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` KUKATPALLY CENTREKUKATPALLY CENTREKUKATPALLY CENTREKUKATPALLY CENTRE

IMPORTANT QUESTIONS

FOR

INTERMEDIATE PUBLIC EXAMINATIONS

IN

MATHS-IA



Points to Remember

• Mathematics question paper contains 24 questions divided into three sections A, B and C. In Section

A 10 very short answer questions of 2 marks each and you have to answer all questions. In Section B,

7 short answer questions of 4 marks each and you have to answer 5 questions. In Section C

7 long answer questions of 7 marks each and you have to answer 5 questions.

• Mathematics is a subject where you learn by doing things. The more you practice the better you will

be able to understand the concepts.

• It’s not just the answer that counts in Mathematics. Much of your scoring is based on the

intermediate steps in getting to the answer. So show all your steps. Your objective should be to

convince your examiner that you know how to solve the problem.

• Don’t just aim for 90 percent. Aim high and go for 100 percent. This is a subject where you can score

cent percent.

• No additional sheet provided except the main answer booklet which contains 24 ruled pages, you can

use the last few pages in main answer book to do the rough work. Cancel those rough pages at last

• If you are not in a position to solve any particular question, then it is better not to waste time on that

question. Skip and head to the next question, then solve those questions afterwards

• You must write the formula whenever it is required as marks for the formula may be given in the

marking scheme and chances of errors are less

• Underline important points in the answer to make them more prominent.

• Mark the attempted questions so that you can come back to the unattempted at the end of the paper

• Always attempt all the questions including choice because you may get full marks if question is mis-

printed or wrong.

• Clarity in each step and error free calculations is the key to success in the mathematics exam.

• As far as possible, maintain the sequence of answers.

• Bright students should ideally look at attempting the paper from Section A and then Section B and

then Section C

• If you have finished the paper it is better not to submit the answer sheet. Utilize the remaining time

for revision.

• First complete booklet which contains all the important questions and then do all examples and

exercise problems in Telugu Academy Text book.

All the best

Mathematics Dept.



Points to Remember

For all those who wish to score 150 out of 150, the time to start the preparation

has come. The wise man finishes the things well before the D-day while the fool

wastes the plenty of time available and rushes up the things in the end, and achieves

nothing. So better you make a well thought out plan of study. A good plan is the

road-map to success, it shows you the final destination. Reduce your plan to writing.

The moment you complete this, you will have definitely given a concrete shape to

your desire. The reason that most people never reach this goal is that they don’t

define them, learn about them, or seriously consider them as achievable. Remember

this famous quote: “you were born to win, but to be a winner, you must plan to win,

prepare to win, and expect to win.” Once you have put a plan of study, stick to it

completely. Initially you will find it somewhat difficult to follow, but it would

become a routine after sometime paving the way of success for you.

We, at FIITJEE will always help you, motivate you, guide you in achieving

your dream of scoring 100%. In this direction we are providing you with important

questions and answers booklet-1 which will ensure you at least 50 marks out of 75 if

you are thorough with all these questions and answers. In this series we will provide

booklet-2 which will ensure you at least 70 if you do everything in this booklet-2.To

score 75 out of 75 you need to do the entire telugu academy text book which will help

to achieve good ranks in JEE (Mains), BITSAT and other engineering entrance exams.

For your reference, we are providing March 2014 Intermediate public

examination paper in the next page.



INTERMEDIATE PUBLIC EXAMINATION, MAY 2014

Total No. of Questions - 24 Reg.

Total No. of Printed Pages - 2 No.

Part - III

MATHEMATICS, Paper-I (A) (English Version)

Time : 3 Hours] [Max. Marks : 75

SECTION - A 10 × 2 = 20 M I. Very Short Answer Type questions:

1. If the function f is defined by ( ) 2

3 2, 3

2, 2 2

2 1, 3

x x

f x x x

x x

− >= − − ≤ ≤ + <−

then find the values of ( )4f and

( )2.5f .

2. Find the domain of the real valued function ( )( )( )2

1

1 3f x

x x=

− +

3. If 2 4

1A

k

= −

and 2 0A = , then find the value of k .

4. If ω is a complex (nonreal) cube root of 1, then show that

2

2

2

1

1 0

1

ω ω

ω ω

ω ω

=

5. Find the unit vector in the direction of vector 2 3a i j k= + + .

6. If the vectors 3 4i j kλ− + + and 8 6i j kµ + + are collinear vectors, then find λ and µ .

7. Find the angle between vectors 2 3i j k+ + and 3 2i j k− + .

8. Find the period of the function 4 9

cos5

x + .

9. Find the minimum and maximum values of 3sin 4cosx x− .

10. If 3

sinh6

x= , find ( )cosh 2x and ( )sinh 2x .

SECTION – B 5 × 4 = 20 M

II. Short Answer Type questions: (i) Attempt any five questions (ii) Each question carries four marks

11. If

1 2 2

3 2 1 2

2 2 1

A

= − − −

, then show that 1 TA A− =



12. If , ,i j k are unit vectors along with positive directions of coordinate axes, then show that the four

points 4 5 ,i j k+ + ,j k− − 3 9 4i j k+ + and 4 4 4i j k− + + are coplanar.

13. Find the area of the triangle whose vertices are ( )1,2,3A , ( )2,3,1B and ( )3,1,2C .

14. Prove that tan sec 1 1 sin

tan sec 1 cos

θ θ θ

θ θ θ

+ − +=

− +

15. Solve the equation, ( )2cot 3 1 cot 3 0x x− + + = ; 02

xπ

< <

16. Prove that 1 1 13 5 3232sin cos cos

5 13 325− − − − =

17. In ABC∆ , show that 2 2 2cot cot cotabc

a A b B c CR

+ + =

SECTION – C 5 × 7 = 35 M

III. Long Answer Type questions: (i) Attempt any five questions (ii) Each question carries seven marks

18. If : ,f A B→ :g B C→ are bijective functions, then prove that ( ) 1 1 1gof f og− − −= .

19. Using mathematical induction, prove that statement for all n N∈ :

3 3 3 3 3 31 1 2 1 2 3

.......1 1 3 1 3 5

+ + ++ + +

+ + + upto n terms = 22 9 13

24

nn n + +

.

20. If

2 3

2 3

2 3

1

1 0

1

a a a

b b b

c c c

+

+ =

+

and

2

2

2

1

1 0

1

a a

b b

c c

≠ , show that 1abc =− .

21. Solve the following equations by the Gauss-Jordan method. 2 3 9,x y z− + = 6,x y z+ + =

2x y z− + = .

22. If ( ) ( ) ( )1, 2, 1 , 4,0, 3 , 1,2, 1A B C= − − = − = − and ( )2, 4, 5D= − − then find the distance between AB

and CD lines.

23. If , ,A B C are the angles in a triangle, then prove that

sin sin sin 1 4sin sin sin2 2 2 4 4 4

A B C A B Cπ π π − − − + + = +

24. In ,ABC∆ if 13, 14, 15a b c= = = , then show that 65

, 4,8

R r= = 1 221

, 122

r r= = and 3 14r = .



VERY SHORT ANSWER QUESTIONS

1.

A. If { }: 0f R− → ℝ is defined by ( )1

f x xx

= + then prove that ( )( ) ( ) ( )2 2 1f x f x f= +

Sol. Given ( )1

f x xx

= +

Then ( ) ( )2 22

1 11 1

1f x f x

x+ = + + +

( )( )2

222

1 12x x f x

xx

= + + = + =

B. If { }: 0f − →ℝ ℝ is defined by ( ) 33

1f x x

x= − then show that ( )

10f x f

x

+ =

Sol. Given ( ) 33

1f x x

x= −

∴ ( ) 3 33 3

1 1 10f x f x x

x x x

+ = − + − =

C. If :f →ℝ ℝ is defined by ( )2

2

1

1

xf x

x

−=

+ , then show that ( )tan cos2f θ = θ

Sol. Given ( )2

2

1

1

xf x

x

−=

+

⇒

2

2 2 22

2 2 2 2

2

sin1

1 tan cos sincos(tan ) cos21 tan sin cos sin

1cos

f

θ−

− θ θ− θθθ = = = = θ+ θ θ θ+ θ

+θ

D. If { }: 1f − ± →ℝ ℝ is defined by ( )1

log ,1

xf x

x

+=

− then show that ( )

2

22

1

xf f x

x

= +

Sol. Given ( )1

log1

xf x

x

+=

−

Then ( )222

2 2

2

21

2 1 2 1 11log log log 2log 22 1 11 1 21

1

x

x x x x xxf f xx x xx x xx

+ + + + ++ = = = = = − −+ + −−

+

E. If { }0, , , ,6 4 3 2

Aπ π π π

= and :f A B→ is a surjection defined by ( ) cosf x x= , then find B

Sol. Given :f A B→ defined by ( ) cosf x x= is a surjection

⇒ Co domain B = range of f = ( )f A

( )0 , , , ,6 4 3 2

f f f f f π π π π =

{ }cos0 ,cos ,cos ,cos ,cos6 4 3 2

π π π π= °

3 1 11, , , ,0

2 22

=



F. If { }2 , 1 , 0 , 1 , 2A= − − and :f A B→ is a surjection defined by ( ) 2 1f x x x= + + , then find B .

Sol. Given :f A B→ defined by ( ) 2 1f x x x= + + is a surjection

⇒ Co domain B = range of f = ( )f A

( ) ( ) ( ) ( ) ( ){ }2 , 1 , 0 , 1 , 2f f f f f= − − { }3,1,1,3,7=

range of f { }1,3,7=

G. If { }1,2,3, 4A= and :f A→ ℝ is a function defined by ( )2 1

1

x xf x

x

− +=

+, then find the range

of f

Sol. Given :f A B→ defined by ( )2 1

1

x xf x

x

− +=

+

⇒ Range of f = ( )f A ( ) ( ) ( ) ( ){ }1 , 2 , 3 , 4f f f f=

{ }1 3 7 13, , ,

2 3 4 5= { }1 7 13

,1, ,2 4 5

=

H. If ( )2 4

2 4

cos sin

sin cos

x xf x x

x x

+= ∀ ∈

+ℝ then show that ( )2012 1f =

Sol. Given ( )2 4

2 4

cos sin

sin cos

x xf x

x x

+=

+

2 4

2 4

1 sin sin

1 cos cos

x x

x x

− +=

− +

( )( )

2 2

2 2

1 sin 1 sin

1 cos 1 cos

x x

x x

− −=

− −

2 2

2 2

1 sin cos

1 cos sin

x x

x x

−=

− 1=

∴ ( )2012 1f =

2.

