ∫1

𝑥2− 𝑎2 𝑑𝑥

Consider 1

𝑥2− 𝑎2 =1

(𝑥−𝑎)(𝑥+𝑎)

=1

2𝑎[(𝑥 + 𝑎) − (𝑥 − 𝑎)

(𝑥 − 𝑎)(𝑥 + 𝑎)]

=1

2𝑎[

1

(𝑥 − 𝑎)]

−1

2𝑎[

1

(𝑥 + 𝑎)]

∴ 𝐼 = ∫1

2𝑎[

1

(𝑥−𝑎)] 𝑑𝑥 − ∫

1

2𝑎[

1

(𝑥−𝑎)] 𝑑𝑥

=1

2𝑎[log|𝑥 − 𝑎| − log|𝑥 + 𝑎|] + 𝑐

=1

2𝑎log |

𝑥−𝑎

𝑥+𝑎| + 𝑐

Consider∫𝑥

𝑥4− 16 𝑑𝑥 = ∫

𝑥

(𝑥2)2− 16 𝑑𝑥 Let

𝑥2 = 𝑡 then 𝑥 𝑑𝑥 =𝑑𝑡

2

∴ 𝐼 =1

2∫

𝑑𝑡

(𝑡)2− 42 =1

2×

1

2×4log |

𝑡−4

𝑡+4| + 𝑐 .



𝒂𝟐− 𝒙𝟐 with respect

to x and hence evaluate ∫𝟏

𝟏𝟔− (𝟐𝒙+𝟑)𝟐 𝒅𝒙.

Solution: Let 𝐼 = ∫1

𝑎2− 𝑥2 𝑑𝑥

Consider 1

𝑎2− 𝑥=

1

(𝑎−𝑥)(𝑎+𝑥)

=1

2𝑎[(𝑎 + 𝑥) + (𝑎 − 𝑥)

(𝑎 − 𝑥)(𝑎 + 𝑥)]

=1

2𝑎[

1

(𝑎 − 𝑥)]

+1

2𝑎[

1

(𝑎 + 𝑥)]

∴ 𝐼 = ∫1

2𝑎[

1

(𝑎−𝑥)] 𝑑𝑥 + ∫

1

2𝑎[

1

(𝑎+𝑥)] 𝑑𝑥

=1

2𝑎[−log|𝑎 − 𝑥| + log|𝑎 + 𝑥|] + 𝑐

=1

2𝑎log |

𝑎+𝑥

𝑎−𝑥| + 𝑐

Consider𝐼 = ∫1

16−(2𝑥+3)2 𝑑𝑥

Let (2𝑥 + 3) = 𝑡 then 𝑑𝑥 =𝑑𝑡

2

∴ 𝐼 =1

2∫

𝑑𝑡

42− 𝑡2 =1

2×

1

2×4log |

4+𝑡

4−𝑡| + 𝑐 .

=1

16log |

4+(2𝑥+3)

4−(2𝑥+3)| + 𝑐


1

x awith

respect to 𝒙 and hence evaluate

2

1dx

5x 2x

Substituting 𝑥 = 𝑎 sec 𝜃

𝑑𝑥 = asec 𝑡𝑎𝑛𝑑𝜃

∫1

√𝑥2−𝑎2𝑑𝑥=∫

𝑎𝑠𝑒𝑐.𝑡𝑎𝑛

√𝑎2 sec2 −𝑎2𝑑

=∫𝑎𝑠𝑒𝑐.𝑡𝑎𝑛

√𝑎2(sec2 −1)𝑑 =

∫𝑎𝑠𝑒𝑐.𝑡𝑎𝑛

√𝑎2 tan2 𝑑

=∫ 𝑠𝑒𝑐 𝑑

=log|sec+tan|+C1

=log|𝑥

𝑎+ √

𝑥2

𝑎2 − 1|+C1

= log|𝑥 + √𝑥2 − 𝑎2|-log|a|+C1

=log|𝑥 + √𝑥2 − 𝑎2|+C

where C=C1 –log|a|

Now ∫1

√𝑥2 + 2𝑥 + 2𝑑𝑥

=x2+2x+1=(x2+2x+1)+1

=(x+1)2+(1)2

∫1

√𝑥2 + 2𝑥 + 2𝑑𝑥=∫

1

√(𝑥+1)2+1𝑑𝑥

=log|𝑥 + 1 + √𝑥2 + 2𝑥 + 2| + 𝑐


1

x awith

respect to 𝒙 and hence evaluate

∫𝟏

𝒙𝟐+𝟐𝒙+𝟐𝒅𝒙

Let I= 2 2

1dx;

x aPut x=atan,

then dx=a sec2 ;

x2+a2 =a2 tan2 +a2

=a2(tan2 +1)=a2sec2

2

2 2 2 2

1

1 asec d 1I dx d

x a a sec a

1 1 xtan c

a a a

Consider ∫𝟏

𝒙𝟐+𝟐𝒙+𝟐𝒅𝒙

7.Find the integral of √𝒙𝟐 + 𝒂𝟐 with

respect to x and evaluate ∫ √𝒙𝟐 + 𝟗 𝒅𝒙.

Solution: Let 𝐼 = ∫ √𝑥2 + 𝑎2 𝑑𝑥 applying


We get 𝐼 = 𝑥 √𝑥2 + 𝑎2 −1

2∫

2𝑥2

√𝑥2+𝑎2 𝑑𝑥

= 𝑥 √𝑥2 + 𝑎2 − ∫𝑥2+𝑎2−𝑎2



= 𝑥 √𝑥2 + 𝑎2 − 𝐼 + ∫𝑎2


2𝐼 = 𝑥 √𝑥2 + 𝑎2 + 𝑎2 log|𝑥 + √𝑥2 + 𝑎2| + 𝑐

𝐼 =𝑥

2 √𝑥2 + 𝑎2 +

𝑎2

2log|𝑥 + √𝑥2 + 𝑎2| + 𝑐

Consider ∫ √𝑥2 + 9 𝑑𝑥 =𝑥

2 √𝑥2 + 9

+9

2log |𝑥 + √𝑥2 + 9| + 𝑐

8. Find the integral of √𝒂𝟐 − 𝒙𝟐 with

respect to x and evaluate

∫ √𝟏 + 𝟒𝒙 − 𝒙𝟐 𝒅𝒙.

Solution: Let 𝐼 = ∫ √𝑎2 − 𝑥2 𝑑𝑥 applying


We get 𝐼 = 𝑥 √𝑎2 − 𝑥2 + ∫𝑥2

√𝑎2−𝑥2 𝑑𝑥

= 𝑥 √𝑎2 − 𝑥2 − ∫𝑎2−𝑥2−𝑎2

√𝑎2−𝑥2 𝑑𝑥

𝐼 = 𝑥 √𝑎2 − 𝑥2 − ∫𝑎2

√𝑎2−𝑥2 𝑑𝑥 − 𝐼

2𝐼 = 𝑥 √𝑎2 − 𝑥2 − 𝑎2 sin−1 (𝑥

𝑎) + 𝑐

𝐼 =𝑥

2 √𝑎2 − 𝑥2 +

𝑎2

2sin−1 (

𝑥

𝑎) + 𝑐

Consider ∫ √1 + 4𝑥 − 𝑥2 𝑑𝑥

= ∫ √5 − (𝑥 − 2)2 𝑑𝑥

=(𝑥−2)

2 √5 − (𝑥 − 2)2 −

5

2sin−1 (

𝑥−2

√5) + 𝑐

Find the integral of √𝒙𝟐 − 𝒂𝟐 with

respect to x and hence evaluate ∫ 𝒙𝟐 +

𝟐𝒙 + 𝟓 𝒅𝒙

Solution: Let 𝐼 = ∫ √𝑥2 − 𝑎2 𝑑𝑥 applying


We get 𝐼 = 𝑥 √𝑥2 − 𝑎2 −1

2∫

2𝑥2

√𝑥2−𝑎2 𝑑𝑥

= 𝑥 √𝑥2 − 𝑎2 − ∫(𝑥2−𝑎2)+𝑎2


= 𝑥 √𝑥2 − 𝑎2 − ∫ [√𝑥2 − 𝑎2 +𝑎2

√𝑥2−𝑎2] 𝑑𝑥

= 𝑥 √𝑥2 − 𝑎2 − 𝐼 − ∫𝑎2


2𝐼 = 𝑥 √𝑥2 − 𝑎2 − 𝑎2 log|𝑥 + √𝑥2 − 𝑎2| + 𝑐

𝐼 =𝑥

2 √𝑥2 + 𝑎2 −

𝑎2

2log|𝑥 + √𝑥2 + 𝑎2| + 𝑐

Now consider 𝐼 = ∫ 𝒙𝟐 + 𝟐𝒙 + 𝟓 𝒅𝒙

x2+2x+5=(x2+2x+1)+4=(x+1)2+4

I=∫[(x + 1)2 + 22] dx

Put x+1=t dx=dt

𝐼 = √𝑡2 + 22𝑑𝑡

=𝑡

2√𝑡2 + 22+

4

2 log |𝑡 + √𝑡2 + 22|

=𝑥+1

2√𝑥2 + 2𝑥 + 5

+ 2log| x+1+√𝑥2 + 2𝑥 + 1|

Note: The above questions is for 5 mark

questions in part D of the question paper for

second PUC.

Note: In this chapter “Indefinite Integrals”

Some of the solved examples given in the text

book are not included in the question bank.

The students are advised to go through the

those questions also.

*******


Detail of the concepts to be mastered by every student of class second PUC

with exercises and examples of NCERT Text Book.

Indefinit

e

Integrals

(i) Integration by substitution * Text book , Vol. II Examples 5&6

Page 300, 302,301,303

(ii) ) Application of trigonometric

function in integrals

** Text book , Vol. II Ex 7 Page 306,

Exercise 7.3 Q13&Q24

(iii) Integration of some particular

function

22 ax

dx , ,

, ,

, ,

*** Text book , Vol. II Exp 8, 9, 10

Page 311,312,313, Exercise 7.4 Q

3,4,8,9,13&23

(iv) Integration using Partial Fraction *** Text book , Vol. II Exp 11&12

Page 318 Exp 13 319,Exp 14 & 15

Page320

(v) Integration by Parts ** Text book , Vol. II Exp 18,19&20

Page 325 Exs 7.6 QNO ,10,11,

17,18,20

(vi)Some Special Integrals

,

*** Text book , Vol. II Exp 23 &24

Page 329

(vii) Miscellaneous Questions ** Text book , Vol. II Solved Ex. 40,

41

viii)Some special integrals Text book Supplimentary material

Page 614,615

SYMBOLS USED :

* : Important Questions, ** :Very Important Questions, *** : Very-Very Important Questions

22 ax

dx

dx

xa

1

22 cbxax

dx2

cbxax

dx

2

cbxax

dx)qpx(2

cbxax

dx)qpx(

2

dxxa 22

dxax 22

Definite Integrals One marks questions :

Evaluate the following integrals :-

(1). ∫ sin5 𝑥 𝑑𝑥𝜋 2⁄

−𝜋2⁄

𝑆𝑜𝑙𝑛: ∫ sin5 𝑥 𝑑𝑥𝜋 2⁄

−𝜋2⁄

= 0 ∵ sin5 𝑥 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.

(2). ∫ 2𝑥

1+𝑥2 𝑑𝑥.1

0

𝑆𝑜𝑙𝑛: ∫ 2𝑥

1+𝑥2 𝑑𝑥

1

0 = [log(1 + 𝑥2)]0

1 = log 2

(3). ∫1

√1−𝑥2

1√2

⁄

0 𝑑𝑥 .

𝑆𝑜𝑙𝑛: ∫1

√1−𝑥2

1√2

⁄

0 𝑑𝑥 = (sin⊣ 𝑥)

0

1√2

⁄ = sin⊣ 1

√2− sin⊣ 0 =

𝜋

4 .

(4). ∫ (3 sin 𝑥 − 4 sin3 𝑥) 𝜋

3⁄

0 𝑑𝑥.