A. Find the domain of definition of the function ( )y x , given by 2 2 2yx + =

Sol. Given 2 2 2yx + =

⇒ 2 2 2y x= − ⇒ ( )2log 2 2xy = −

( ) 2 2 0xf x ∈ ⇔ − >ℝ 12 2x⇔ < ( ),1x⇔ ∈ −∞

∴ Domain ( ),1= −∞

B. Find the domain of the real valued function ( )2

1

6 5f x

x x=

− −

Sol. ( ) 26 5 0f x x x∈ ⇔ − − ≠ℝ

2 6 5 0x x⇔ − + ≠ ( )( )1 5 0x x⇔ − − ≠ 1 , 5x⇔ ≠

∴ Domain of { }1,5f = −ℝ

C. Find the domain of the real valued function ( ) ( )( )2 3f x x x= + −

Sol. ( ) ( )( 2) 3 0f x x x∈ ⇔ + − ≥ℝ 2 3x or x⇔ ≤− ≥ ( ] [ ), 2 3,x⇔ ∈ −∞ − ∪ ∞



∴ Domain of f is ( ] [ ), 2 3,−∞ − ∪ ∞

D. Find the domain of the real valued function ( ) 2 1f x x x= − + +

( ) 2 0f x x∈ ⇔ − ≥ℝ and 1 0x+ ≥

⇔ 2x ≤ and 1x ≥ − ⇔ [ ]1,2x∈ −

∴ Domain of f is [ ]1,2−

E. Find the domain of the real valued function ( )( )

1

log 2f x

x=

−

( ) 2 0f x x∈ ⇔ − >ℝ and 2 1x− ≠

⇔ 2x< and 1x ≠ ⇔ ( ) { },2 1x∈ −∞ −

∴ Domain of f is ( ) ( ),1 1,2−∞ ∪

F. Find the domain of the real valued function ( ) 24f x x x= −

Sol. ( ) 24 0f x x x∈ ⇔ − ≥ℝ

⇔ 2 4 0x x− ≤ ⇔ ( )4 0x x− ≤ ⇔ [ ]0,4x ∈

∴ Domain of f is [ ]0,4

G. Find the domain of the real valued function ( ) 2 3 2f x x x= − +

Sol. ( ) 2 3 2 0f x x x∈ ⇔ − + ≥ℝ

2 3 2 0x x⇔ − + ≥ ( )( )1 2 0x x⇔ − − ≥ 1 2x or x⇔ ≤ ≥

∴ Domain of f is ( ] [ ),1 2,−∞ ∪ ∞

H. Find the domain of the real valued function ( ) ( )2log 4 3f x x x= − +

Sol. ( ) 2 4 3 0f x x x∈ ⇔ − + >ℝ

⇔ ( )( )1 3 0x x− − > ⇔ 1 3x or x< >

∴ Domain of f is ( ) ( ),1 3,−∞ ∪ ∞

I. Find the domain of the real valued function ( )2 2x x

f xx

+ + −=

Sol. ( )f x R∈ ⇔ 2 0, 2 0, 0x x x+ ≥ − ≥ ≠

2, 2, 0x x x⇔ ≥− ≤ ≠

⇔ [ ) ( ]2,0 0,2x∈ − ∪

∴ Domain of f is [ ) ( ]2,0 0,2− ∪

J. Find the domain of the real valued function ( )( )10

12

log 1f x x

x= + +

−

Sol. ( ) 2 0 , 1 0 , 1 1f x x x x∈ ⇔ + ≥ − > − ≠ℝ

⇔ 2 1 , 0x x x≥− < ≠ ⇔ [ ) ( )2,0 0,1x∈ − ∪



∴ Domain of f is [ )2,0) (0,1− ∪

K. If ( ) ( )1

11

xf x x

x

+= ≠±

− then find ( )0 0f f f x and ( )0 0 0f f f f x

Sol. Given ( )1

1

xf x

x

+=

−

( ) ( )( )

11

1 1 1 2111 1 1 211

xx x x xxfof x f f x f x

xx x xx

++ + + + −−= = = = = = + − + − +−

−

( ) ( )( ) ( )1

1

xfofof x f fof x f x

x

+= = =

−

( ) ( )( ) ( )fofofof x fof fof x fof x x= = =

L. If ( ) ( ) ( ){ }1,2 , 2, 3 , 3, 1f = − − then find

(i) 2 f (ii) 2 f+ (iii) 2f (iv) f

Sol. Given ( )( ) ( ){ }1,2 2, 3 , 3, 1f = − −

Domain of { }1,2,3 ,f = Range of { }2, 3, 1f = − −

(i) ( ) ( ) ( ){ }2 1,4 , 2, 6 , 3, 2f = − −

(ii) ( ) ( ) ( ){ }2 1,4 , 2, 1 , 3,1f+ = −

(iii) ( ) ( ) ( ){ }2 1,4 , 2,9 , 3,1f =

(iv) ( ){ }1, 2f =

M. Let ( ) ( ) ( ) ( ){ }1, , 2, , 4, , 3,f a c d b= and ( ) ( ) ( ) ( ){ }1 2, , 4, , 1, , 3,g a b c d− = then show that

( )1 1 1gof f o g

− − −=

Sol. Given ( ) ( ) ( ) ( ){ }1, , 2, , 4, , 3,f a c d b= and ( ) ( ) ( ) ( ){ }1 2, , 4, , 1, , 3,g a b c d− =

⇒ ( ) ( ) ( ) ( ){ }1 ,1 , ,2 , , 4 , ,3f a c d b− = and ( ) ( ) ( ) ( ){ },2 , , 4 , ,1 , ,3g a b c d=

Now ( ) ( ) ( ) ( ){ }1,2 , 2,1 , 3,4 , 4,3gof =

( ) ( ) ( ) ( ) ( ){ }1

2,1 , 1,2 , 4,3 , 3,4gof−

=

( ) ( ) ( ) ( ){ }1 1 1,2 , 2,1 , 3,4 , 4,3f o g− − =

∴ ( )1 1 1gof f o g

− − −=

N. If : , :f g→ →ℝ ℝ ℝ ℝ defined by ( ) ( ) 23 2 , 1f x x g x x= − = + then find

(i) ( )1 2gof − (ii) ( )1gof x−

Sol. Let ( ) ( )1 2 2 3 2 2 4/3f a f a a a− = ⇒ = ⇒ − = ⇒ =

∴ ( )1 2 4/3f − =



Now ( ) ( )( ) ( )1 1 16 252 2 4/3 1

9 9gof g f g− −= = = + =

( ) ( )( ) ( )( )1 1 3 1 2gof x g f x g x− = − = − −

( ) ( )2 23 5 3 5 1 9 30 26g x x x x− = − + = − +

O. If ( ) ( )1

2 1 ,2

xf x x g x

+= − = for all x∈ℝ , then find ( )gof x

Sol. ( ) ( )( ) ( )2 1gof x g f x g x= = − 2 1 1

2

xx

− += =

P. If ( ) ( ) ( )22 , , 2f x g x x h x x= = = for all x∈ℝ , then find ( )( )fo goh x

Sol. ( )( ) ( )( )( ) ( )( )( ) ( )( )2fo goh x f goh x f g h x f g x= = = ( )24 2f x= =

3.

A. If 1 2 3 8

A ,B3 4 7 2

= =

and 2X A B+ = , then find X .

Sol: Given 1 2 3 8

A ,B3 4 7 2

= =

Now 2X A B 2X B A+ = ⇒ = −

3 8 1 2 2 6

2X7 2 3 4 4 2

⇒ = − = −

Hence 2 6 1 31

X4 2 2 12

= = − −

B. If x 3 2y 8 5 2

z 2 6 2 a 4

− − = + − − −

then find the values of x,y,z and a

Sol: Given x 3 2y 8 5 2

z 2 6 2 a 4

− − = + − − −

x 3 5,2y 8 2,z 2 2⇒ − = − = + = − and 6 a 4= − − x 8,y 5,z 4⇒ = = = − and a 10= −

C. If 2 4

A1 k

= −

and 2A 0= , then find the value of k .

Sol: Given 2 4

A1 k

= −

and 2A 0=

2 4 2 4 0 0

1 k 1 k 0 0

⇒ = − −

2

0 8 4k 0 0

0 02 k 4 k

+ ⇒ = − − − +

8 4k 0⇒ + = and 2 k 0− − = and 2k 4 0− =

k 2⇒ = − and k 2− and k 2= ±

k 2⇒ = −



D. If

3 0 0

A 0 3 0

0 0 3

=

, then find 4A .

Sol: 23 0 0 3 0 0 9 0 0

A 0 3 0 0 3 0 0 9 0

0 0 3 0 0 3 0 0 9

= =

and 4 2 29 0 0 9 0 0 81 0 0

A A A 0 9 0 0 9 0 0 81 0

0 0 9 0 0 9 0 0 81

= ⋅ = =

E. If 2 0 1

A1 1 5

= −

and 1 1 0

B0 1 2

− = −

then find ( )11AB

Sol: ( )11 12 1

1 1 0 2 2AB BA 0 1

0 1 2 2 111 5

− − − = = = = −

F. If 2 4

A5 3

− = −

then find 1A A+ and 1AA

Sol: 1 2 4 2 5 4 9A A

5 3 4 3 9 6

− − − + = + = − − −

1 2 4 2 5 20 22AA

5 3 4 3 22 34

− − − = = − − −

G. If

1 2 3

A 2 5 6

3 x 7

− =

is a symmetric matrix, then find x .

Sol: Given A is symmetric matrix 1A A⇒ =

1 2 3 1 2 3

2 5 6 2 5 x

3 x 7 3 6 7

− − ⇒ =

Comparing we get x 6=

4.

A. If

1 0 0

A 2 3 4

5 6 x

= −

and det A 45= then find x .

Sol: det A 45=

( )

1 0 0

2 3 4 45

5 6 x

1 3x 24 45

3x 69

x 23

−⇒ =

−

⇒ − =⇒ =⇒ =

B. Find the rank of the matrix

1 0 0

0 0 1

0 1 0



Sol: ( )1 0 0

0 0 1 1 0 1 1 0

0 1 0

= − = − ≠

Since determinant of 3 3× matrix is non zero its rank = 3

C. Find the rank of the matrix

1 1 1

1 1 1

1 1 1

.

Sol:

1 1 1

1 1 1 0

1 1 1

=

determinant of 3 3× matrix is zero

1 1

01 1

=

determinant of any 2 2× sub matrix is zero

⇒ rank of matrix =1, since it is non zero matrix.

D. Find the trace of

1 3 5

2 1 5

2 0 1

− −

.

Sol: The trace of a square matrix is defined as some of its principle diagonal elements.

∴ trace of given matrix = 1 1 1 1− + =

5.

A. Find unit vector in the direction of vector a 2i 3j k= + + .

Sol: The unit vector in the direction of a vector a is given by 1

a aa

=ɵ

Now a 4 9 1 14= + + =

Therefore, ( )1a 2i 3j 4k

14= + +ɵ

B. Find a vector in the direction of vector a i 2j= − that has magnitude 7 units.

Sol: The unit vector in the direction of the given vector a is ( )1 1a a i 2 j

5a= = −ɵ

Therefore, the vector having magnitude equal to 7 and in the direction of a is ( )17a 7 i 2 j

5= ⋅ −ɵ

C. Find the unit vector in the direction of the sum of the vectors a 2i 2j 5k= + − and b 2i j 3k= + +

Sol: The sum of the given vectors c a b 4i 3j 2k= + = + − and c 16 9 4 29= + + =

( )1 1c c 4i 3j 2k

29c∴ = = + −ɵ

D. If the vectors 3i 4j kλ− + + and i 8j 6kµ + + are collinear vectors, then find λ and µ .



Sol: Since the vectors 3i 4j kλ− + + and i 8j 6kµ + + are collinear

we have 3 4

6, 38 6

λ µ λµ− = = ⇒ = − =

E. If the position vectors of the points A, B and C are 2i j k+ − , 4i 2j 2k− + + and 6i 3j 3k− −

respectively and AB ACλ= then find the value of λ .

Sol: Given OA 2i j k,OB 4i 2j 2k,OC 6i 3j 13k= − + − = − + + = − −��

then

AB OB OA 2i j 3k= − = − + +��

AC OC OA 8i 4j 12k= − = − −��

Now AB ACλ=��

( ) ( )2i j 3k 8i 4j 12kλ⇒ − + + = − −

1

8 24

λ λ⇒ = − ⇒ = −

F. If OA i j k= + +��

, AB 3i 2j k= − +��

, BC i 2j 2k= + −��

and CD 2i j 3k= + +��

then find the vector OD��

.

Sol: Given OA i j k= + +��

, AB 3i 2j k= − +��

, BC i 2j 2k= + −��

and CD 2i j 3k= + +��

Now OD OA AB BC CD 7i 2j 3k= + + + = + +��

G. Let a 2i 4j 5k= + − , b i j k= + + and c j 2k= + . Find the unit vector in the opposite direction of

a b c+ + .

Sol: Given a 2i 4j 5k= + − , b i j k= + + and c j 2k= +

a b c 3i 6j 2k⇒ + + = + −

Unit vector in the opposite direction of ( ) ( )1 1a b c a b c 3i 6 j 2k

7a b c+ + = − + + = − + −

+ +

6.

A. If ,α β and γ are the angles made by the vector 3i 6j 2k− + with the positive directions of the

coordinate axes then find cos ,cos ,cosα β γ

Sol: Given vector a 3i 6j 2k= − +

Unit vector in the direction of 3 6 2

a a i j k7 7 7

= = − +ɵ

Then 3 6 2

cos ,cos ,cos7 7 7

α β γ= = − =

B. In the two dimensional plane, prove by using vector methods, the equation of the line whose

intercepts on the axes are ‘a’ and ‘b’ is x y

1a b

+ =

Sol: Let ( )A a,0= and ( )B 0,b=

A ai∴ = , B bj=



Equation of line AB��

is ( ) ( )r 1 t ai t b j= − +

If r xi y j= + then ( )x 1 t a= − and y tb=

x y

1 t t 1a b

∴ + = − + =

C. Find the equation of the line parallel to the vector 2i j 2k− + and which passes through the point

A whose position vector is 3i j k+ − . If P is a point on this line such that AP 15= , find the

position vector of p.

Sol: The vector equation of the given line is

( ) ( )r 3i j k t 2i j 2k= + − + − + , t ∈ℝ

Since ( )AP t 2i j 2k= − + , we have

2 2 215 AP 4t t 4t 3t t 5= = + + = ± ⇒ = ±

( ) ( )op 3i j k 5 2i j 2k 13i 4j 9k∴ = + − ± − + = − +

D. Find the vector equation of the line passing through the point 2i 3j k+ + and parallel to the

vector 4i 2j 3k− + .

Sol: Let a 2i 3j k= + + and b 4i 2j 3k= − +

The vector equation of line passing through point a and parallel to vector b is r a tb, t= + ∈ℝ

Therefore required equation is ( ) ( )r 2i 3j k 4i 2 j 3k= + + + − +

E. OABC is a parallelogram. If OA a= and OC c= , find the vector equation of the side BC.

Sol:

Given OC c=��

and OA a=��

Then OB OA AB OA OC a c= + = + = +��

Vector equation of side BC is equation of the line passing

through points ( )B a c+ and ( )C c

∴ equation is ( ) ( )r t a c 1 t c, t= + + − ∈ℝ A

BC

Oa

c�

F. If a,b,c are position vectors of the vertices A, B and C respectively of ABC∆ , then find the vector

equation of the median through the vertex A.

Sol: Let D be the mid point of BC.

Then position vector of D, OB OC b c

OD2 2

+ += =��

��

∴ the vector equation of AD is the line passing through ( )A a and b c

D2

+

is



( ) b cr ta 1 t , t

2

+= + − ∈

ℝ .

G. Find the vector equation of the line joining the points 2i j 3k+ + and 4i 3j k− + −

Sol: Let a 2i j 3k= + + , b 4i 3j k= − + −

The vector equation of line passing through points a and b is ( )r 1 t a tb= − + .

∴ required equation is ( )( ) ( )r 1 t 2i j 3k t 4i 3j k= − + + + − + − .

H. Find the vector equation of the plane passing through the points i 2j 5k, 5j k− + − − and 3i 5j− +

Sol: Let a i 2j 5k= − + , b 5j k= − − , c 3i 5j= − + .

The vector equation of the plane passing through the points a,b and c is

( )r 1 t s a tb sc= − − + + , s, t∈ℝ

∴ required equation of plane is

( )( ) ( ) ( )r 1 t s i 2 j 5k t 5j k s 3i 5j= − − − + + − − + − + where s and t ∈ℝ .

7.

A. If 6 2 3a i j k= + + and 2 9 6 ,b i j k= − + then find .a b and the angle between a and b

Sol: Given 6 2 3a i j k= + + and 2 9 6b i j k= − +

Then . 12 18 18 12a b= − + =

2 2 26 2 3 7a = + + = and ( )22 22 9 6 11b = + − + =

. 12

cos77

a b

a bθ∴ = =

Angle between vectors, 1 12cos

77θ − =

B. If 2 3a i j k= + − and 3 2 ,b i j k= − + then show that a b+ and a b− are perpendicular to each

other

Sol: Given 2 3a i j k= + − and 3 2b i j k= − +

4a b i j k∴ + = + − and 2 3 5a b i j k− =− + −

( ) ( ) ( ) ( ) ( )( ). 4 2 1 3 1 5 0a b a b∴ + − = − + + − − =

a b∴ + and a b− are at right angles

C. Let a and b are non zero, non collinear vectors.

If ,a b a b+ = − then find the angle between a and b

Sol: Given a b a b+ = −

2 2

a b a b⇒ + = −



( ) ( ) ( ) ( ). .a b a b a b a b⇒ + + = − −

( ) ( )2 2 2 22 . 2 .a b a b a b a b⇒ + + = + −

( )4 . 0a b⇒ =

. 0a b⇒ =

∴ Angle between a and b is 90°

D. If a i j k= − − and 2 3 ,b i j k= − + then find the projection vector of b on a and its magnitude.

Sol: Given a i j k= − − and 2 3b i j k= − +

Now projection vector of b on a = 2

.b aa

a

( ) ( )