𝑆𝑜𝑙𝑛: ∫ (3 sin 𝑥 − 4 sin3 𝑥) 𝜋

3⁄

0 𝑑𝑥 = ∫ sin 3𝑥 𝑑𝑥 =

𝜋3⁄

0−

(cos 3𝑥)0

𝜋3⁄

3 =

−1

3 [−1 − 1] =

2

3

(5). ∫𝑥3

1+𝑥2 1

−1𝑑𝑥.

𝑆𝑜𝑙𝑛: ∫𝑥3

1+𝑥2 1

−1𝑑𝑥 = 0 because

𝑥3

1+𝑥2 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.

Two marks questions :

(1) Evaluate ∫ |𝑥 + 2|5

−5𝑑𝑥

Soln: Let 𝑥 + 2 = 𝑡 ⇒ 𝑑𝑥 = 𝑑𝑡 𝑤ℎ𝑒𝑛 𝑥 = 5 ⇒ 𝑡 = 7; 𝑤ℎ𝑒𝑛 𝑥 = −5 ⇒ 𝑡 = −3

∫ |𝑥 + 2|5

−5𝑑𝑥 ⇒ ∫ |𝑡|

7

−3𝑑𝑡 ⇒

1

2(𝑡|𝑡|)−3

7 ⇒1

2(7|7| − (−3)|−3|) ⇒ 29

(2) Evaluate ∫𝑥

𝑥2+1

3

2 𝑑𝑥

Soln: Let 𝑥2 + 1 = 𝑡 ⇒ 2𝑥 𝑑𝑥 = 𝑑𝑡 ⇒ 𝑥𝑑𝑥 =𝑑𝑡

2 𝑤ℎ𝑒𝑛 𝑥 = 2 ⇒ 𝑡 = 5; 𝑤ℎ𝑒𝑛 𝑥 = 3 ⇒ 𝑡 = 10

∫𝑥

𝑥2+1

3

2 𝑑𝑥 = ∫

1

𝑡

10

5

𝑑𝑡

2=

1

2(log 𝑡)5

10 =1

2 (log 10 − log 5) =

1

2log 2.

(3) Evaluate ∫ cos 2𝑥𝜋

20

𝑑𝑥.

Soln : ∫ cos 2𝑥𝜋

20

𝑑𝑥 =1

2(sin 2𝑥)

0

𝜋

2 =1

2(0 − 0) = 0

(4) Evaluate ∫sin 𝑥

1+cos2 𝑥

𝜋

20

𝑑𝑥

Soln: Let cos 𝑥 = 𝑡 ⇒ sin 𝑥 𝑑𝑥 = −𝑑𝑡; 𝑤ℎ𝑒𝑛 𝑥 = 0 ⇒ 𝑡 = 1; 𝑤ℎ𝑒𝑛 𝑥 =𝜋

2⇒ 𝑡 = 0

∫sin 𝑥

1+cos2 𝑥

𝜋

20

𝑑𝑥 = − ∫𝑑𝑡

1+𝑡2

0

1= −(tan−1 𝑡)1

0 = − (0 −𝜋

4) =

𝜋

4

(5) ∫ 𝑥 𝜋

0sin 𝑥 cos2 𝑥 𝑑𝑥.

𝑆𝑜𝑙𝑛: 𝐿𝑒𝑡 𝐼 = ∫ 𝑥 𝜋

0sin 𝑥 cos2 𝑥 𝑑𝑥 = ∫ (𝜋 − 𝑥) sin 𝑥 cos2 𝑥

𝜋

0𝑑𝑥.

∴ 2 𝐼 = ∫ (𝑥 + 𝜋 − 𝑥)𝜋

0 sin 𝑥 cos2 𝑥 𝑑𝑥 = 𝜋 ∫ cos2 𝑥.

𝜋

0 sin 𝑥 𝑑𝑥

= 𝜋 ∫ 𝑡2 −1

1 (−𝑑𝑡) here cos 𝑥 = 𝑡 ⇒ − sin 𝑥 𝑑𝑥 = 𝑑𝑡 when 𝑥 = 0 ⇒ 𝑡 = 1; 𝑥 = 𝜋 ⇒ 𝑡 = −1

2𝐼 = 𝜋 ∫ 𝑡2 𝑑𝑡

1

−1

= 𝜋.(𝑡3)−1

1

3 =

𝜋

3 (1 + 1) =

2𝜋

3 ⇒ 𝐼 =

𝜋

3 .

Six marks questions :

(1) Prove that ∫ 𝒇(𝒙)𝒃

𝒂 𝒅𝒙 = ∫ 𝒇(𝒂 + 𝒃 − 𝒙)

𝒃

𝒂𝒅𝒙 and evaluate ∫

𝒅𝒙

𝟏+√𝐭𝐚𝐧 𝒙

𝝅

𝟑𝝅

𝟔

.

Soln: Consider ∫ 𝑓(𝑎 + 𝑏 − 𝑥)𝑏

𝑎𝑑𝑥

𝑝𝑢𝑡 𝑎 + 𝑏 − 𝑥 = 𝑡

⤇ − 𝑑𝑥 = 𝑑𝑡. 𝐴𝑡 𝑥 = 𝑎. , 𝑡 = 𝑎 + 𝑏 – 𝑎 = 𝑏 𝑎𝑛𝑑 𝑎𝑡 𝑥 = 𝑏. , 𝑡 = 𝑎 + 𝑏 – 𝑏 = 𝑎

∴ 𝑅𝐻𝑆 = ∫ 𝑓(𝑎 + 𝑏 − 𝑥) 𝑑𝑥𝑏

𝑎= ∫ 𝑓(𝑡) (−𝑑𝑡)

𝑎

𝑏 [∵ ∫ 𝑓(𝑥) 𝑑𝑥

𝑏

𝑎= − ∫ 𝑓(𝑥)

𝑎

𝑏 𝑑𝑥 ]

= ∫ 𝑓(𝑡) 𝑑𝑡 𝑏

𝑎= ∫ 𝑓(𝑥) 𝑑𝑥

𝑏

𝑎= 𝐿𝐻𝑆. [Definite integrals are independent of variable]

Consider ∫𝑑𝑥

1+√tan 𝑥

𝜋

3𝜋

6

= ∫√cos 𝑥

√cos 𝑥+√sin 𝑥

𝜋 3⁄

𝜋 6⁄ 𝑑𝑥

Let I = ∫√cos 𝑥


𝜋 3⁄

𝜋 6⁄ 𝑑𝑥 = ∫

√cos(𝜋

6+

𝜋

3−𝑥)

√cos(𝜋

6+

𝜋

3−𝑥) + √sin(

𝜋

6+

𝜋

3−𝑥)

𝜋 3⁄

𝜋 6⁄ 𝑑𝑥 ..........(1)

⇒ 𝐼 = ∫√sin 𝑥

√sin 𝑥 + √cos 𝑥

𝜋 3⁄

𝜋 6⁄ 𝑑𝑥....................(2)

Adding (1) and (2)

2 I = I + I = ∫√cos 𝑥


𝜋 3⁄

𝜋 6⁄ 𝑑𝑥 + ∫

√sin 𝑥

√sin 𝑥 + √cos 𝑥

𝜋 3⁄

𝜋 6⁄ 𝑑𝑥 = ∫



𝜋 3⁄

𝜋 6⁄ 𝑑𝑥

2𝐼 = (𝑥)𝜋6⁄

𝜋3⁄

= 𝜋

3−

𝜋

6=

𝜋

6 ∴ I =

𝜋

12

(2) Prove that ∫ 𝒇(𝒙)𝒂

𝟎 𝒅𝒙 = ∫ 𝒇(𝒂 − 𝒙)

𝒂

𝟎 𝒅𝒙 and hence evaluate ∫ 𝐥𝐨𝐠(𝟏 + 𝐭𝐚𝐧 𝒙)

𝝅

𝟒𝟎

𝒅𝒙.

Soln : Consider ∫ 𝑓(𝑎 − 𝑥)𝑎

0 𝑑𝑥

Let 𝑎 − 𝑥 = 𝑡 ⇒ 𝑑𝑥 = −𝑑𝑡; 𝑤ℎ𝑒𝑛 𝑥 = 0 ⇒ 𝑡 = 𝑎; 𝑤ℎ𝑒𝑛 𝑥 = 𝑎 ⇒ 𝑡 = 0

∫ 𝑓(𝑎 − 𝑥)𝑎

0 𝑑𝑥 = ∫ 𝑓(𝑡)(−𝑑𝑡)

0

𝑎 [∵ ∫ 𝑓(𝑥) 𝑑𝑥

𝑏

𝑎= − ∫ 𝑓(𝑥)

𝑎

𝑏 𝑑𝑥 ]

= ∫ 𝑓(𝑡)𝑎

0𝑑𝑡 = ∫ 𝑓(𝑥)

𝑎

0𝑑𝑥 [Definite integrals are independent of variable]

Let I = ∫ log(1 + tan 𝑥) 𝜋

4⁄

0 𝑑𝑥 = ∫ log (1 + tan ⟨

𝜋

4− 𝑥⟩)

𝜋4⁄

0 𝑑𝑥 = ∫ log (1 +

1−tan 𝑥

1+tan 𝑥)

𝜋4⁄

0 𝑑𝑥

= ∫ log (2

1+tan 𝑥)

𝜋4⁄

0 𝑑𝑥

∴ 2I = I + I = ∫ log(1 + tan 𝑥)

𝜋4⁄

0

𝑑𝑥 + ∫ log (2

1 + tan 𝑥)

𝜋4⁄

0

𝑑𝑥 = ∫ log 2 𝑑𝑥 =

𝜋4⁄

0

log 2 (𝑥)0

𝜋4⁄

=𝜋

4 log 2

∴ I = 𝜋

8 log 2.

(3) Prove that ∫ 𝒇(𝒙) 𝒅𝒙𝟐𝒂

𝟎= {

𝟐. ∫ 𝒇(𝒙)𝒂

𝟎 𝐝𝐱 , 𝐢𝐟 𝐟(𝟐𝐚 − 𝐱) = 𝒇(𝒙),

𝟎 , 𝒊𝒇 𝒇(𝟐𝒂 − 𝒙) = −𝒇(𝒙) and hence evaluate

∫𝟏

𝒂𝟐 𝐜𝐨𝐬𝟐 𝒙+𝒃𝟐 𝐬𝐢𝐧𝟐 𝒙

𝝅

𝟎𝒅𝒙.

Soln : ∫ 𝑓(𝑥) 𝑑𝑥 =2𝑎

0 ∫ 𝑓(𝑥) 𝑑𝑥

𝑎

0+ ∫ 𝑓(𝑥) 𝑑𝑥

2𝑎

𝑎… . . (1)

𝑃𝑢𝑡 2𝑎 − 𝑥 = 𝑡 ⤇ 𝑑𝑥 = −𝑑𝑡

Consider ∫ 𝑓(𝑥) 𝑑𝑥.2𝑎

𝑎 𝐴𝑡 𝑥 = 𝑎. , 𝑡 = 𝑎, 𝑎𝑛𝑑 𝑎𝑡 𝑥 = 2𝑎, 𝑡 = 0

= ∫ 𝑓(2𝑎 − 𝑡)(−𝑑𝑡)0

𝑎

= ∫ 𝑓(2𝑎 − 𝑡) 𝑑𝑡 𝑎

0= ∫ 𝑓(2𝑎 − 𝑥) 𝑑𝑥

𝑎

0… . (2)

𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 (2) 𝑖𝑛 (1)., 𝑤𝑒 𝑔𝑒𝑡 ∶ ∫ 𝑓(𝑥) 𝑑𝑥 =2𝑎

0∫ 𝑓(𝑥) 𝑑𝑥

𝑎

0+ ∫ 𝑓(2𝑎 − 𝑥)

𝑎

0 𝑑𝑥

𝐼𝑓 𝑓(2𝑎 − 𝑥) = 𝑓(𝑥) the above result ∫ 𝑓(𝑥) 2𝑎

0𝑑𝑥 = 2 ∫ 𝑓(𝑥) 𝑑𝑥. ,

𝑎

0

𝐴𝑛𝑑 𝑖𝑓 𝑓(2𝑎 − 𝑥) = −𝑓(𝑥)., it reduces to: ∫ 𝑓(𝑥) 2𝑎

0𝑑𝑥 = 0.