( )2 22

2 3 1

1 1 1i j k

+ −= − −

+ − + −

( )4

3i j k= − −

Magnitude = . 4

3

b a

a=

E. If the vectors 3 5i j kλ − + and 2 i j kλ λ− − are perpendicular to each other, find λ

Sol: Given 3 5i j kλ − + and 2 i j kλ λ− − are perpendicular

( ) ( )3 5 . 2 0i j k i j kλ λ λ∴ − + − − =

22 3 5 0λ λ⇒ + − =

( )( )2 5 1 0λ λ⇒ + − =

5

2λ⇒ =− or 1

F. Find the angle between the plane 2 3 6 5x y z− − = and 6 2 9 4x y z+ − =

Sol: Angle between the planes = angle between their normals

Here normals to the planes are 1 2 3 6n i j k= − − and 2 6 2 9n i j k= + −

Then angle between normals, 1 21

1 2

.cos

n n

n nθ −

=

1 12 6 54cos

49 121

− − + = = 1 60

cos77

−

G. If 1e and 2e be unit vectors making angle θ . If 1 21

sin ,2

e e λθ− = then find λ

Sol: Given 1 21

sin2

e e λθ− =



Squaring both sides, 2 2 2

1 2 1 21

2 cos sin4

e e e e θ λθ + − =

( ) 212 2cos sin

4θ λθ⇒ − =

2 214sin sin

4 2

θλθ

⇒ =

1

2λ⇒ =

H. Find the equation of the plane through the point ( )3, 2,1− and perpendicular to the vector

( )4,7, 4−

Sol: Equation of plane through point ( )3 2a i j k− + and perpendicular to vector ( )4 7 4b i j k+ − is

( ). 0r a b− =

( )( ) ( )3 2 . 4 7 4 0r i j k i j k∴ − − + + − =

( ) ( ) ( ). 4 7 4 3 2 . 4 7 4r i j k i j k i j k∴ + − = − + + −

( ). 4 7 4 6r i j k∴ + − =−

I. If 2 3 5a i j k= − + and 4 2b i j k=− + + then find a b× and unit vector perpendicular to both a

and b

Sol: Given 2 3 5a i j k= − + and 4 2b i j k=− + +

Then 2 3 5 26 9 5

1 4 2

i j k

a b i j k× = − =− − +

−

The unit vector perpendicular to both a and b

( )1

26 9 5782

a bi j k

a b

×=± =± − − +

×

J. Find the area of the parallelogram for which the vectors 2 3a i j= − and 3b i k= − are adjacent

sides

Sol: The vector area of parallelogram with sides a and b is

2 3 0 3 2 9

3 0 1

i j k

a b i j k× = − = + +

−

∴ Area = 94a b× =

K. If 2 3a i j k= + + and 3 5b i j k= + − are two sides of a triangle, then find its area

Sol: Vector area of the triangle with a and b as two sides



= ( )1

2a b×

11 2 3

23 5 1

i j k

=

−

( )1

17 102

i j k= − + −

∴ Area of the triangle = 1 390

2 2a b× =

L. Let 2a i j k= − + and 3 4b i j k= + − . If θ is the angle between a and b , then find sinθ

Sol: sina b

a bθ

×= and given 2 , 3 4a i j k b i j k= − + = + −

Where 2 1 1 3 5 11

3 4 1

i j k

a b i j k× = − =− + +

−

155 , 6 , 26a b a b⇒ × = = =

Now 155 155

sin1566 26

a b

a bθ

×= = =

8.

A. Prove that 2 3 7

cot .cot .cot ....cot 116 16 16 16

π π π π=

B. If 3sin 4cos 5,θ θ+ = then find the value of 4sin 3cosθ θ− .

C. If cos sin 2 cos ,θ θ θ+ = prove that cos sin 2 sinθ θ θ− =

D. Prove that ( )2 2 2 2 2tan cot sec cos sec .cosec ecθ θ θ θ θ θ+ = + = .

E. If tan 20o λ= , then show that 2tan160 tan110 1

21 tan160 .tan110

o o

o o

λ

λ

− −=

+.

F. Prove that tan sec 1 1 sin

tan sec 1 cos

θ θ θ

θ θ θ

+ − +=

− +.

G. Prove that ( )( )1 cot cos 1 tan sec 2ecθ θ θ θ+ − + + =

H. Prove that ( ) ( ) ( )4 2 6 63 sin cos 6 sin cos 4 sin cos 13θ θ θ θ θ θ− + + + + =

I. Prove that ( ) ( ) ( )2 2 2 2sin cos cos sec tan cot 7ecθ θ θ θ θ θ+ + + − + =

J. Prove that 3 5 7 9

cot .cot .cot .cot .cot 120 20 20 20 20

π π π π π=

9.

A. Draw the graph of 2cosy x= in ( )0,π .

B. Draw the graph of sin2y x= in ( ),π π− .



C. Find the period of the function defined by ( ) ( )= + + + + 2tan 4 9 ...f x x x x n x

D. Find the periods of the functions:

a) ( ) tan 5f x x= b) ( )4 9

cos5

xf x

+ = c) ( ) sinf x x=

E. Prove that 2 21 1 3 1sin 52 sin 22

2 2 4 2

o o + − =

F. Prove that tan70 tan 20 2 tan50o o o− =

G. If 45oA B+ = , then prove that

i) ( )( )1 tan 1 tan 2A B+ + = ii) ( )( )cot 1 cot 1 2A B− − =

H. Prove that cos9 sin9

cot 36cos9 sin9

o oo

o o

+=

−

I. Draw the graph of the tanx between 0 and 4

π.

J. Draw the graph of the cos2x in the interval [ ]0,π .

K. Find the extreme values of 5cos 3cos 83

x xπ + + +

over R.

L. Find the range of

i) 7 cos 24sin 5x x− + ii) 13cos 3 3 sin 4x x+ −

M. Find the minimum and maximum values of

i) 3cos 4sinx x+ ii) sin 2 cos2x x−

N. If θ is not an odd multiple of 2

π and if tan 1θ≠ , then show that

1 sin2 cos2tan

1 sin 2 cos2

θ θθ

θ θ

+ −=

+ +.

O. Prove that 1 3

4sin10 cos10o o

− =

P. Prove that ( )4 cos66 sin84 3 15o o+ = +

Q. Prove that ( ) ( )cos cos 120 cos 240 0o oθ θ θ+ + + + =

10.

A. Prove that, for any x R∈ , ( ) 3sinh 3 3sinh 4sinhx x x= +

B. If 5

cosh2

x= , find the values of (i) ( )cosh 2x and (ii) ( )sinh 2x

C. If sinh 5,x= show that ( )log 5 26e

x= +

D. Show that 1 1 1tanh log 3

2 2 e

− =

E. If 3

sinh ,4

x= find ( )cosh 2x and ( )sinh 2x



F. Prove that (i) ( ) ( ) ( )cosh sinh cosh sinh ,nx x nx nx− = − for any n R∈

(ii) ( ) ( ) ( )cosh sinh cosh sinh ,nx x nx nx+ = + for any n R∈

G. If =sinh 3x then show that ( )= +log 3 10e

x

SHORT ANSWER QUESTIONS

11.

A. If

1 2 2

A 2 1 2

2 2 1

=

then show that 2A 4A 5I 0− − =

Sol: 21 2 2 1 2 2 9 8 8

A A A 2 1 2 2 1 2 8 9 8

2 2 1 2 2 1 8 8 9

= ⋅ = =

1 2 2 4 8 8

4A 4 2 1 2 8 4 8

2 2 1 8 8 4

= =

1 0 0 5 0 0

5I 5 0 1 0 0 5 0

0 0 1 0 0 5

= =

Hence 29 8 8 4 8 8 5 0 0 0 0 0

A 4A 5I 8 9 8 8 4 8 0 5 0 0 0 0 0

8 8 9 8 8 4 0 0 5 0 0 0

− − = − − = =

B. Show that ( )( )( )

2

2

2

1

1

1

a a

b b a b b c c a

c c

= − − −

Sol. LHS =

2 2

2 2 12 2 2

2 2 2

3 3 1

1 1

1 0

1 0

a a a aR R R

b b b a b a

c c c a c a R R R

→ −= − −

− − → −

( )( )

21

0 1

0 1

a a

b a c a b a

c a

= − − +

+

by taking b a− , c a− common in 2R and 3R

( )( ) ( ) ( )( ). 1b a c a c a b a= − − + − +

by expanding along first column

( )( )( )b a c a c b= − − −

( )( )( )a b b c c a= − − −



C. Without expanding the determinant show that

2

b c c a a b a b c

c a a b b c b c a

a b b c c a c a b

+ + +

+ + + =

+ + +

Sol. LHS

b c c a a b

c a a b b c

a b b c c a

+ + +

= + + +

+ + +

1 1 2 3R R R R→ + +

( ) ( ) ( )2 2 2a b c a b c a b c

c a a b b c

a b b c c a

+ + + + + +

= + + +

+ + +

2

a b c a b c a b c

c a a b b c

a b b c c a

+ + + + + +

= + + +

+ + +

2 2 1 3 3 1;R R R R R R→ − → −

2

a b c a b c a b c

b c a

c a b

+ + + + + +

= − − −

− − −

1 1 2 3R R R R→ + +

2

a b c

b c a

c a b

= − − −

− − −

2

a b c

b c a

c a b

= ( by taking ' '− common 2 3,R R rows )

D. If ' 'w is complex (non real) cube root of 1 then show that

2

2

2

1

1 0

1

w w

w w

w w

=

Sol. LHS

2

2

2

1

1

1

w w

w w

w w

=

1 1 2 3R R R R→ + +

2 2 2

2

2

1 1 1

1

1

w w w w w w

w w

w w

+ + + + + +

=



( )2 2

2

0 0 0

1 1 0

1

w w w w

w w

= + + =∵

0=

E. Find the value of x , if

2 2 3 3 4

4 2 9 3 16 0

8 2 27 3 64

x x x

x x x

x x x

− − −

− − − =

− − −

Sol. 0

2 2 3 3 4

4 2 9 3 16

8 2 27 3 64

x x x

x x x

x x x

− − −

= − − −

− − −

2 2 1 3 3 1;R R R R R R→ − → −

2 2 3 3 4

2 6 12

6 24 60

x x x− − −

= − − −

− − −

0 ( )( )