Let I = ∫1

𝑎2 cos2 𝑥+ 𝑏2 sin2 𝑥

𝜋

0 𝑑𝑥 = 2 ∫

1

𝑎2 cos2 𝑥+ 𝑏2 sin2 𝑥

𝜋2⁄

0 𝑑𝑥

= 2 ∫sec2 𝑥

a2 + 𝑏2 tan2 𝑥

𝜋2⁄

0 𝑑𝑥

Put b tan 𝑥 = 𝑡, ⤇ 𝑏 sec2 𝑥 𝑑𝑥 = 𝑑𝑡, ∴ sec2 𝑥 𝑑𝑥 = 𝑑𝑡

𝑏, 𝐴𝑡 𝑥 = 0., 𝑡 = 0, 𝐴𝑡 𝑥 =

𝜋

2, 𝑡 → ∞.

𝐼 = 2 ∫𝑑𝑡

𝑏(𝑎2+𝑡2)

∞

0= 2

1

𝑏

1

𝑎 (tan⊣

𝑡

𝑎)

0

∞

= 2

𝑎𝑏 (

𝜋

2− 0) =

𝜋

𝑎𝑏

(4) Prove that ∫ 𝒇(𝒙) 𝒅𝒙𝒃

𝒂= ∫ 𝒇(𝒙)

𝒄

𝒂 𝒅𝒙 + ∫ 𝒇(𝒙)

𝒃

𝒄 𝒅𝒙 and hence evaluate ∫ |𝒙𝟑𝟐

−𝟏− 𝒙|𝒅𝒙

Soln: 𝑒𝑡 𝐹(𝑥)𝑏𝑒 𝑡ℎ𝑒 𝑎𝑛𝑡𝑖𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑓(𝑥).

𝑇ℎ𝑒𝑛. , 𝑅𝐻𝑆 = ∫ 𝑓(𝑥)

𝑐

𝑎

𝑑𝑥 + ∫ 𝑓(𝑥)𝑑𝑥

𝑏

𝑐

= 𝐹 (𝑐) − 𝐹 (𝑎) + 𝐹 (𝑏) – 𝐹 (𝑐)

∫ f(x) dxb

a= F(b) = F(a) = ∫ f(x)

b

adx = LHS.

𝐶𝑙𝑒𝑎𝑟𝑙𝑦, 𝑥3 − 𝑥 𝑥(𝑥 − 1) (𝑥 + 1) will be positive between −1 & 0, negative between 0 & 1 and again positive between 1 & 2.

∴ ∫(𝑥3 − 𝑥) 𝑑𝑥 =

2

−1

∫(𝑥3 − 𝑥) 𝑑𝑥 +

0

−1

∫(𝑥 − 𝑥3)

1

0

𝑑𝑥 + ∫(𝑥3 − 𝑥)

2

1

𝑑𝑥.

= (𝑥4

4−

𝑥2

2)

−1

0

+ (𝑥2

2−

𝑥4

4)

0

1

+ (𝑥4

4−

𝑥2

2)

1

2

= 0 − [1

4−

1

2] + [

1

2−

1

4] + {(4 − 2) − (

1

4−

1

2)}

= + 1

4 +

1

4 + 2 +

1

4= 2

3

4=

11

4


−𝒂 𝒅𝒙 = {

𝟐 ∫ 𝒇(𝒙)𝒂

𝟎 𝐝𝐱. , 𝐢𝐟 𝐟 𝐢𝐬 𝐚𝐧 𝐞𝐯𝐞𝐧 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧

𝟎 , 𝒊𝒇 𝒇 𝒊𝒔 𝒂𝒏 𝒐𝒅𝒅 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 and hence evaluate

∫ 𝐬𝐢𝐧𝟕 𝒙𝝅

𝟐

–𝝅

𝟐

𝒅𝒙.

Soln: consider ∫ 𝑓(𝑥) 𝑑𝑥𝑎

−𝑎= ∫ 𝑓(𝑥)

0

−𝑎 𝑑𝑥 + ∫ 𝑓(𝑥) 𝑑𝑥

𝑎

0… . (1)

𝑃𝑢𝑡 𝑥 = −𝑡 ⤇ 𝑥 = −𝑑𝑡

Consider ∫ 𝑓(𝑥) 𝑑𝑥0

−𝑎 𝐴𝑡 𝑥 = −𝑎. , 𝑡 = 𝑎 𝐴𝑛𝑑, 𝑎𝑡 𝑥 = 0., 𝑡 = 0.

= ∫ 𝑓(−𝑡)(−𝑑𝑡)0

𝑎 = ∫ 𝑓(−𝑡) 𝑑𝑡

𝑎

0= ∫ 𝑓(−𝑥) 𝑑𝑥

𝑎

0= {

∫ 𝑓(𝑥)𝑎

0dx if f is an even function, and,

− ∫ 𝑓(𝑥)𝑑𝑥𝑎

0 𝑖𝑓 𝑓 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.

… . (2)

𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑠 𝑒𝑞. (2) 𝑖𝑛 𝑒𝑞. (1). , 𝑤𝑒 𝑔𝑒𝑡 ∶

∫ 𝑓(𝑥) 𝑑𝑥

𝑎

−𝑎

= (∫ 𝑓(𝑥) 𝑑𝑥, 𝑖𝑓 𝑓 𝑖𝑠 𝑒𝑣𝑒𝑛

𝑎

0

− ∫ 𝑓(𝑥) 𝑑𝑥, 𝑖𝑓 𝑓 𝑖𝑠 𝑜𝑑𝑑𝑎

0

) + ∫ 𝑓(𝑥) 𝑑𝑥

𝑎

0

= {2. ∫ 𝑓(𝑥)

𝑎

0

dx , if f is an even function, and,

0 , 𝑖𝑓 𝑓 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.

Hence the value of ∫ sin7 𝑥 𝑑𝑥𝜋

2⁄−𝜋

2

= 0 [∴ sin7(−𝑥) = − sin7 𝑥 ⤇ 𝑖𝑡 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛]


𝟎 𝒅𝒙 = ∫ 𝒇(𝒂 − 𝒙)

𝒂

𝟎 𝒅𝒙 and hence evaluate ∫ 𝐥𝐨𝐠 𝐬𝐢𝐧 𝒙

𝝅𝟐⁄

𝟎𝒅𝒙.

Soln : Consider ∫ 𝑓(𝑎 − 𝑥)𝑎

0 𝑑𝑥


∫ 𝑓(𝑎 − 𝑥)𝑎

0 𝑑𝑥 = ∫ 𝑓(𝑡)(−𝑑𝑡)

0

𝑎 [∵ ∫ 𝑓(𝑥) 𝑑𝑥

𝑏

𝑎= − ∫ 𝑓(𝑥)

𝑎

𝑏 𝑑𝑥 ]

= ∫ 𝑓(𝑡)𝑎

0𝑑𝑡 = ∫ 𝑓(𝑥)

𝑎


𝐿𝑒𝑡 𝐼 = ∫ log sin 𝑥 𝑑𝑥 =

𝜋 2⁄

0

∫ log sin (𝜋

2− 𝑥) 𝑑𝑥 =

𝜋 2⁄

0

∫ log cos 𝑥

𝜋 2⁄

0

𝑑𝑥

2 I = ∫ log sin 𝑥𝜋 2⁄

0 𝑑𝑥 + ∫ log cos 𝑥 𝑑𝑥

𝜋 2⁄

0= ∫ log(sin 𝑥 cos 𝑥)

𝜋 2⁄

0 𝑑𝑥

= ∫ log (sin 2𝑥

2) 𝑑𝑥 =

𝜋 2⁄

0∫ log sin 2𝑥 𝑑𝑥 −

𝜋 2⁄

0∫ log 2 𝑑𝑥

𝜋 2⁄

0… . (1)

Consider ∫ log sin 2𝑥 𝑑𝑥𝜋 2⁄

0 𝑃𝑢𝑡 2𝑥 = 𝑡 ⤇ 𝑑𝑥 =

𝑑𝑡

2

𝐴𝑡 𝑥 = 0, 𝑡 = 0. 𝐴𝑛𝑑,

= ∫ log sin 𝑡 𝑑𝑡

2

𝜋

0 𝐴𝑡 𝑥 = 𝜋

2,⁄ 𝑡 = 𝜋.

= ∫ log sin 𝑡 𝑑𝑡

𝜋 2⁄

0

[∴ ∫ 𝑓(𝑥)𝑑𝑥 = 2 ∫ 𝑓(𝑥)𝑑𝑥, 𝑖𝑓 𝑓(2𝑎 − 𝑥) = 𝑓(𝑥)

𝑎

0

2𝑎

0

]

= ∫ log sin 2𝑥

𝜋 2⁄

0

𝑑𝑥 = 𝐼 … … (2)

𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑠 𝑒𝑞 (2) 𝑖𝑛 𝑒𝑞. (1)., 𝑤𝑒 𝑔𝑒𝑡 ∶ 2 I = I − ∫ log 2 𝑑𝑥 𝜋 2⁄

0∴ I = − log 2 (𝑥)

0

𝜋2⁄

= − 𝜋

2 log 2


𝟎𝒅𝒙 = 𝟐 ∫ 𝒇(𝒙)

𝒂

𝟎𝒅𝒙 when 𝒇(𝟐𝒂 − 𝒙) = 𝒇(𝒙) and hence evaluate ∫ |𝐜𝐨𝐬 𝒙|

𝝅

𝟎𝒅𝒙 .

Soln : ∫ 𝑓(𝑥) 𝑑𝑥 =2𝑎

0 ∫ 𝑓(𝑥) 𝑑𝑥

𝑎

0+ ∫ 𝑓(𝑥) 𝑑𝑥

2𝑎

𝑎… . . (1)

𝑃𝑢𝑡 2𝑎 − 𝑥 = 𝑡 ⤇ 𝑑𝑥 = −𝑑𝑡

Consider ∫ 𝑓(𝑥) 𝑑𝑥.2𝑎

𝑎 𝐴𝑡 𝑥 = 𝑎. , 𝑡 = 𝑎, 𝑎𝑛𝑑 𝑎𝑡 𝑥 = 2𝑎, 𝑡 = 0

= ∫ 𝑓(2𝑎 − 𝑡)(−𝑑𝑡)0

𝑎

= ∫ 𝑓(2𝑎 − 𝑡) 𝑑𝑡 𝑎

0= ∫ 𝑓(2𝑎 − 𝑥) 𝑑𝑥

𝑎

0… . (2)

𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 (2) 𝑖𝑛 (1)., 𝑤𝑒 𝑔𝑒𝑡 ∶ ∫ 𝑓(𝑥) 𝑑𝑥 =2𝑎

0∫ 𝑓(𝑥) 𝑑𝑥

𝑎

0+ ∫ 𝑓(2𝑎 − 𝑥)

𝑎

0 𝑑𝑥

𝐼𝑓 𝑓(2𝑎 − 𝑥) = 𝑓(𝑥) the above result becomes ∫ 𝑓(𝑥) 2𝑎

0𝑑𝑥 = 2 ∫ 𝑓(𝑥) 𝑑𝑥

𝑎

0

Consider 𝐼 = ∫ |cos 𝑥|𝜋

0𝑑𝑥 = 2 ∫ |cos 𝑥|

𝜋

20

𝑑𝑥 because |cos(𝜋 − 𝑥)| = |− cos 𝑥| = |cos 𝑥|

∴ 𝐼 = 2 ∫ cos 𝑥𝜋

20

𝑑𝑥 because cos 𝑥 is positive in 1st quadrant.

𝐼 = 2 (sin 𝑥)0

𝜋

2 = 2(1 − 0) = 2.


−𝒂𝒅𝒙 = {

𝟐 ∫ 𝒇(𝒙)𝒂

𝟎 𝐝𝐱. , 𝐢𝐟 𝐟 𝐢𝐬 𝐚𝐧 𝐞𝐯𝐞𝐧 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧

𝟎 , 𝒊𝒇 𝒇 𝒊𝒔 𝒂𝒏 𝒐𝒅𝒅 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 and hence evaluate

∫ (𝒙𝟑 + 𝒙 𝐜𝐨𝐬 𝒙 + 𝐭𝐚𝐧𝟓 𝒙 + 𝟏)𝝅

𝟐

−𝝅

𝟐

𝒅𝒙

Soln: consider ∫ 𝑓(𝑥) 𝑑𝑥𝑎

−𝑎= ∫ 𝑓(𝑥)

0

−𝑎 𝑑𝑥 + ∫ 𝑓(𝑥) 𝑑𝑥

𝑎

0… . (1)

𝑃𝑢𝑡 𝑥 = −𝑡 ⤇ 𝑥 = −𝑑𝑡

Consider ∫ 𝑓(𝑥) 𝑑𝑥0

−𝑎 𝐴𝑡 𝑥 = −𝑎. , 𝑡 = 𝑎 𝐴𝑛𝑑, 𝑎𝑡 𝑥 = 0., 𝑡 = 0.