2 2 3 3 4

2 6 1 3 6

1 4 10

x x x− − −

= − −

⇒

2 2 3 3 4

1 3 6 0

1 4 10

x x x− − −

=

Expanding the determinant along first row, we get

( )( ) ( )( ) ( )( )2 30 24 2 3 10 6 3 4 4 3 0x x x− − − − − + − − =

⇒ 6 12 8 12 3 4 0x x x− − + + − =

⇒ 4 0x− =

⇒ 4x=

F. If

3 3 4

2 3 4

0 1 1

A

− = − −

then show that 1 3A A− =

Sol: We have to prove 1 3A A− =

We have to prove 1 4AA A− =

We have to prove 4A I=

Now 23 3 4 3 3 4

. 2 3 4 2 3 4

0 1 1 0 1 1

A A A

− − = = − − − −

9 6 0 9 9 4 12 12 4

6 6 0 6 9 4 8 12 4

0 2 0 0 3 1 0 4 1

− + − + − − + = − + − + − − + − + + − − +



3 4 4

0 1 0

2 2 3

− = − − −

4 2 23 4 4 3 4 4

. 0 1 0 0 1 0

2 2 3 2 2 3

A A A

− − = = − − − − − −

9 0 8 12 4 8 12 0 12

0 0 0 0 1 0 0 0 0

6 0 6 8 2 6 8 0 9

− − − + + − − = + + + + + + − + + − − − + +

1 0 0

0 1 0

0 0 1

I

= =

G. Show that ( )

2

3

2 2 1 1

2 1 2 1 1

3 3 1

a a a

a a a

+ +

+ + = −

Sol. LHS =

2 2 2 1 1

2 1 2 1

3 3 1

a a a

a a

+ +

+ +

1 1 2 2 2 3;R R R R R R→ − → −

2 1 1 0

2 2 1 0

3 3 1

a a

a a

− −

= − −

( )( )

1 1 0

1 1 2 1 0

3 3 1

a

a a

+

= − −

By taking 1a− common in 1R and 2R

( )( ) ( )1 1 . 1. 1 2a a a= − − + −

By expanding along third column

( )3

1a= −

H. If

1 2 2

3 2 1 2

2 2 1

A

= − −

, then show that 1 1A A− =

Sol. We have to prove 1 1A A− =

i.e., We have to prove 1 1. .A A A A− −=

We have to prove 1AA I=



Now 1

1 2 2 1 2 21 1

2 1 2 2 1 23 3

2 2 1 2 2 1

AA

− = − = − − − −

1 4 4 2 2 4 2 4 21

2 2 4 4 1 4 4 2 29

2 4 2 4 2 2 4 4 1

+ + + − − + − = + − + + − + + − + − − + − + +

9 0 010 9 0

90 0 9

=

1 0 0

0 1 0

0 0 1

I

= =

12.

A. Show that the points ( ) ( ) ( )A 2i j k ,B i 3j 5k ,c 3i 4j 4k− + − − − − are the vertices of a right angled

triangle.

Sol: Given OA 2i j k= − +��

, OB i 3j 5k= − −��

, OC 3i 4j 4k= − −��

AB OB OA i 2 j 6k AB 41∴ = − = − − − ⇒ =��

BC OC OB 2i j k BC 6= − = − + ⇒ =��

CA OA OC i 3j 5k CA 35= − = − + + ⇒ =��

2 2 2AB BC AC⇒ = +

∴ the points A, B, C are the vertices of right angled triangle.

B. If a b c dα+ + = , b c d aβ+ + = and a,b,c are non-coplanar vectors, then show that

a b c d O+ + + = .

Sol: Given a b c dα+ + = ……………….(1)

and b c d aβ+ + = …………………(2)

substituting the value of d from (2) in (1), we get

( )a b c a c bα β+ + = − −

( ) ( ) ( )1 a 1 b 1 c 0αβ α α⇒ − + + + + =

Since a,b,c are non coplanar, we get

1 0αβ− = , 1 0α+ = , 1 0α+ = 1, 1α β⇒ = − = −

Substituting ' 'α value in (1), we get a b c d o+ + + =

C. a,b,c are non-coplanar. Prove that the four points a 4b 3c− + − , 3a 2b 5c+ − , 3a 8b 5c− + − ,

3a 2b c− + + are coplanar

Sol: Given four points A, B, C, D with position vectors are



OA a 4b 3c= − + −��

, OB 3a 2b 5c= + −��

, OC 3a 8b 5c= − + −��

, OD 3a 2b c= − + +��

Now AB OB OA 4a 2b 2c= − = − −��

AC OC OA 2a 4b 2c= − = − + −��

AD OD OA 2a 2b 4c= − = − − +��

( )3 3 1 2

4 2 2 4 2 2

AB AC AD 2 4 2 2 4 2 R R R R 0

2 2 4 0 0 0

− − − − ∴ = − − = − − → − + =

− −

��

A,B,C,D∴ are coplanar.

D. Show that the points whose position vectors are 2a 3b 5c, a 2b 3c, 7a c− + + + + − are collinear

when a,b,c are non coplanar vectors.

Sol: Let A, B, C be the given points

Then OA 2a 3b 5c= − + +��

, OB a 2b 3c= + +��

, OC 7a c= −��

Now AB OB OA 3a b 2c= − = − −��

BC OC OB 6a 2b 4c= − = − −��

BC 2AB∴ =��

Hence A, B, C are collinear.

E. Let ABCDEF be a regular hexagon with centre ‘O’ . Show that AB+AC+AD+AE+AF=3AD=6AO

Sol:

From the figure,

AB AC AD AE AF+ + + +

( ) ( )AB AE AD AC AF= + + + +

( ) ( )ED AE AD AD CD= + + + +

( )AD AD AD AB ED,AF CD= + + = =∵

( )3AD 6AO AD 2AO= = =∵

A B

C

DE

F i

13.

A. Find the areas of the triangle whose vertices are ( ) ( )1,2,3 , 2,3,1A B and ( )3,1,2C

Sol: Given point are ( ) ( )1,2,3 , 2,3,1A B and ( )3,1,2C

2 , 2AB i j k AC i j k∴ = + − = − −��

Area of triangle = 1 1

1 1 22 2

2 1 1

i j k

AB AC× = −

− −

��

1

3 3 32

i j k= − − − 1 3 3

9 9 92 2

= + + =



B. If 2, 3P Q= = and ( ), ,6

P Qπ

= then find 2

P Q×

Sol: ( )2 2 2 2sin ,P Q P Q P Q× =

1

4 9 94

= × × =

C. Prove that the vectors 2 , 3 5a i j k b i j k= − + = − − and 3 4 4c i j k= − − are coplanar.

Sol: Given 2 , 3 5a i j k b i j k= − + = − − and 3 4 4c i j k= − −

Now

2 1 1

1 3 5

3 4 4

a b c

−

= − − − −

( ) ( ) ( )2 12 20 1 4 15 4 9= − + − + + − + 16 11 5 0=− + + =

, ,a b c∴ are coplanar

D. Find the volume of the parallelepiped whose coterminus edges are represented by the vectors

2 3 , 2i j k i j k− + − + and 2i j k+ −

Sol: Let 2 3 , 2a i j k b i j k= − + = − + and 2c i j k= + −

Volume of the parallelepiped = [ ]a b c

2 3 1

1 1 2 2 15 3 14 14

2 1 1

−

= − = − − + = − =

−

E. If the vectors 2 ,a i j k= − + 2 3b i j k= + − and 3 5c i Pj k= + + are coplanar, then find P.

Sol: The vectors , ,a b c are coplanar if and only if 0a b c =

2 1 1

0 1 2 3

3 5

a b c

P

−

∴ = = −

( ) ( ) ( )2 10 3 1 5 9 6P P= + + + + −

20 6 14 6P P= + + + −

7 28P= +

4P⇒ =−

F. Prove that for any three vectors , ,a b c 2b c c a a b a b c + + + =

Sol:

0 1 1

1 0 1

1 1 0

b c c a a b a b c + + + =

( ) ( )1 1 1 1 a b c =− − +

2 a b c =



G. For any three vectors, , ,a b c prove that 2

b c c a a b a b c × × × =

Sol: ( ) ( ) ( )( ).b c c a a b b c c a a b × × × = × × × ×

( ) ( ).b c c a b a c a a b = × −

( ).b c c a b a = ×

= b c a c a b

2

a b c =

H. Find the volume of the tetrahedron having the edges , , 2i j k i j i j k+ + − + +

Sol:

Let , , 2a i j k b i j c i j k= + + = − = + +

The volume of the tetrahedron having adjacent sides , ,a b c

1

6a b c =

( ) ( ) ( )11 1 1 1 1 3

6= − − +

1

6=

14.

A. If 2

sec tan3

θ θ+ = , find the value of sinθ and determine the quadrant in which θ lies.

B. Find the value of ( ) ( )6 6 4 42 sin cos 3 sin cosθ θ θ θ+ − +

C. If cos 0θ> , tan sin mθ θ+ = and tan sin nθ θ− = , then show that 2 2 4m n mn− =

D. Let ABC be a triangle such that cot cot cot 3A B C+ + = , then prove that ABC is an

equilateral triangle.

E. Find the maximum and minimum values of

i) 3sin 4cosx x− ii) cos 2 2 sin 33 3

x xπ π + + + −

F. If θ is not an integral multiple o f2

π, prove that tan 2 tan2 4tan 4 8cot8 cotθ θ θ θ θ+ + + =

G. For A R∈ , prove that

a) ( ) ( )1

sin .sin 60 sin 60 sin34

A A A A+ − =

b) ( ) ( )1

cos .cos 60 cos 60 cos34

A A A A+ − = and hence deduce that

c) 3

sin 20 sin 40 sin60 sin8016

o o o o = d) 2 3 4 1

cos cos cos cos9 9 9 9 16

π π π π=



H. If 3A is not an odd multiple of 2

π, prove that ( ) ( )tan .tan 60 .tan 60 tan 3A A A A+ − = and

hence find the value of tan6 tan 42 tan66 tan78o o o o .

I. If ,α β are the solutions of the equation cos sina b cθ θ+ = ( , ,a b c are non-zero numbers) then

show that

i) 2 2

2sin sin

bc

a bα β+ =

+ ii)

2 2

2 2sin .sin

c a

a bα β

−=

+

J. If A is not an integral multiple of π , prove that sin16

cos .cos2 .cos4 .cos816sin

AA A A A

A= and

hence deduce that 2 4 8 16 1

cos .cos .cos .cos15 15 15 15 16

π π π π=

K. If 1

sin sin4

x y+ = and 1

cos cos3

x y+ = , then show that

a) 3

tan2 4

x y+= ii) ( )

7cot

24x y+ =

L. Prove that 4cos12 cos 48 cos72 cos36o o o o=

M. If 4

cos cos5

x y+ = and 2

cos cos7

x y− = , find the value of 14 tan 5cot2 2

x y x y− ++

N. If , ,x y z are non zero real numbers and if 2 4

cos cos cos3 3

x y zπ π

θ θ θ = + = +

for some

Rθ∈ , then show that 0xy yz zx+ + =

O. Prove that 4 4 4 43 5 7 3sin sin sin sin

8 8 8 8 2

π π π π+ + + =

P. Prove 2 3 4 5 1

cos .cos .cos .cos .cos11 11 11 11 11 32

π π π π π=

Q. Prove that 3 7 9 1

1 cos 1 cos 1 cos 1 cos10 10 10 10 16

π π π π + + + + =

15.

A. If x is acute and ( ) ( )sin 10 cos 3 68x x+ ° = − ° , find x in degrees

B. Solve 22cos 3 sin 1 0θ − θ + =

C. Find all values of 0x ≠ in ( ),−π π satisfying the equation 21 cos cos ..... 38 4x x+ + + =

D. Solve tan 3cot 5secθ + θ = θ

E. Solve 21 sin 3sin cos+ θ = θ θ

F. Solve ( )2 sin cos 3x x+ =

G. Solve 4sin sin2 sin 4 sin 3x x x x=

H. Solve sin2 cos2 sin cosx x x x− = −

I. Solve 2 22sin sin 2 2x x+ =



J. Solve ( )2cot 3 1 cot 3 0,x x− + + = 02

xπ< <

16.

A. Prove that 1 14 1sin 2 tan

5 3 2− − π + =

B. If 1 1 1sin sin sin ,x y z− − −+ + = π then prove that ( )4 4 4 2 2 2 2 2 2 2 2 24 2x y z x y z x y y z z x+ + + = + +

C. Solve arc 5

sinx

+

arc 12

sinx

= ( )02

xπ >

D. Solve 1 1sin sin 23

x x− − π+ =

E. Prove that 1 1 11 1 1tan tan tan

2 5 8 4− − − π+ + =

F. Find the value of 1 14 2tan cos tan

5 3− − +

G. If 1 1sin cos6

x x− − π− = , then find x .

H. 1 1 13 8 36sin sin cos

5 17 85− − −+ =

I. 1 1 14 3 27cos sin tan

5 1134

− − −+ =

J. If 1 1 1cos cos cos ,p q r− − −+ + = π then prove that 2 2 2 2 1p q r pqr+ + + =

K. If 1 1 1sin sin sin ,x y z− − −+ + = π then prove that 2 2 21 1 1 2x x y y z z xyz− + − + − =

L. If 1 1 1tan tan tan ,x y z− − −+ + = π then prove that x y z xyz+ + =

M. Solve ( )1 1sin 1 2sin2

x x− − π− − =

N. Solve 1 1cos sin2 6

xx− − π+ =

O. Show that ( ) ( )2 1 2 1sec tan 2 cos cot 2 10ec− −+ =

17.

A. In ABC∆ if 3, 4a b= = and 3

sin ,4

A = find angle B.

B. If the lengths of the sides of a triangle are 3,4,5. Find the circum radius of the triangle.

C. If 6, 5, 9a b c= = = then find angle A

D. In ,ABC∆ show that ( )cos 25b c A+ =∑

E. In a ,ABC∆ if ( )( ) 3 ,a b c b c a bc+ + + − = find A

F. If 4, 5, 7a b c= = = find cos2

B



G. In ,ABC∆ find 2 2cos cos2 2

C Bb c+

H. If 5

tan2 6

A = and 2

tan2 5

C = , determine the relation between , ,a b c

I. Prove that ( ) 2 2cos cosa b C c B b c− = −

J. If 26 ,a cms= 30b cms= and 63

cos65

C = then find c

K. In a ABC∆ , prove that 1 2 3

1 1 1 1

r r r r+ + =

L. Show that 21 2 3rr r r = ∆

M. In an equilateral triangle find r

R

N. If 90 ,A = ° show that ( )2 r R b c+ = +

O. In a ,ABC∆ show that the sides , ,a b c are in A.P if and only if 1 2 3, ,r r r are in H.P.

P. In a ,ABC∆ if 1 1 3

a c b c a b c+ =

+ + + + , show that 60c = °

Q. If : : 7 : 8 : 9a b c = , find cos : cos : cosA B C

R. If 3 ,b c a+ = then find the value of cot .cot2 2

B C

S. If cot : cot : cot 3 : 5 : 72 2 2

A B C = . Show that : : 6 : 5 : 4a b c =

T. Prove that ( )22 2 2

cot cot cot2 2 2

cot cot cot

A B Ca b c

A B C a b c

+ + + +=

+ + + +

LONG ANSWER QUESTIONS

18.

A. Let : , :f A B g B C→ → are bijections. Then prove that :gof A C→ is also a bijection.

Sol. Given : , :f A B g B C→ → are bijections.

Then :gof A C→ is a function

To prove gof one-one :

Let 1 2,a a A∈ be such that ( )( ) ( )( )1 2gof a gof a=

⇒ ( )( ) ( )( )1 2g f a g f a=

⇒ ( ) ( )1 2f a f a= [ Since g is one-one ]

⇒ 1 2a a= [ Since f is one-one ]

⇒ :gof A C→ is one-one



To prove gof onto :

Let c C∈ , Since :g B C→ is onto there exists b B∈ such that ( )g b c=

Now b B∈ and :f A B→ is onto therefore there exists a A∈ such that ( )f a b=

Now ( ) ( )( ) ( )c g b g f a gof a= = =

∴ For each c C∈ there exists a A∈ such that ( )gof a c=

⇒ gof is onto

∴ :gof A C→ is a bijection

B. Let :f A B→ , :g B C→ be bijections. Then( )1 1 1gof f og

− − −=

Sol. Given :f A B→ and :g B C→ are bijections

∴ :gof A C→ is also a bijection

and ( )1:gof C A

−→ is also a bijection

Further, 1 :g C B− → and 1 :f B A− → are also bijections

and 1 1 :f o g C A− − → is also a bijection

∴ The functions ( )1

gof−

and 1 1f og− − have same domain C

Let c C∈ , :g B C→ is a bijection, there exists a unique b B∈ such that ( )g b c= i.e., ( )1g c b− =

Now , :b B f A B∈ → is a bijection, there exists a unique a A∈ such that ( )f a b= i.e., ( )1f b a− =

∴ ( ) ( )( ) ( )( )c g b g f a gof a= = =

⇒ ( ) ( )1

gof c a−

=

and ( ) ( )( ) ( )1 1 1 1 1f o g c f g c f b a− − − − −= = =

⇒ ( )( )1 1f o g c a− − =

Hence ( )1 1 1gof f o g

− − −=

C. Let :f A B→ , AI and BI be identity functions on A and B respectively. Then

A Bf o I f I o f= =

Sol. Since :AI A A→ and :f A B→ are functions

⇒ Af o I is a function from A B→

Hence Af o I and f have same domain A

Let ,a A∈ then ( ) ( )( ) ( ) [ ]SinceA A Af o I a f I a f a I a a A= = = ∀ ∈

∴ Af o I f= ………………(1)

Since :f A B→ and :BI B B→ are functions



⇒ BI o f is a function from A B→

Hence BI o f and f have same domain A

Let ,a A∈ then ( ) ( )( ) ( ) ( ) ( )andB B BI o f a I f a f a I b b b B f a B = = = ∀ ∈ ∈ ∵

∴ BI o f f= ……………….(2)

From (1) and (2) A Bf o I f I o f= =

D. Let :f A B→ be a bijection then 1Bfof I− = and 1

Af of I− =

Sol. Given :f A B→ is a bijection

⇒ 1 :f B A− → is also a bijection

Hence 1 :fof B B− → is also a bijection

Where as :BI B B→ is also a bijection

∴ 1fof − and BI have same domain B

Let b B∈ , since :f A B→ is a bijection there exists a unique a A∈ such that

( ) ( )1. .,f a b i e f b a−= =

Thus ( ) ( )( ) ( ) ( )1 1Bfof b f f b f a b I b− −= = = =

∴ 1Bfof I− =

Similarly 1 :fof A A− → is also a bijection

and :AI A A→ is also a bijection

∴ 1fof − and AI have same domain A

Let a A∈ , since :f A B→ is a bijection there exists aunique b B∈ such that

( ) ( )1. .,f a b i e f b a−= =

Thus ( )( ) ( ) ( )1 1 1Afof a f f a f b a I a− − −= = = =

∴ 1Af of I− =

E. Let :f A B→ be a function then f is a bijection if and only if there exists a function :g B A→

such that Bfog I= and Agof I= and in this case 1g f −=

Sol. ( )⇒ Let :f A B→ is a bijection

We have to prove that there exists a function :g B A→ such that Bfog I= and Agof I=

Since :f A B→ is a bijection

1 :f B A− → is also a bijection and let 1g f −=

Then ( ) ( ) ( ) ( )1Bfog b fof b f a b I b−= = = = where ( )f a b=

⇒ Bfog I=



and ( ) ( )( ) ( ) ( ) ( )1Agof a g f a g b f b a I a−= = = = =

⇒ Agof I=

( )⇐ Let there exists a function :g B A→ such that Bfog I= and Agof I=

We have to prove f is bijection and 1g f −=

Now Agof I= and it is one-one ⇒ f is one-one

and Bfog I= and it is onto ⇒ f is onto

⇒ f is a bijection from A B→

⇒ 1f − is also a bijection from B A→

We already have :g B A→

∴ 1f − and g are defined on same domain B

Let b B∈ , since :f A B→ is bijection, there exists a unique a A∈ such that ( )f a b=

i.e., ( )1f b a− =

Now ( ) ( ) ( ) ( )( ) ( )1Af b a I a gof a g f a g b− = = = = =

⇒ 1f g− =

F. Let :f A B→ , :g B C→ and :h C D→ . Then ( ) ( )ho gof hog of= , that is , composition of functions

is associative.