= ∫ 𝑓(−𝑡)(−𝑑𝑡)0

𝑎 = ∫ 𝑓(−𝑡) 𝑑𝑡

𝑎

0= ∫ 𝑓(−𝑥) 𝑑𝑥

𝑎

0= {

∫ 𝑓(𝑥)𝑎

0dx if f is an even function, and,

− ∫ 𝑓(𝑥)𝑑𝑥𝑎

0 𝑖𝑓 𝑓 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.

… . (2)

𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑠 𝑒𝑞. (2) 𝑖𝑛 𝑒𝑞. (1). , 𝑤𝑒 𝑔𝑒𝑡 ∶

∫ 𝑓(𝑥) 𝑑𝑥

𝑎

−𝑎

= (∫ 𝑓(𝑥) 𝑑𝑥, 𝑖𝑓 𝑓 𝑖𝑠 𝑒𝑣𝑒𝑛

𝑎

0

− ∫ 𝑓(𝑥) 𝑑𝑥, 𝑖𝑓 𝑓 𝑖𝑠 𝑜𝑑𝑑𝑎

0

) + ∫ 𝑓(𝑥) 𝑑𝑥

𝑎

0

= {2. ∫ 𝑓(𝑥)

𝑎

0

dx , if f is an even function, and,

0 , 𝑖𝑓 𝑓 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.

Consider 𝐼 = ∫ (𝑥3 + 𝑥 cos 𝑥 + tan5 𝑥 + 1)𝜋

2

−𝜋

2

𝑑𝑥 since 𝑥3, 𝑥 cos 𝑥 , tan5 𝑥 odd functions

𝐼 = ∫ 1𝜋

2

−𝜋

2

𝑑𝑥 = (𝑥)−

𝜋

2

𝜋

2 =𝜋

2− (

𝜋

2) = 𝜋.


𝟎 𝒅𝒙 = ∫ 𝒇(𝒂 − 𝒙)

𝒂

𝟎 𝒅𝒙 and hence evaluate ∫

𝐬𝐢𝐧𝟑𝟐 𝒙

𝐬𝐢𝐧𝟑𝟐 𝒙+𝐜𝐨𝐬

𝟑𝟐 𝒙

𝝅𝟐⁄

𝟎𝒅𝒙.

Soln: Consider ∫ 𝑓(𝑎 − 𝑥)𝑎

0 𝑑𝑥


∫ 𝑓(𝑎 − 𝑥)𝑎

0 𝑑𝑥 = ∫ 𝑓(𝑡)(−𝑑𝑡)

0

𝑎 [∵ ∫ 𝑓(𝑥) 𝑑𝑥

𝑏

𝑎= − ∫ 𝑓(𝑥)

𝑎

𝑏 𝑑𝑥 ]

= ∫ 𝑓(𝑡)𝑎

0𝑑𝑡 = ∫ 𝑓(𝑥)

𝑎


Consider 𝐼 = ∫sin

𝟑𝟐 𝑥

sin𝟑𝟐 𝑥+cos

𝟑𝟐 𝑥

𝜋2⁄

0𝑑𝑥 ..................(1)

= ∫sin

𝟑𝟐(

𝜋

2−𝑥)

sin𝟑𝟐(

𝜋

2−𝑥)+cos

𝟑𝟐(

𝜋

2−𝑥)

𝜋2⁄

0𝑑𝑥

𝐼 = ∫cos

𝟑𝟐 𝑥

cos𝟑𝟐 𝑥+sin

𝟑𝟐 𝑥

𝜋2⁄

0𝑑𝑥 ....................(2)

Adding (1) and (2)

2𝐼 = ∫sin

𝟑𝟐 𝑥


𝟑𝟐 𝑥

𝜋2⁄

0+

cos𝟑𝟐 𝑥

cos𝟑𝟐 𝑥+sin

𝟑𝟐 𝑥

𝑑𝑥 = ∫sin

𝟑𝟐 𝑥+cos

𝟑𝟐 𝑥


𝟑𝟐 𝑥

𝜋2⁄

0 𝑑𝑥 = ∫ 𝑑𝑥

𝜋2⁄

0= (𝑥)

0

𝜋

2 =𝜋

2− 0 =

𝜋

2

Hence 𝐼 =𝜋

4.


𝟎 𝒅𝒙 = ∫ 𝒇(𝒂 − 𝒙)

𝒂

𝟎 𝒅𝒙 and hence evaluate ∫ (𝟐 𝐥𝐨𝐠 𝐬𝐢𝐧 𝒙 − 𝐥𝐨𝐠 𝐬𝐢𝐧 𝟐𝒙)

𝝅𝟐⁄

𝟎𝒅𝒙.

Soln: Consider ∫ 𝑓(𝑎 − 𝑥)𝑎

0 𝑑𝑥


∫ 𝑓(𝑎 − 𝑥)𝑎

0 𝑑𝑥 = ∫ 𝑓(𝑡)(−𝑑𝑡)

0

𝑎 [∵ ∫ 𝑓(𝑥) 𝑑𝑥

𝑏

𝑎= − ∫ 𝑓(𝑥)

𝑎

𝑏 𝑑𝑥 ]

= ∫ 𝑓(𝑡)𝑎

0𝑑𝑡 = ∫ 𝑓(𝑥)

𝑎


Consider 𝐼 = ∫ (2 log sin 𝑥 − log sin 2𝑥)𝜋

2⁄

0𝑑𝑥 = ∫ (log sin2 𝑥 − log sin 2𝑥)

𝜋2⁄

0𝑑𝑥

= ∫ log (sin2 𝑥

sin 2𝑥 )

𝜋2⁄

0 𝑑𝑥 = ∫ log (

sin2 𝑥

2sin 𝑥 cos 𝑥 )

𝜋2⁄

0𝑑𝑥 = ∫ log (

1

2tan 𝑥)

𝜋2⁄

0𝑑𝑥 ..........(1)

𝐼 = ∫ log (1

2(tan (

𝜋

2− 𝑥)) )

𝜋2⁄

0𝑑𝑥 = ∫ log (

1

2cot 𝑥)

𝜋2⁄

0...............(2)

Adding (1) and (2)

2𝐼 = ∫ log (1

2tan 𝑥 ) + log (

1

2cot 𝑥)

𝜋2⁄

0

𝑑𝑥 = ∫ log (1

2tan 𝑥 .

1

2cot 𝑥 )

𝜋2⁄

0

𝑑𝑥 = ∫ log (1

4)

𝜋2⁄

0

𝑑𝑥 = log1

4(𝑥)

0

𝜋2⁄

2𝐼 =𝜋

2log

1

4 ⇒ 𝐼 =

𝜋

4log

1

4.

1

CHAPTER 8:

APPLICATION OF INTEGRALS 3 mark questions

Question 1:

Find the area of the region bounded by the

curve y2 = x and the lines x = 1, x = 4 and the

x-axis.

Answer :

The area of the region bounded by the curve,

y2 = x, the lines, x = 1 and x = 4, and the x-

axis is the area ABCD.

Question 2:

Find the area of the region bounded by y2 =

9x, x = 2, x = 4 and the x-axis in the first

quadrant.

Answer :


y2 = 9x, x = 2, and x = 4, and the x-axis is the

area ABCD.

2

Question 3:

Find the area of the region bounded by x2 =

4y, y = 2, y = 4 and the y-axis in the first

quadrant.

Answer :


x2 = 4y, y = 2, and y = 4, and the y-axis is the

area ABCD.

Question 4:


curve y2 = 4x, y-axis and the line y = 3 is

Answer : The area bounded by the curve, y

2 = 4x, y-

axis, and y = 3 is represented as

Question 5:

Find the area lying between the curve y2 =

4x and y = 2x is

Answer :

The area lying between the curve, y2 = 4x

and y = 2x, is represented by the shaded area

OBAO as

3

The points of intersection of these curves are

O (0, 0) and A (1, 2).

We draw AC perpendicular to x-axis such

that the coordinates of C are (1, 0).

∴ Area OBAO = Area (OCABO) – Area

(ΔOCA)

square units

Question 5. Find area enclosed by the Parabola y

2=4ax

and its latus rectum by integration Solution: y

2 = 4ax ---- (1) and

the equation of the Latus rectum is given by x = a ……… (2)

From (2) and (1) y2 =4a

2 y = 2a

Required area A = 2 [area of OSP

= 2 ∫

= 2 ∫ √

. √ . dx

= 4 √ (

)

= √

[ a√ ] =

Sq. units

5 MARK QUESTIONS:

Question 1:


ellipse

Answer : The given equation of the ellipse,

, can be represented as

It can be observed that the ellipse is

symmetrical about x-axis and y-axis.

4

∴ Area bounded by ellipse = 4 × Area of

OAB

Therefore, area bounded by the ellipse = 4 ×

3π = 12π units

Question 2:


ellipse

Answer : The given equation of the ellipse can be

represented as

It can be observed that the ellipse is

symmetrical about x-axis and y-axis.

∴ Area bounded by ellipse = 4 × Area OAB

Therefore, area bounded by the ellipse =

Question 3:

Find the area of the region in the first

quadrant enclosed by x-axis, line

and the circle

Answer : The area of the region bounded by the circle,

, and the x-axis is the

area OAB.

5

The point of intersection of the line and the

circle in the first quadrant is .

Area OAB = Area ΔOCA + Area ACB

Area of OAC

Area of ABC

Therefore, area enclosed by x-axis, the line

, and the circle in the

first quadrant =

Question 7:

Find the area of the smaller part of the circle

x2 + y

2 = a

2 cut off by the line

Answer :

The area of the smaller part of the circle, x2

+ y2 = a

2, cut off by the line, , is the

area ABCDA.

It can be observed that the area ABCD is

symmetrical about x-axis.

∴ Area ABCD = 2 × Area ABC

6

Therefore, the area of smaller part of the

circle, x2 + y

2 = a

2, cut off by the line,

, is units.

Question 8:

The area between x = y2 and x = 4 is divided

into two equal parts by the line x = a, find

the value of a.

Answer :

The line, x = a, divides the area bounded by

the parabola and x = 4 into two equal parts.

∴ Area OAD = Area ABCD

It can be observed that the given area is

symmetrical about x-axis.

⇒ Area OED = Area EFCD

From (1) and (2), we obtain

Therefore, the value of a is .

Question 9:


parabola y = x2 and

Answer : The area bounded by the parabola, x

2 =

y,and the line, , can be represented as

7

The given area is symmetrical about y-axis.

∴ Area OACO = Area ODBO

The point of intersection of parabola, x2 = y,

and line, y = x, is A (1, 1).

Area of OACO = Area ΔOAM – Area

OMACO

Area of ΔOAM

Area of OMACO

⇒ Area of OACO = Area of ΔOAM – Area

of OMACO

Therefore, required area = units

Question 10:

Find the area bounded by the curve x2 = 4y

and the line x = 4y – 2

Answer : The area bounded by the curve, x

2 = 4y, and

line, x = 4y – 2, is represented by the shaded

area OBAO.

Let A and B be the points of intersection of

the line and parabola.

Coordinates of point .

Coordinates of point B are (2, 1).

We draw AL and BM perpendicular to x-

axis.

It can be observed that,

Area OBAO = Area OBCO + Area OACO

… (1)

Then, Area OBCO = Area OMBC – Area

OMBO

Similarly, Area OACO

= Area OLAC – Area OLAO

8

Therefore, required area =

Question 11:


curve y2 = 4x and the line x = 3

Answer : The region bounded by the parabola, y

2 =

4x, and the line, x = 3, is the area OACO.