Sol. Given :f A B→ , :g B C→ , :h C D→ are functions

⇒ :gof A C→ , :h C D→ are functions

⇒ ( ) :ho gof A D→ is also a function

Further :f A B→ , :hog B D→ are functions

⇒ ( ) :hog of A D→ is also a function

∴ ( )ho gof and ( )hog of have the same domain A

Let ' 'a be any element of A then

( ) ( ) ( )( )ho gof a h gof a = ( )( )( )h g f a= ( ) ( )( )hog f a= ( ) ( )hog of a=

∴ ( ) ( )ho gof hog of=

19.

A. Use Mathematical induction to prove that 2 1 3 13.5 2n n+ ++ is divisible by 17.

Sol: Let the given statement is

( ) :S n 2 1 3 12.4 3n n+ ++ is divisible by 11

Step-1: We prove ( )1s is true

( ) 3 41 : 2.4 3s + is divisible by 11



: 128 81+ is divisible by 11

: 209 is divisible by 11

( )1s is true since 209 = 19 11×

Step-2: Let ( )s k is true

2 1 3 12.4 3k k+ ++ is divisible by 11

2 1 3 12.4 3 11k k m+ +∴ + = ( )1→

Step-3: We prove ( )1s k + is true

Now ( ) ( )2 1 1 3 1 12.4 3k k+ + + ++

2 1 2 3 1 32.4 .4 3 .3k k+ += +

2 1 3 116.2.4 27.3k k+ += +

( )3 1 3 116 11 3 27.3k km + += − + (by (1))

3 116 11 11.3 km += × +

( )3 111 16 3 11k M+= + =

( )1s k∴ + is true.

∴ By Mathematical Induction ( )s n is true n N∀ ∈

B. Use Mathematical Induction to prove that 3 3 3 3 3 31 1 2 1 2 3

.....1 1 3 1 3 5

+ + ++ + + ++ + +

upto n terms =

22 9 1324

nn n + +

Sol: Here nth term = ( )

3 3 3 31 2 3 ...

1 3 5 ... 2 1

n

n

+ + ++ + + + −

( )( )

22

2

2

1144

n nn

n

++

= =

The give statement is

( ) ( )23 3 3 3 3 3211 1 2 1 2 3

: ... 2 9 131 1 3 1 3 5 4 24

n ns n n n

++ + + + + + + = + + + + +

Step-1: We prove ( )1s true

( ) [ ]31 1

1 2 9 131 24

s = = + +

1 1=

( )1s∴ = true

Step-2: Let ( )s k is true



( )23 3 3

211 1 2.... 2 9 13

1 1 3 4 24

k kk k

++ ∴ + + + = + + +

( )1→

Step-3: We prove ( )1s k + true

By ( )1 , ( )23 3 3 3 3 3

211 1 2 1 2 3.... 2 9 13

1 1 3 1 3 5 4 24

k kk k

++ + + + + + + = + + + + +

Adding ( )22

4

k + both sides, we get

( ) ( ) ( )2 2 23 3 3

21 2 21 1 2.... 2 9 13

1 1 3 4 4 24 4

k k kkk k

+ + ++ + + + + = + + + +

( )3 2 22 9 13 6 4 4

24

k k k k k+ + + + +=

3 22 15 37 24

24

k k k+ + +=

( )( )21 2 13 24

24

k k k+ + +=

( ) ( )212 1 9 1 13

24

kk k

+ = + + + +

( )1s k⇒ + is true

∴ By Mathematical Induction, ( )s n is true for all n N∈

C. Use Mathematical Induction to prove the statement, ( )22

3 3 3 3 11 2 3 ... ,

4

n nn n N

++ + + + = ∀ ∈

D. Use Mathematical Induction to prove that statement, ( ) ( )( )2

1

2 1 2 12 1 ,

3

n

k

n n nk n N

=

− +− = ∀ ∈∑

E. Use Mathematical Induction to prove the statement, 22 3.2 4.2 ...+ + + + upto n terms = .2 ,nn

n N∀ ∈

F. Show that, n N∀ ∈ , 1 1 1

.....1.4 4.7 7.10

+ + + + upto n terms = 3 1

n

n +

G. Show that 49 16 1n n+ − is divisible by 64 for all positive integers n

H. Use Mathematical Induction to prove that 2 1 3 12.4 3n n+ ++ is divisible by 11, n N∀ ∈

I. Use Mathematical Induction, prove that statement, 2.3 3.4 4.5 .....+ + + + upto n terms =

( )2 6 11

3

n n n+ +

J. Use Mathematical Induction, prove that statement, 1.2.3 2.3.4 3.4.5 ... n+ + + + terms =

( )( )( )1 2 3

4

n n n n+ + +



K. Use Mathematical Induction, prove that statement, ( ) ( )2 2 2 2 2 31 1 2 1 2 3 .....+ + + + + + + upto n

terms = ( ) ( )21 2

12

n n n+ +

L. If x and y are natural numbers and x y≠ , using mathematical induction, show that n nx y− is

divisible by x y− , for all n N∈

20.

A. Show that ( )3

2

2 2

2

a b c a b

c b c a b a b c

c a c a b

+ +

+ + = + +

+ +

Sol. LHS =

2

2

2

a b c a b

c b c a b

c a c a b

+ +

+ +

+ +

1 1 2 3C C C C→ + +

2 2 2

2

2

a b c a b

c b c a b

c a c a b

+ +

= + +

+ +

( )

1

2 0 2

0 2

a b

a b c b c a b

a c a b

= + + + +

+ +

2 2 1 3 3 1;R R R R R R→ − → −

( )

1

2 0 2 0

0 0 2

a b

a b c b c a

c a b

= + + + +

+ +

Expanding along first column

( )( ) ( )2 3

2 2a b c a b c a b c= + + + + = + +

B. Show that ( )

2 2 2 2

22 2 2 3 3 3

2 2 2

2

2 3

2

bc a c ba b c

b c a c ac b a a b c abc

c a b b a ab c

−

= − = + + −

−

Sol.

2a b c a b c a b c

b c a b c a b c a

c a b c a b c a b

=

a b c

b c a

c a b

=

a b c

c a b

b c a

− − −

By interchanging 2 3&R R and multiplying 1R with – 1 in second determinant



2 2 2

2 2 2

2 2 2

a bc bc ab ab c ac b ac

ab c ab b ac ac bc bc a

ac ac b bc a bc c ab ab

− + + − + + − + +

= − + + − + + − + +

− + + − + + − + +

2 2 2

2 2 2

2 2 2

2

2

2

bc a c b

c ac b a

b a ab c

−

= −

−

Again

a b c

b c a

c a b

( ) ( ) ( )2 2 2a bc a b b ac c ab c= − − − + −

3 3 33abc a b c= − − −

∴ ( )

2

23 3 33

a b c

b c a abc a b c

c a b

= − − −

C. Show that ( )3

2 2

2 2

2 2

a b c a c

b b c a b a b c

c c c a b

− −

− − = + +

− −

Sol. LHS

2 2

2 2

2 2

a b c a c

b b c a b

c c c a b

− −

= − −

− −

1 1 2 3R R R R→ + +

2 2

2 2

a b c a b c a b c

b b c a b

c c c a b

+ + + + + +

= − −

− −

( )

1 1 1

2 2

2 2

a b c b b c a b

c c c a b

= + + − −

− −

1 1 3 2 2 3,C C C C C C→ − → −

( ) ( )

0 0 1

0 2a b c a b c b

a b c a b c c a b

= + + − + +

+ + + + − −

D. Show that ( )( )( )

2

2 4

2

a a b c a

a b b b c a b b c c a

c a c b c

− + +

+ − + = + + +

+ + −

Sol. Let , ,a b C b c A c a B+ = + = + =

Then 2 , 2 , 2a A B C b B C A c C A B− = − − − = − − − = − −



∴ LHS

2

2

2

a a b c a A B C C B

a b b b c C B A C A

c a c b c B A C A B

− + + − −

= + − + = − −

+ + − − −

1 1 2 2 2 3;R R R R R R→ + → +

A B B A B A

B C B C C B

B A C A B

− − +

= + − −

− −

1 1 2 2 2 3,C C C C C C→ + → +

0 2

2 0

B B A

B C B

B A C B C A B

+

= −

+ − − −

3 3 1C C C→ +

0 2

2 0

B B A

B B C

B A C B C

+

= +

+ −

Expanding along first row

( )( ) ( ) ( )( )2 2 2B BC B A B C B A B C B = − − + + + + −

[ ] ( )( ) ( )( )2 2 2 2B BC B B A B C B B A C B= − + + + + + −

2 24 4 4B C B C ABC= − + +

4 ABC=

( )( )( )4 b c c a a b= + + +

E. If 1 1 1

2 2 2

3 3 3

a b c

A a b c

a b c

=

is a son-singular matrix then A is invertible then 1

det

adj AA

A− =

Sol. By definition 1 2 3

1 2 3

1 2 3

A A A

Adj A B B B

C C C

=

Now ( )1 1 1 1 2 3

2 2 2 1 2 3

3 3 3 1 2 3

a b c A A A

A adj A a b c B B B

a b c C C C

=

1 1 1 1 1 1 1 2 1 2 1 2 1 3 1 3 1 3

2 1 2 1 2 1 2 2 2 2 2 2 2 3 2 3 2 3

3 1 3 1 3 1 3 2 3 2 3 2 3 3 3 3 3 3

a A b B c C a A b B c C a A b B c C



+ + + + + + = + + + + + + + + + + + +

0 0 1 0 0

0 0 0 1 0

0 0 10 0

A

A A A I

A

= = =



∴ ( ) ( )detA adj A A I=

Since A is non singular, det 0A≠

⇒ det

adj AA I

A

=

Similarily det

adj AA I

A

=

Let ,det

adjAB

A= then AB BA I= =

Hence A is invertible and 1

det

adjAB A

A−= =

21.