The area OACO is symmetrical about x-

axis.

∴ Area of OACO = 2 (Area of OAB)

Therefore, the required area is units.

Question 12:

Find the area of the circle 4x2 + 4y

2 = 9

which is interior to the parabola x2 = 4y

Answer : The required area is represented by the

shaded area OBCDO.

Solving the given equation of circle, 4x

2 +

4y2 = 9, and parabola, x

2 = 4y, we obtain the

point of intersection as

.

It can be observed that the required area is

symmetrical about y-axis.

9

∴ Area OBCDO = 2 × Area OBCO

We draw BM perpendicular to OA.

Therefore, the coordinates of M are .

Therefore, Area OBCO = Area OMBCO –

Area OMBO

Therefore, the required area OBCDO is

units

Question :13

Using integration finds the area of the region

bounded by the triangle whose vertices are

(–1, 0), (1, 3) and (3, 2).

Answer : BL and CM are drawn perpendicular to x-

axis.

It can be observed in the following figure

that,

Area (ΔACB) = Area (ALBA) + Area

(BLMCB) – Area (AMCA) … (1)

Equation of line segment AB is

Equation of line segment BC is

Equation of line segment AC is

Therefore, from equation (1), we obtain

Area (ΔABC) = (3 + 5 – 4) = 4 units

Question 14:

Using integration find the area of the

triangular region whose sides have the

equations y = 2x +1, y = 3x + 1 and x = 4.

Answer : The equations of sides of the triangle are y =

2x +1, y = 3x + 1, and x = 4.

On solving these equations, we obtain the

vertices of triangle as A(0, 1), B(4, 13), and

C (4, 9).


10

Area (ΔACB) = Area (OLBAO) –Area

(OLCAO)

Question 15:

Find the smaller area enclosed by the circle

x2 + y

2 = 4 and the line x + y = 2 is

Answer :

The smaller area enclosed by the circle, x2 +

y2 = 4, and the line, x + y = 2, is represented

by the shaded area ACBA as


Area ACBA = Area OACBO – Area

(ΔOAB)

1

Differential Equations

1 Mark questions:

1. Define a differential equation.

It is an equation containing derivatives.

2. Define order of a differential equation.

It is highest order of derivative appearing in the given equation.

3. Define degree of a differential equation.

It is the highest power of highest ordered derivative appearing in the given

equation.

4. Define general solution of a differential equation.

It is solution of given differential equation and it contains arbitrary constants.

5. Define particular solution of a differential equation.

It is that solution of given differential equation and is free from arbitrary constants.

2 Marks questions:

1. Form a differential equation of family of

i. Straight lines with slope = m and passing through origin.

Consider a straight line with slope = m and passing through origin

i.e. y = mx -----(1)

y1 = m -------(2) y = x y

1

ii. Circles with centre on y-axis and passing through origin.

022

0222102

122

1

1

1122

yyxyxfy

yyx

fyyyxfyyxConsider

2. Solve the following by using separation of variables.

i. dydxdxydyx

cyxxyyxdxydyxdxyd

ii. 01 2

2

x

ya

dx

dy

cxyx

dx

y

dy

11

22sinsin0

11

iii. 011 22 ydx

dyx

2

cxyx

dx

y

dy

11

22tantan0

11

iv. 2

1 1.1 yxy

cx

xydxxy

dy

2tan1

1

21

2

v. 03212 dyxdxy

cyxy

dy

x

dx

12log

2

132log

2

10

1232

vi. 22 1 ydx

dyx

cx

yx

dx

y

dx

1

tan01

1

22

vii. 23 2 xdx

dy

cxxxx

ydxxdy 223

323 3

22

viii. 11 xey

cxeydxedy xx 1

ix. 0y

dy

x

dx

cyxy

dy

x

dx loglog0

x. 0 dyydyx

cx

yxyd

x

dxydyx

00

2

xi. 0sincos dyydxx

cyxdyydxx cossin0sincos

xii. 0cos2 dxdyxy

cyy

dxxdyy tan2

0sec2

2

xiii. 011

y

y

x

x

e

dye

e

dxe

cee

e

ed

e

ed yx

y

y

x

x

1log1log0

1

1

1

1

xiv. 011 22 dyyxdyxy

3

cxy

x

dxy

y

dyydxxxdy

y

y

22

22

2

2

1log1log

01

2

1

201

1

xv. yxedx

dy

ceedxedyee

e

dx

dy xyxy

y

x

Problems on homogenous equations 3 mark question:

1. yx

yx

dx

dySolve

cyxxy

xx

yxxyx

V

dVV

V

dV

x

dx

V

dVV

V

VV

V

V

dx

dVx

V

V

Vxx

Vxx

dx

dVxV

dx

dVxV

dx

dy

xyx

yx

dx

dyGiven

221

22

1

22

2

2

logtan

loglogtanlog1

2.

2

1

1

1

1

1

1

1

1

1

11

Vyput,1,

Linear Differential Equations:

Solve the following:

1. yy

x

dy

dx2

cyy

xydydyyyxyisSolution

yePdyefI

yydyy

dyp

yQy

P

y

2222

..

loglog1

2;1

11

1log

1

1

2. 2

1 2 xyxy

4

cx

yxx

dxxdxxxxyisSolution

xeeisfIxxPdx

xQx

Pxx

yy

xPdx

44..

.loglog2

;22

42

4322

2log2

1

2

3. 02;cot2cot 2

1 yxxxxyy

2222

22

22

2

sinlog

2

sinsin02,

sinsinsin2sinsin2

sinsin2cossin2

sincotsin2sin.

sinsinlogcot

cot2;cot

xxxyccQygiven

cxxxydxxxxxdxxx

xdxdxxxdxxxdxxx

dxxxxxxxyisSolution

xeexdxxPdx

xxxQxP

xPdx

4. xeyy 2

1 3

ceeyedxedxeeeyisSolution

eefIxdxPdx

dxeQP

xxxxxxx

xPdx

x

3233

3

2

..

..33

;3

5. 211

2

30,3

yx

yyyyyyx

1

2

loglog3;1

33

yyPdyyQy

P

yy

x

dy

dx

y

yx

dy

dx

cyy

xdyy

yyxisSolution

yefI y

333.11

.

1..

1log

6. xxyy sintan21

cxxyxdx

dxxxxyisSolution

xefI

xxdxxPdx

xQxP

x

322

22

2coslog

2

coscoscoscos

cos.sincos.

cos..

coslogcoslog2tan2

sin;tan2

2

5

7. yxdy

dx

dx

dyyx 1

cyeexeyeeye

edydyyeexisSol

eefIydyPdy

yxQPyxdy

dx

yyyyyy

yyy

yPdy

1

...

..1

;1

8. 21

2

1

121

xxyyx

cxxy

dxx

dxx

xxyisSol

xeefIxx

xPdx

xQ

x

xP

xy

x

xy

xPdx

12

22

22

21log2

2

222221

tan1

1

1

1

1.11..

1..1log1

2

1

1;

1

2

1

1

1

2

2

9. xxyy 2sincot31

xdxxxx

dxx

dxxxxxyisSol

xeefI

xxxPdx

xQxP

xPdx

sec2tansec2sin

cos

cossin2.sinsin..

sin..

sinlogsinlog3cot3

2sin;cot3

2

113

3sinlog

3

3

10. 2

1

2 1

tan

1 y

y

y

x

dy

dx

ceyxeetedteteedt

ytdtetdyy

yeexisSol

eefIydyy

Pdy

ya

yQ

yP

yyttttt

tyy

yPdx

11

11

1

tan1tan

1

2

1tantan

tan1

2

2

1

2

tan

tan,1

tan..

..tan1

1

tan;

1

1

6

Statement problems:

1. Find equation of a curve which passes through origin given that slope at any point

on it = sum of coordinates.

11110000

...

..1

;1

00&

xx

xxxxx

xxx

xPdx

exyeccygiven

cexeyedxexe

edxyedxxeyisSol

eefIxdxPdx

xQPxydx

dy

yyxdx

dygiven

2. Find the equation of a curve which passes through (0, 2) given that sum of

coordinates at any point exceeds slope at that point by 5.

xdxPdxxQP

xydx

dy

dx

dyyxGiven

15;1

55,

86

81.510220,

5

55

5...

xeey

ccygiven

ceexeye

edxexedxeedx

dxexyeisSoleefI

xx

xxxx

xxxxx

xxxPdx

3. Find the equation of curve which passes through (0, 1) given that slope of that at

any point = sum of abscissa and product of coordinates.

2

8,110,

....

2;

,

22

00

222

222

2

22

222

222

xx

xxx

xxx

eey

cceeygiven

ceede

dxeeyisSolefI

xdxxPdxxQxP

xxydx

dyxyx

dx

dyGiven

7

4. Find the equation of the curve which passes through (0, 1) given that slope at any

point on it

dx

dy satisfies (x – y) (dx + dy) = dx –dy.

12log11log1010,

log

,

xyxccygiven

cyxyxyx

yxddydx

dydxdydxyxGiven

5. Find the equation of curve which passes through (0, 2) given that product of

ordinate and slope at that point = abscissa.

04222

202

220,

22

,

2222

2

22

yxxy

ccygiven

cxy

dxxdyy

xdx

ydyGiven

6. Find the equation of a curve which passes through (1, 1).

22, yxdx

dyxGiven

13loglog22log13log

log213log11,

log22

12log

2

222,

xxyc

cxygiven

cxxdxx

y

dxx

x

y

dyyx

dx

dyxgiven

7. At any point P on a curve slope =2 (slope segment joining P & A (-4, -3). Find its

equation if it passes through (-2, 1)

4log23log

02log2log12,

4log23log

42

34

32,

xy

ccyygiven

cxy

x

dx

y

dy

x

x

dx

dyGiven

8. In a bank principal, increases continuously at the rate of 5% per year. In how many years

of Rs.100 doubles itself? Use 6931.0log 2 e

8

862.136931.0202log20

202log2100020001000

1000100010000

.

20log20.

100

5

202020

0

2020

t

teeeP

kkePgiven

KePeeP

ct

PdtP

dpP

dt

dpGiven

ttt

tct

9. Find the equation of a curve whose differential equations is y1= ex sinx given that it

passes through origin.

2

1cossin

22100cossin

2

sinsinsin1

xxe

ycyButcxxe

y

dxxedyxedx

dyxey

xx

xxx

10. Find equation of a curve, whose differential equation 22 11 yxdx

dy

given that it passes through A (0, ½)

3log2

1

31

1log

2

13log

2

1

00211

211log

2

1210

31

1log

2

1

11

11,

3

3

2

2

22

xx

y

yc

cygiven

cx

xy

y

dxxy

dyyx

dx

dygiven

1

MATHEMATICS II PUC

VECTOR ALGEBRA QUESTIONS & ANSWER

I One Mark Question

k.3j2i ofdirection in ther unit vecto theFind1) ++

kjia

kjiaaa

141

143

142ˆ

1432ˆ

14194132a

k 3j2i aLet222

++=∴

++==∴

=++=++=

++=

r

r

rr

r

r

equal? b & a vectors theAre .a If j.2i b & 2ji aLet 2)rrrrrr b=+=+=

ba

abaarr

rr

=∴

=+==+= 52,52 2222

But vectors are not equal since the corresponding components are distinct i.e. directions are different.

3) Find the values of x & y so that vectors 2i+3j and xi+4j are equal.

3,2

32

32

==∴+=+∴=

+=+=

yxyixijibaGiven

yjxibjiarr

rr

( ) ( ) ( ) ( ) ( ) 3322131221 kj-2i . 3k - 2j i k.j-2i &3k - 2j i vectorsofproduct dot or scalar theFind4)

−=−−=−−+=++++

2

r.unit vecto a is a 1a

121

21

21

21 a

j2

1-i2

1 2j - i a

r.unit vecto a is 2j-i that Show )5

22

rr

r

r

∴=∴

=+=

−+

=

==

II Two Marks Questions:

1) Find the vector parallel to the vector i - 2j and has magnitude 10 units.

( )

52

aˆ

5a

21a

j2i aLet22

jiaa −==∴

=∴

−+=

−=

r

rr

r

r

r

520

510b vector Reqd

52.10

ˆ bNow

10

units 10 magnitude having a toparallel vectors thebeLet

ji

ji

ab

b

b

−=∴

−=

=

=∴

r

rrr

r

rr

2k- j ia vector theof cosinesdirection and ratiosdirection theFind2) +=r

.

( )

kjia

kjiaaa

a

a

kjia

62

61

61ˆ

62ˆ

6

211

2222

−+=∴

−+==

=∴

−++=

−+=

r

r

rr

r

r

r

QIf vectors are parallel then unit vector along & parallel vectors are same

3

( )

−62,

61,

61 i.e.a of components are cosinesdirection

2- 1, 1,i.e. a of components are ratiosdirection Here

r

r

3) Show the vectors 2i-3j+4k and -4i+6j-8k are collinear.

( )

collinear. areb &a

another of interms expressed becan vector One24322864

432

rr

rr

r

∴

∴−=+−−=−+−=

+−=

akjikjib

kjia

( ) ( )

( ) ( )

13x

13x

121 x

12a x

12ax.a-x1aGiven

12ax.a-x ,ar unit vecto afor if ,x Find)4

2

22

22

=∴

=

=−

=−

=+

=

=+

r

r

r

rr

rrrr

r

rrrrrr

( )( )

5

5

9843422

.2a...

4b. a and3,2aGiven

4b. a and 3,2asuch that are b and a vectors twoif ,b-a Find)5

2

22

222

=−∴

=−

+−=+−=

+−=−

===

===

ba

ba

bbabatkw

b

b

rr

rr

rrrrrr

rrrr

rrrrrrrr

4

bba

b

ba

b

ba

bbatkw

bba

rrrr

rr

rr

rr

rr

rr

rr

rrrrrr

a.

1a

.

1cos

1cos1-, of valuesallfor cosa

.

a.cos...

a that prove b and a vectorsany twoFor )6

≤∴

≤∴

≤∴

≤≤=

=

•≤•

θ

θθθ

θ

( )

( )

{ }bba

b

babb

bababba

bbaba

bba

rrrr

rr

rrrrrrrr

rrrrQ

rrrr

rrrrrr

rrrrrr

+≤+∴

+≤

≤++≤

≤++≤

++=+

+≤+

a

a

b.a properties previous Froma2a

...2a

.2a

equalityin Triangle a that prove b and a vectorsany twoFor )7

2

22

22

222

( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( )

22

22

22

35.116

35.116

35.10.216

.35.10.21.6b7a2.b5-a3

b7a2.b5-a3 Evaluate)8

bbaa

bbaa

bbabaa

bbabbaaa

rrrr

rrrr

rrrrrr

rrrrrrrrrrrr

rrrr

−+=

−+=

−−+=

−−+=+

+

( ) ( )( ){ } { } { }( ) { } ( )

kjba

kjikji

kjiba

baba

1919

191914142122127227

223771

and find2k 2j-3ib &7k 7j-ia If )9

+=×∴

+−−+−=+−+−−−−−=

−−=×

××+=+=

rr

rr

rrrrrr

5

( ) ( )( ) ( )

( ) ( ) ( )

3and 2

2706-2027-20,27-6

scofficient Equating62272276

12762

27k6j2iGiven 27k6j2i if & Find)10

==∴

===

=−+−−−

=∴

=++×++=++×++

λµ

λµλµ

λµλµµλ

µλµλµλ

okji

okji

okjiokji

r

r

r

r

)(2)()(

)()(

)()( LHSConsider

)(2)()( that Show )11

bababa

obabao

bbabbaaa

baba

bababa

rrrrrr

rrrrrr

rrrrrrrr

rrrr

rrrrrr

×=+×−∴

−×+×+=

×−×−×+×=

+×−

×=+×−

( ) ( ) ( )2042

11311211111111-1-321

product pleScalar tri

kji andk j-i- 3k,2ji vectorsofproduct plescalar tri theFind)12

=++−=+−+−−−−−=

=

+++++

( ) ( ) ( )

40123

06433032231211

0123

11211

coplanar are vectorsGiven thatcoplanar. arek -2j-3i&kj-i 2k,ji vectors theif Find)13

=∴=+−=

=+−++==+−+−−−+=

=−−

−∴

+++

λλ

λλλλ

λ

λλ

6

III Three Marks Questions:

ly.respective externally and internally 1:2 ratio in the Q and P points the joining line divides which Rpoint a ofector position v theFind

. ba and 23 ectorsposition v with Q and P points heConsider t1)rrrr

+=−= OQbaOP

( ) ( )

( ) ( )

abOR

bbaa

babanm

OPnOQmOR

aOR

bbaa

babanm

OPnOQmOR

nmbaOQbaOP

rr

rr

rrrr

r

rr

rrrr

rrrr

−=

++−=

−−−+

=

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THREE DIMENSIONAL GEOMETRY.

One mark questions:

1) If a line makes angles 900, 1350 and 450 with the x, y and z axes respectively.

Find its direction cosines.

Solution:

Let = 900, = 135, = 450

Let l, m, n are the direction cosines of a line

l = cos =cos 900 = 0

m=cos = cos 1350 = 1

2 ,

n = cos = cos 450 1

2

2) If a line has direction ratio’s -18, 12, -4. Then what are its direction cosines.

Solution:

x = - 18 y = 12 z = -4

2 2 2

2 2 2 18 12 4 324 144 16 484 22r x y z

Direction cosines are 18 9

22 11

xl

r

12 6

22 114 2

22 11

ym and

rz

nr

3) Find the direction cosines of x, y and z axis.

Solution:

The x – axis makes angles 00, 900, 900 with the positive direction of

x, y and z – axis.

Direction cosines of x – axis are cos 00, cos 900, cos 900 i.e. 1, 0, 0.

Similarly direction cosines of y axis are cos 900, cos 00, cos 900 i.e. 0, 1, 0

and direction cosines of z – axis are cos 900, cos 900, cos 00, i.e. 0, 0, 1

4) Find the direction cosines of a line which makes equal angles with the co-

ordinate axes.

Solution:

Let , , be the angles made by the line with the positive direction of x-axis,

y –axis and z – axis

Also = = and cos2 + cos2 + cos2 = 1

cos2 + cos2 + cos2 = 1

3 cos2 = 1 cos2 = 1

3 cos =

1

3

The direction cosines are 1 1 1

, ,3 3 3

5) Find the equation of the plane having intercept 3 on the y-axis and parallel to

ZOX plane

Solution:

Y – intercept = b = 3

Any plane parallel to ZOX is y = b

The equation of the plane is y = 3

6) Find the distance of the plane 2x – 3y + 4z - 6 = 0 from the origin.

Solution:

Consider 2x – 3y + 4z – 6 = 0

2x – 3y + 4z = 6 – (1)

The Direction ratios are (2, -3, 4) = (x1, y1, z1)

22 22 3 4 4 9 16 29r

The Direction cosines are 1 2

29

xl

r

1 3

29

ym

r

1 4

29

zn

r

Divide equation (1) by 29

2 3 6

29 29 29 29

yx y z

and is of the form lx + my + nz = d

The distance of the plane from origin is 6

29d

7) Find the equation of the plane which makes intercepts 1, -1 and 2 on the x, y and

z axes respectively.

Solution:

a = x – intercept = 1, b = y – intercept = -1 and c = z – intercept = 2

The equation of the line is 1 . . 11 1 2

y yx z x zi e

a cb

8) Determine the direction cosines of the normal to the plane and the distance from

the origin is x + y + z = 1

Solution:

Consider x + y + z = 1 - (1)

Direction ratio’s of the plane are 1, 1, 1

2 2 2 1 1 11 1 1 1 1 1 3

3 3 3r l m n

Divide equation (1) by 1

33 3 3 3

yx z

It is of the form lx + my + nz = p

P = distance from origin = 1

3 9) Find the intercepts cut off by the plane 2x + y – z = 5

Solution:

Consider 2x + y – z = 5

2

1 . . 15 5 5 5 55/2

y yx z x zi e

a = x– intercept = 5/2 b = y – intercept = 5 c = z – intercept = -5

10) Show that the planes 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0 are perpendicular.

Solution:

Consider 2x + y + 3z - 2 = 0 i.e. 2x + y + 3z = 2

And x – 2y + 5 = 0 i.e. x – 2y + 0.z = - 5

The normals to the plane are

1 22 3 2andP i j k P i j

1 2. 2 1 1 2 3 0 2 2 0 0P P

The planes 1 2P Pand are perpendicular

11) Show that the planes 2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0 are parallel.

Solution:

Consider 2x – y + 3z – 1 = 0 i.e. 2x – y + 3z = 1

And 2x – y + 3z + 3 = 0 i.e. 2x – y + 3z = -3

The normals to the plane are

1 22 3 2 3P i j k and P i j k

1 1 1

1 21

2 1 31, 1 1

2 1 3

a b cand

a cb

1 1 1

2 22

1a b c

a cb

The planes P1 and P2 are parallel.

12) Find the equation of the plane parallel to x – axis and passing through the origin.

Solution:

The direction ratio’s of x-axis is 1, 0, 0

The equation of the line through origin and parallel to x-axis

is 0 0 0

. .1 0 0 1 0 0

x y z x y zi e

13) Find the vector equation of the straight line passing through (1.2.3) and

perpendicular to the plane . 2 5 9 0 r i j k

Solution:

The required line passes through (1, 2, 3) and perpendicular to the plane

. 2 5 9 0r i j k is

2 3 2 5r i j k i j k

14) Find the equation of the plane passing through (a, b, c) and parallel to the plane

. 2 r i j k

Solution:

Consider . 2 2r i j k x y z

Any plane parallel to the given plane is x + y + z =

and is pass through (a, b, c) a + b + c =

Hence the equation of the plane parallel to the given plane is

x + y + z = a + b + c

15) Find the distance between the two planes 2x+3y+4z=4 and 4x+6y+8z= 12.

Solution:

Consider 2x + 3y + 4z = 4 - (1)

And 4x + 6y + 8z = 12

i.e. 2x + 3y + 4z – 6 = 0 - (2)

Distance from the point to the plane (2) = 2 2 2

2 3 4 6

2 3 4

x y z

4 6 2 2

4 9 16 29 29

Two mark questions:

1) Show that the points (2, 3, 4) (-1, -2, 1) and (5, 8, 7) are collinear.

Solution:

A = (2, 3, 4) B = (-1, -2, 1) and C = (5, 8, 7)

Direction ratio’s of the line joining A & B are, 2+1, 3+2, 4-1, i.e. 3, 5, 3

Direction ratio’s of the line joining B & C are -1-5, -2-8, 1-7, i.e. -6, -10, -6

The direction ratio’s of AB & BC are proportional & B is the common point

of AB & BC

The points A, B, C are collinear

2) Show that the line through the points (1, -1, 2) (3, 4, -2) is perpendicular to the

line through the points (0, 3, 2) and (3, 5, 6)

Solution:

Let A = (1, -1, 2) B = (3, 4, -2) C = (0, 3, 2) and D = (3, 5, 6)

Direction ratio’s of AB are, a1 = 3-1=2, b1 = 4- (-1) = 4+1=5 & C1 = -2-2 = -4

Direction ratio’s of CD are a2 = 3-0=3, b2 = 5-3=2, C2 = 6-2=4

Now a1a2 + b1b2 + c1c2 = 2 (3) + 5 (2) + (-4) 4

= 6+10 - 16 = 0

AB is perpendicular to CD

3) Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line

through the points (-1, -2, 1) (1, 2, 5).

Solution:

Let A = (4, 7, 8) B = (2, 3, 4) C = (-1, -2, 1) D = (1, 2, 5)

Direction ratio’s of AB are a1 = 2 – 4 = -2, b1 = 3-7=-4, c1 = 4 – 8 = -4

Direction ratio’s of CD are a2 = 1- (-1) =1+1=2, b2 = 2-(-2) = 2+2=4, c2 = 5-1 =4

1 1 1

2 22

2 4 41, 1, 12 4 2

a b c

a cb

1 1 1

2 22

a b c

a cb Hence AB is parallel to CD

4) The Cartesian equation of a line is 45 6

73 2yx z . Write its equation in

vector form.

Solution:

Consider 45 6

73 2yx z

a = (5, -4, 6) and b = (3, 7, 2) are the direction ratio’s

Vector equation of the line is r a b

5 4 6 3 7 2r i j k i j k

5) Find the distance of the point (2, 3, -5) from the plane 2 2 9r i j k

Solution:

Consider . 2 2 9 2 3 5r i j k and a i j k

and 2 2 9N i j k and d

. 2 1 3 2 5 2 2 6 10 18a N

2

2 21 2 2 1 4 4 9 3N

Distance of a point from the plane = d = .

18 9 93

3 3

a N d

N

6) Find the equation of the plane passing through the line of intersection of the

plane x + y + z = 6 and 2x + 3y + 4z – 5 = 0 and the point (1, 1, 1)

Solution:

Consider x + y + z = 6 x + y + z -6 = 0 and 2x + 3y + 4z – 5 = 0

The equation of the plane passing through the intersection of the two planes

is x + y + z – 6 + (2x + 3y + 4z – 5) = 0 and is pass through (1, 1, 1)

1 + 1 + 1 – 6 + (2+3+4-5) = 0

- 3 + 4 = 0 4 = 3 = ¾

The equation is (x + y + z – 6) + 3

4 (2x + 3y + 4z – 5) = 0 (multiply by 4)

4x + 4y + 4z – 24 + 3 (2x + 3y + 4z – 5) = 0

4x + 4y + 4z – 24 + 6x + 9y + 12z – 15 = 0

10x + 13y + 16z – 39 = 0

7) Derive the direction cosine of a line passing through two points.

Solution:

Let l, m, n be the direction cosines of a line PQ and the line PQ makes ,

and with positive directions of x, y and z axes respectively. Draw the

perpendiculars from P and Q to xy – plane to meet at R & S and draw PN

perpendicular to QS.

From the le PNQ, ˆPQN

2 1cosZ ZQN ON OQ

PQ PQ PQ

2 1 2 1cos cosx x y y

Similarly andPQ PQ

Where 2 2 2

2 1 2 1 2 1PQ x x y y z z

8) The Cartesian equation of a line is 53 6

2 4 2yx z

Find the vector equation of the line

Solution:

Consider 53 6

2 4 2yx z

3 652 4 2

x zy

x1 = -3 y1 = 5 z1 = -6 and a = 2 b = 4 and c = 2

1 1 1, , 3, 5, 6a x y z

, , 2, 4, 2b a b c are direction ratio’s

The vector equation of a line is r a b

3 5 6 2 4 2r i j k i j k

9) Find the vector equation of the plane which is at a distance of 7 units from the

origin and normal to the vector 3i + 5j – 6k

Solution:

2

2 23 5 6 3 5 6 9 25 36 70let n i j k and n

3 5 6 3 5 6ˆ

70 70 70 70

n i j kand n i j k

n

The equation of the plane ˆ. 7r n d and d

3 5 6. 7

70 70 70r i j k

10) Find the distance of the point (3, -2, 1) from the plane 2x – y + 2z + 3 = 0

Solution:

Consider 2x – y + 2z + 3 = 0

1 1 1

2 2 2 22 2

2 3 1 2 2 1 3

2 1 2

ax by cz dd

a b c

6 2 2 3 13

34 1 4

Three mark questions

1) Find the vector and Cartesian equations of the line that passes through the

points (3, -2, -5) and (3, -2, 6)

Solution:

Let A = (3, -2, -5) B = (3, -2, 6)

Direction ratio’s of AB are, a = 3 – 3 = 0

b = -2 – (-2) = - 2+2 = 0

c = 6 – (-5) = 6 + 5 = 11

0. 0. 11 11 3, 2, 5b ai bj ck i j k k and a = 3i – 2j – 5k

Vector equation of a line passing through two points is r a b

3 2 5 11r i j k k

Cartesian equation of a line is 1 1 1x x y y z z

a b c

2 53 3 2 5. .

0 0 11 0 0 11

y zx x y zi e

2) Show that three lines with direction cosine 12 3 4

, , ;13 13 13

4 12 3 3 4 12, , ; , ,

13 13 13 13 13 13

are mutually perpendicular.

Solution:

L1, L2, L3 are three lines.

The direction cosine of the line 1 1 1 1

12 3 4, , , ,

13 13 13L l m n

Direction cosines of the line 2 2 2 2

4 12 3, , , ,

13 13 13L l m n

Direction cosines of the line 3 3 3 3

3 4 12, , , ,

13 13 13L l m n

1 2 1 2 1 2

12 4 3 12 4 3 48 36 120

13 13 13 13 13 13 169l l m m n n

1 2L is perpendicular to L

2 3 2 3 2 3

4 3 12 4 3 12 12 48 36 48 48. . 0

13 13 13 13 13 13 169 169l l m m n n


3 1 3 1 3 1

3 12 4 3 12 4 36 12 48 48 48. 0

13 13 13 13 13 13 169 169l l m m n n


Hence the three lines are mutually perpendicular

3) Find the angle between the pair of lines 3 5r i j k i j k and

7 4 2 2 2r i k i j k

Solution:

Consider 3 5r i j k i j k 1b i j k

7 4 2 2 2r i k i j k 2 2 2 2b i j k

1 2

2 2 21 2

. 1 2 1 2 1 2 2 2 2 6

1 1 1 3 4 4 4 12 2 3

b b

b b

1 2 0

1 2

0

. 6 6 6cos 1 cos0

2 3 62 3. 3

0

b b

b b

4) Find the equation of the line which passes through the point (1, 2, 3) and is

parallel to the vector 3i + 2j – 2k, both in vector form and Cartesian form.

Solution:

Let 1, 2, 3 2 3 3 2 2a i j k and b i j k

The vector equation of the line is r a b

2 3 3 2 2r i j k i j k

Let r be the position vector of the point and r xi yj zk

2 3 3 2 2xi yj zk i j k i j k

2 3 3 2 2i j k i j k

1 3 2 2 3 2i j k

1 3 2 2 3 2

1 3 2 2 3 2

1 2 3

3 2 2

x y and z

x y z

x y z

1 2 3

3 2 2

x y z

is the equation of the line in Cartesian form.

5) Find the distance between parallel lines 2 4 2 3 6r i j k i j k and

3 3 5 2 3 6r i j k i j k

Solution:

Consider 2 4 2 3 6r i j k i j k

And 3 3 5 2 3 6r i j k i j k

1 12 4 2 3 6a i j k b i j k

and 2 23 3 5 2 3 6a i j k b i j k

1 2b b The lines are parallel

2 2 21 2 2 3 6 2 3 6b b b i j k and b

2 1 2 10 4 9 36 49 7a a i j k

2 1 2 3 6 3 6 2 12 2 6 9 14 4

2 1 1

i j k

b a a i j k i j k

2 2 22 1 9 4 4 81 196 16 293b a a

Distance between parallel lines = d = 2 1 293

7

b a a

b

6) Find the angle between the pair of lines 3 1 3

3 5 4

x y z

1 4 5

1 1 2

x y zand

Solution:

Consider 1

3 1 31 3, 5, 4

3 5 4

x y zDirection ratios of b

2

1 4 52 ' 1, 1, 2

1 1 2

x y zand Direction ratio s of b

1 2. 3 1 5 1 4 2 3 5 8 16b b

2 2 21

2 2 22

3 5 4 9 25 16 50 25 2 5 2

1 1 2 1 1 4 6

b

b

1 2

1 2

1

. 16 16 16 16 8cos

5 2 6 5 12 5 4 3 5 2 3 5 3

8cos

5 3

b b

b b

7) Find the shortest distance between the lines

1 1 1 3 5 7

7 6 1 1 2 1

x y z x y zand

Solution:

Consider 1 1 1 3 5 7

7 6 1 1 2 1

x y z x y zand

2

1 2

1

1 1 1. . 3 5 7

7 6 1

2

7 6

x y zi e a i j k

a i j k b i j k

b i j k

2 1 3 5 7 4 6 8a a i j k i j k i j k

1 2 7 6 1 6 2 7 1 14 6

1 2 1

4 6 8

i j k

b b i j k

i j k

2 2 2

1 2 4 6 8 16 36 64 116b b

Shortest distance = 1 2 2 1

1 2

. 16 36 64

116

b b a ad

b b

116

116 4 29 2 29116

8) Find the equation of the planes passing through three points (1, 1, 0)

(1, 2, 1) and (-2, 2, -1)

Solution:

Let a = (1, 1, 0) b = (1, 2, 1) and c = (-2, 2, -1) and r xi yj zk

1 1 0r a x i y j z k

0, 1, 1 3, 1, 1AB b a and AC c a

The vector equation of the plane is . 0r a AB AC

1 1

0 1 1 0

3 1 1

x y z

(x-1) (-1-1) – (y-1) (0+3) + z (0 + 3) = 0

-2(x-1) -3 (y-1) + 3z = 0

-2x + 2 – 3y + 3 + 3z = 0

-2x – 3y + 3z + 5 = 0

2x + 3y - 3z - 5 = 0

2x + 3y – 3z = 5 is the equation of the plane

9) Find the angle between the pair of lines given by 3 2 4 2 2r i j k i j k

and 5 2 3 2 6r i j i j k .

Solution: 1 22 2 3 2 6b i j k b i j k

1 2 1

1 2

. 3 4 12 19 19cos cos

21 219 49

b b

b b

10) Prove that if a plane has intercepts a, b, c and is at a distance of p units from the

origin then 2

2 2 2

1 1 1 1

a b c p

Solution:

Let a, b, c, are the intercepts of the plane

And the equation is 1 1x y z

a b c

P = The distance of the plane (1) from (0, 0, 0)

2 2 2 2 2 2 2 2 2

0 0 0 1 1 1

1 1 1 1 1 1 1 1 1P

a b c a b c a b c

2

2 2 2 2

2 2 2

1 1 1 1 1

1 1 1P

p a b c

a b c

Five mark questions:

1) Derive the equation of the line in space passing through a point and parallel to a

vector, both in the vector form and Cartesian form.

Solution:

Let a be the position vector of the given point A. w.r. to

the origin O of the rectangular co-ordinate system. Let l

be the line which passes through the point A and is

parallel to the given vector b . Let r be the position

vector of an arbitrary point P on the line. Then AP is

parallel to b .

i.e. AP b where is a real number

OP OA b

r a b

r a b is the vector equation of the line

Let A = (x1, y1, z1) be the co-ordinates of the given point and the direction ratio’s of the

line are a, b, c.

Let P = (x, y, z) be the co-ordinate of any point

Then 1 1 1r xi yj zk and a x i y j z k and b ai bj ck and r a b

xi + yj + zk = (x1i + y1j + z1k) + (ai + bj + ck)

= x1i + y1j + z1k + ai + bj + ck

= (x1 + a) i + (y1 + b) j + (z1 + c) k

Equating the coefficients of i, j and k we get

x = x1 + a y = y1 + b and z = z1 + c

these are the parametric equations of a line

x – x1 = a y – y1 = b and z – z1 = c

1 1 1x x y y z z

a b c

1 1 1x x y y z z

a b c

. This is the Cartesian equation of the line.

2) Derive the equation of a line in space passing through two given points both

invector form and Cartesian form.

Solution:

Let & &a b r are the position vectors of the two

points A (x1, y1, z1) is (x2, y2, z2) and p (x, y, z)

respectively.

AP OP OA r a and AB OB OA b a

If the point p lien on the line AB if and only if

AP and AB are collinear.

. .AP AB i e r a b a

r a b a is the vector equation of the line passing through two points.

Let 1 1 1 2 2 2, &r xi yj zk a x i y j z k b x i y j z k r a b a

2 1 2 1 2 1

1 2 1 1 2 1 1 2 1

1 2 1 1 2 1 1 2 1

1 2 1 1 2 1 1 2 1

1 1 1

2 1 2 1 2 1

2 1 1 1

, &

xi yj k x i y j z k x x i y y j z z k

x x x i y y y j z z z k

x x x x y y y y z z z z

x x x x y y y y z z z z

x x y y z z

x x y y z z

1 1 1

2 1 2 1 2 1

x x y y z z

x x y y z z

is the Cartesian equation of the line passing through

two points.

3) Derive the shortest distance between two skew lines both in vector form and

Cartesian form.

Proof:

Let l1 and l2 be the skew lines

Let 1 1 2 2r a b and r a b be the skew lines. Let s and T are any two points

on l1 and l2 with position vectors 1 2a and a respectively.

Then the magnitude of the shortest distance is equal to the projection of ST along the

direction of a line.

If PQ is the shortest distance between the lines l1 and l2 then it is perpendicular to

both 1 2b and b and n is the unit vector along PQ .

1 2

1 2

ˆb b

nb b

let be the angle between ST and PQ

Then 2 1

.cos cos

PQ STPQ ST and but PQ d and ST a a

PQ ST

2 1ˆcos

.

d n a a

d ST

1 2 2 1

1 2

cosb b a a

STb b

Shortest distance is d = PQ = 1 2 2 1

1 2

cosb b a a

STb b

is the Shortest distance of skew lines in vector form.

Let 1 :l 1 1 1

1 1 1

x x y y z z

a b c

and 2 2 2

2

2 2 2

x x y y z zl

a b c

be the equations of two

skew lines in Cartesian form.

The shortest distance between two skew lines is

2 2 2

1 2 2 1 1 2 2 1 1 2 2 1

db c b c c a c a a b a b

where

2 1 2 1 2 1

1 1 1

2 2 2

x x y y z z

a b c

a b c

4) Derive the equation of the plane in normal form both in the vector form and

Cartesian form.

Solution:

Consider a plane whose perpendicular distance from the origin is d. If ON is the

normal from the origin to the plane and n is the unit normal vector ON

Then ˆ.ON d n

Let P be any point on the plane then NP is perpendicular to ON

. 0 1NPON

Let r be the position vector of the point P

ˆ.Then NP OP ON r d n

From equation (1) ˆ ˆ. . . 0 0r d n d n But d

ˆ ˆ. 0

ˆ ˆ ˆ ˆ ˆ. . . 0 . 1.1 1

ˆ. 0

r d n n

r n d n n But n n

r n d

ˆ.r n d is the equation of the plane vector form

Let l, m, n be the direction cosines of n

Then n li mj nk and OP r xi yj zk

ˆ.r n d

.xi yj zk li mj nk d

Therefore lx + my + nz = d is the Cartesian equation of the plane in normal form

5) Derive the condition for the coplanarity of two lines in space both in the vector

form and Cartesian form.

Solution:

Let the given lines be 1 1 2 2 2r a b and r a b

The line (1) passes through the point A with the position vector 1a and parallel to 1b

and the line (2) passes through the point B with the position vector 2a and parallel to

2b

Thus 2 1AB B A a a

The given lines are coplanar if and only if AB is perpendicular to 1 2b b

i.e. 21. 0AB b b

1 22 1. 0a a b b is condition for the coplanarity of two lines in vector form.

Let A = (x1, y1, z1) and B = (x2, y2, z2) be the co-ordinates of the points A and B

respectively. Let a1, b1, c1 and a2, b2, c2 be the direction ratio’s of 21b and b

respectively.

Then 2 1 2 1 2 1AB B A x x i y y j z z k

1 21 1 1 2 2 2b a i b j c k and b a i b j c k

The given lines are coplanar if 21. 0AB b b

2 1 2 1 2 1

1 1 1

2 2 2

0

x x y y z z

a b c

a b c

is condition for coplanarity of two lines in cartestion form.

Topic :- Linear Programming

The running of any firm or of a factory involves many constraints like

financial,space,resources,power etc. The objective of any business person would be to make the

profit maximum (in the case of investment to make the cost a minimum) under all the

constraints.The linear programming problem is the problem of optimising an objective function

under a given set of constraints.When the objective function is profit function the optimisation is to

maximise the profit.In the case of cost function the optimisation is to minimise the cost.

All the constaints of the problem are linear inequalities and the objective function is linear.Hence, it

is called LPP.The following are a few illustrations of the LPP by Graphical method .

1. Maximise : P = .

Subject to : ≤ 20

≤ 6

0.

Now we draw the graphs of the equations = 20, and = 6 and recognise the

feasible region.

Y

X

O(0,0)

B(8,6)

C(10,5)

D(15,0)

A(0,6) =600

The shaded region OABCDO (bounded) represents the feasible region in which all the Constraints of

the problem are satisfied. Now, we evaluate the objective function at these corners of the region.

Vertex

0 0 0 0

A 0 6 3X0+5X6=30

B 8 6 3X8+5X6=54

C 10 5 3X10+5X5=55

D 15 0 3X15+5X0=45

From the above table we observe that Maximum value of P is 55 and corresponds to

X = 10 and y = 5

2). Minimise: C =

Subject to : ≥ 6

4.

≥ 6

.

we draw the graphs of the linear equations

O X

Y

B(1,3

)

C(3,1

)

D(

)

The shaded region BCD ( triangular region ,bounded) is the feasible region.

In this region all the constraints of the problem are satisfied.Now we evaluate the objective

function C = at the verticies.

Vertex x y C =

B 1 3 272

C 3 1 304

D 3/2 3/2 216

From the above table we observe that the minimum value of C is 216 and corresponds to

and

.

Note:- The co-ordinates of the points B, C and D can be read from the graph. They may

also be determined by solving the equations of the corresponding pair of lines

(3) Maximise :

Subject to :

Now we draw the graphs of the linear equations and recognise the feasible region.

Y

X

A(0,6) B(3,4)

C(5,0) O

The shaded region OABCO is the feasible region. Now we evaluate at the

vertices.

Vertex

0 0 0 0

A 0 12

B 3 4 23

C 5 0 25

From the table we find that the maximum value of Z = 25 and corresponds to .

Note:The above working is based on the result that the optimal solution of a LPP if exists

will occur at a corner point of the feasible region.

(4) Maximise

Subject to :

We draw the graphs corresponding to the linear equations and shade the feasible region.

O

Y

X

A(0,5)

B(2,4)

C(4,2)

D(4,0)

The shaded region OABCDO(bounded ) is the feasible region.

Now we evaluate at the vertices

.

Vertex

0 0 0 0

A 0 10

B 2 4 14

C 4 2 16

D 4 0 12

From the table we find that the maximum value of p = 16 and corresponds to

(5). Determine the maximum and minimum values of

Z = 4x + y

Subject to: x + 2y 4

x – 2y 0

6

x, y 0

solution: We draw the lines x+ 2y = 4 , x– 2y = 0 and x= 6.

D(6,0)

x+2y=4

x– 2y

x=6

C(6,3)

B(2,1)

X O

Y

A(4,0)

The feasible region is bounded region with A( 4, 0 ) , B ( 2 , 1 ) , C ( 6 , 3 ) and D( 6 , 0 )as

vertices.

Now we find the values of Z = 4x + 2y at these vertices.

Vertex X Y Z = 4x + 2 y

A 4 0 16

B 2 1 10

C 6 3 30

D 6 0 24

From the above table we find that the minimum value of Z is 10 and corresponds to x = 2 and

y = 1. The maximum value of Z is 30 and corresponds to x = 6 and y =3.

APPLICATIONS PROBLEMS

1). A factory manufactures two types of screws, A and B by using two

machines.The time required for the manufacture of one packet of each of the

two types of the screws on the two machines,the total time available of each of

the mechine and the profit obtained on the sale of each packet of the two types

of the screws are given below.

Machine

Screws

Time required in minutes Profit in Rs

Per packet. I II

A 4 6 8

B 6 3 5

Time available

in hours

4 4 --

Formulate this as a Lpp and determine the number of packets of each of the two

types of screws to be manufactured so as to get maximum profit assuming that

all the packets of the screws manufactured are sold.

Solution;-

Let x and y be the number of packets of the type A and type B screws to be

produced.

Then,the total time the mechine I works is 4 x + 6y.

The total time the machine II works is 6x + 3y.

The total profit in Rs. is p = 8x + 5y.

Since, the two machines are available at most for 4 hours = 240 minutes, we

have 4x + 6y 240. and 6x + 3y 240.

.`. the Lpp is to maximise p = 8x + 5y

Subject to:- 4x + 6y 240

6x + 3y 240

x, y 0

Now, we draw the graphs of the linear equations 4x + 6y = 240

and 6x +3y = 240.

The feasible region is bounded region OABCO with O(0,0), A( 0,40 ) , B ( 30,20) ,

and C(40 ,0 ) as vertices.

Now we find the values of p = 8x + 5y at these vertices.

Vertex X Y p = 8x + 5 y

O 0 0 0

A 0 40 200

B 30 20 340

C 40 0 320

Y

X O

B(30,20)

C(40,0)

(0,80)

A(0,40)

D(60,0)


.

`Hence, the maximum profit is Rs. 340 and corresponds to manufacture of 30 packets of A

type.and 20 packets of B type screws.

Suitcase

Machine

Time required in hours Time

available in

hours Type I Type II

A 3 3 18

B 2 4 16

Profit in Rs Per

suitcase

30 42 --

Determine the number of the two types of suitcases to be produced to get maximum profit ..

Solution;-

Let x and y be the number of type I and type II suitcases to be produced.

Then,the total time the mechine A works is 3 x + 3y.

The total time the machine B works is 2x + 4y.

The total profit in Rs. is p = 30x + 42y

Since, the two machines A and B are available at most for 18and 16 hours

respectively, we have 3x + 3y 18. and 2x + 4y 16.

.`. the Lpp is to maximise p = 30x + 42y

Subject to:- 3x + 3y 18

2x + 4y 16

x, y 0

Now, we draw the graphs of the linear equations 3x + 3y = 18

and 2x +4y=16.

2) The production of two types of suitcases requires processing and completion to be done on two machines A and B.The time required for processing and completion of each type of trunk on the two machines, the time available on each machine and profit on each type of the suit case is given below.

E(0,6)

The feasible region is bounded region OABCO with O(0,0), A( 0,4 ) , B ( 4,2) ,

and C(6 ,0 ) as vertices.All the condtions of the problem are satisfied within and

on the boundary of this

Region.

Now we find the values of p = 30 x + 42 y at these vertices.

Vertex x Y p = 30 x + 42 y

O 0 0 0

A 0 4 168

B 4 2 204

C 6 0 180


.

Hence, the maximum profit is Rs 204 and corresponds to manufacture of 4

suitcases of

type I.and 2 suitcases of type II..

Y

X O

B(4,2)

C(6,0)

A(0,4)

D(8,0)

3).At a cattle rearing , it is prescribed that the food for each animal contain at

most 16 units of nutrient A, and at least 24 and 48 units of nutrients B and C

respectively.Two types of fodders are available. The number of units of these

nutrients contained per Kg by the two types of fodders and their cost per Kg is

given below

Content of

nutrient/kg

A B C Cost per kg in

Rs.

Fodder- 1 1 3 2 12

Fodder-2 1 1 6 15

By graphical method determine the number of kg of the two types of the

fodder to be purchased so as to make the cost a minimum,yet meeting the

requirements.

Solution:

Let, x and y kg of fodder-1and fodder-2 be purchased.

Then, the cost in Rs. Is C = 12 x + 15 y.

Total content s of nutrients A, B and C are x + y , 3 x + y and 2 x + 6 y

respectively.

Since the content of A is to at most 16 while it has to be at least 24 for B and

48 for C, the constraints are

x + y , 3 x + y 24 , and 2 x + 6 y 48.

.`.The Lpp is To minimise; C = 12 x + 15 y

Subject to x + y

3 x + y 24

2 x + 6 y 48

X, y 0

Now ,we draw the graphs of the linear equations and recognise the feasible

region

The shaded region MNL ( triangular region ,bounded) is the feasible

region.

In this region all the constraints of the problem are satisfied.Now we evaluate

the objective function C = at the verticies.

Vertex x y C =

M 6 6 162

N 4 12 228

L 1/2 4 206

From the above table we observe that the minimum value of C is 162 and

corresponds to

and .

Therefore, in order to make the cost minimum 6kg each of fodder-1 and

fodder-2 are to be bought and the minimum cost is Rs.162.

Note:- The co-ordinates of the points M, N and L can be read from the graph.

They may also be determined by solving the equations of the corresponding pair

of lines.

O

X

Y

N(4,12

)

L(12,4

) )

A (0,8)

C(0,16

)

E(0,24

)

F (8,0)

D (16,0) B (24,0)

maths chapter wise important questions

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