A. Apply the test of rank to examine whether the following equations are consistent, and if

consistent, find the complete solution. 2 3 8,x y z− + = 2 4x y z− + + = , 3 4 0x y z+ − =

Sol: The augmented matrix is

[ ]

2 1 3 8

1 2 1 4

3 1 4 0

AD

− ≈ − −

2 2 12 ,R R R→ + 3 3 12 3R R R→ −

2 1 3 8

0 3 5 16

0 5 17 24

− = − −

1 1 3 3 2 33 , 5 3R R R R R R→ + → −

6 0 14 40

0 3 5 16

0 0 76 152

=

311 3,

2 76

RRR R→ →

3 0 7 20

0 3 5 16

0 0 1 2

=

1 1 37 ,R R R→ − 2 2 35R R R→ −

3 0 0 6

0 3 0 6

0 0 1 2

=

1 0 0 2

0 1 0 2

0 0 1 2

=



Since ( ) [ ]( ) 3Rank A Rank A D= =

∴ The system is consistent and its solution is

2, 2, 2x y z= = =

B. Show that following system of equations is consistent and solve it completely

3,x y z+ + = 2 2 3,x y z+ − = 1x y z+ − =

Sol: [ ]

1 1 1 3

2 2 1 3

1 1 1 1

A D

≈ − −

2 2 12 ,R R R→ − 3 3 1R R R→ −

1 1 1 3

0 0 3 3

0 0 2 2

≈ − − − −

22 ,

3

RR →

− 3

32

RR →

−

1 1 1 3

0 0 1 1

0 0 1 1

≈

1 1 2 ,R R R→ − 3 3 2R R R→ −

1 1 0 2

0 0 1 1

0 0 0 0

+ ≈

Rank ( )A = Rank [ ]( ) 2A D =

∴ The system is consistent and has infinitely many solutions

Now, the system of equations equivalent to 2x y+ =+ and 1z=

Let ,x k= then 2y k=+ −

Hence [ ] [ ]2 1 ,x y z k k k R= + − ∈ is the solution set.

C. Solve the following simultaneous linear equations by using Cramer’s rule 3 4 5 18,x y z+ + =

2 8 13,x y z− + = 5 2 7 20x y z− + =

Sol: Let

3 4 5

2 1 8 ,

5 2 7

A

= − −

18

13

20

D

=

and

x

X y

z

=

Then the given system of equations can be written as AX D=

( ) ( ) ( )

3 4 5

det 2 1 8 3 7 16 4 14 40 5 4 5

5 2 7

A∆= = − = − + − − + − +

−

27 104 5 136= + + =



( ) ( ) ( )1

18 4 5

13 1 8 18 7 16 13 28 10 20 32 5

20 2 7

∆ = − = − + − + + +

−

162 494 740 408= − + =

( ) ( ) ( )2

3 18 5

2 13 8 3 91 160 18 14 40 5 40 65

5 20 7

∆ = = − − − + −

207 468 125 136=− + − =

3

3 4 18

2 1 13

5 2 20

∆ = −

−

= ( ) ( ) ( )3 20 26 4 40 65 18 4 5− + − − + − +

18 100 18 136= + + =

Hence by Cramer’s rule,

1 4083,

136x

x

∆= = =

2 1361,

136y

∆= = =

∆ 3 136

1136

z∆

= = =∆

D. Solve 3 4 5 18,x y z+ + = 2 8 13x y z− + = and 5 2 7 20x y z− + = by using Matrix Inversion

method.

Sol: Let

3 4 5

2 1 8

5 2 7

A

= − −

,

x

X y

z

=

, and

18

13

20

D

=

Then the given system of equations can be written as

AX D= ( )1→

Now ( ) ( ) ( )

3 4 5

det 2 1 8 3 7 16 4 14 40 5 4 5

5 2 7

A= − = − + − − + − +

−

27 104 5 136 0= + + = ≠

1A−⇒ exists

∴ by ( )1 , AX D= 1 1A AX A D− −⇒ =

1X A D−⇒ =

We have

9 38 37

26 4 14

1 26 11

adjA

− = − − −

1 1.

detX A D adjAD

A−∴ = =

9 38 37 181

26 4 14 13136

1 26 11 20

− − = − − −



4081

136136

136

=

3

1

1

22.

A. If ( ) ( )1, 2, 1 , 4,0, 3A B= − − = − , ( )1,2, 1C = − and ( )2, 4, 5D= − − , find the distance between AB

and CD.

Sol: The given points position vectors w.r.t to origin are

( ) 2 ,OA a i j k= − −��

( ) 4 0 3OB b i j k= − −��

,

( ) 2 ,OC c i j k= + −��

( ) 2 4 5OD d i j k= − −��

The shortest distance between the above lines is

AC AB CD

AB CD

=

×

��

��

Now ( )

0 4 0

3 2 2 4 12 2 40

1 6 4

AC AB CD = − =− − + = − −

��

( ) ( ) ( )3 2 2 8 12 12 2 18 2

1 6 4

i j k

AB CD i j k× = − = − − − − + + − −

− −

��

20 10 20i j k=− + −

( ) ( ) ( )2 2 220 10 20 10 4 1 4 30AB CD× = − + + − = + + =��

∴ Shortest distance =40 4

30 3=

B. Find the shortest distance between the skew lines

( ) ( )6 2 2 2 2r i j k t i j k= + + + − + and

( ) ( )4 3 2 2r i k s i j k= − − + − −

Sol: The line ( ) ( )6 2 2 2 2r i j k t i j k= + + + − + and ( ) ( )4 3 2 2r i k s i j k= − − + − − passes through the

point ( )6 2 2A i j k+ + and parallel to the vector 2 2b i j k= − +

The line ( ) ( )6 2 2 2 2r i j k t i j k= + + + − + and ( ) ( )4 3 2 2r i k s i j k= − − + − − passes through the

point ( )4C i k− − and parallel to the vector 3 2 2d i j k= − −

The shortest distance between the above lines is given by

.AC b d

S Db d

=

×

��



Now ( ) ( ) ( )

10 2 3

1 2 2 10 4 4 2 2 6 3 2 6

3 2 2

AC b d = − = + − − − + − + − −

��

80 16 12 108= + + =

1 2 2 8 8 4

3 2 2

i j k

b d i j k× = − = + +

− −

2 2 2 2 28 8 4 4 2 2 1 12b d× = + + = + + =

∴ Shortest distance = 108

912

=

C. If 2 3 ,a i j k= − − 2b i j k= + − and 3 2c i j k= + − . Prove that ( ) ( )a b c a b c× × ≠ × ×

Sol: Given 2 3 ,a i j k= − − 2b i j k= + − and 3 2c i j k= + −

( ) ( ) ( )2 1 1 2 3 4 1 6 1

1 3 2

i j k

b c i j k× = − = − + − − + + −

−

3 5i j k= + +

( ) ( ) ( ) ( )1 2 3 10 9 5 3 3 2

1 3 5

i j k

a b c i j k× × = − − = − + − + + +

8 5i j k=− − +

( ) ( ) ( ) ( )1 2 3 2 3 1 6 1 4

2 1 1

i j k

a b i j k× = − − = + − − + + +

−

5 5 5i j k= − +

( ) ( ) ( ) ( )5 5 5 10 15 10 5 15 5

1 3 2

i j k

a b c i j k× × = − = − − − − + +

−

5 15 20i j k=− + +

Clearly ( ) ( )a b c a b c× × ≠ × ×

23.

A. All example problems page. 283 – 288

Exercise 6F 2(ii) , 3(i, ii), 4(i,ii), 5(i,ii,iii), 10(i,ii)

24.

A. In a ,ABC∆ prove that 1 2 3 4r r r r R+ + − =

B. In a ,ABC∆ prove that 1 2 3 4 cosr r r r R C+ + − =



C. Prove that 3 2 2

1 2 3

1 1 1 1 1 1 4abc R

r r r r r r r s

− − − = =

∆

D. Show that ( )( )

1 3r

s b s c r=

− −∑

E. Prove that cos cos cosa A b B c C

bc a ca b ab c+ = + = +

F. If 1 2 3, ,P P P are altitudes of the ABC∆ then show that 2 2 21 2 3

1 1 1 cot cot cotA B C

P P P

+ ++ + =∆

G. Show that 2 2 2cos cos cos2 2 2

A B Ca b c s

R

∆+ + = +

H. Prove that ( ) ( ) ( )3 3 3cos cos cos 3a B C b C A c A B abc− + − + − =

I. Show that 2 2 2

2 2 2 2 21 2 3

1 1 1 1 a b c

r r r r

+ ++ + + =∆

J. Show that 31 2 1 1

2

rr r

bc ca ab r R+ + = −

K. In ,ABC∆ if 1 2 38, 12, 24r r r= = = , find , ,a b c

L. Show that 2 3 3 11 2

3 1 2

bc r r ca r rab r r

r r r

− −− = =

M. Show that cos cos cos 1r

A B CR

+ + = +

N. Prove that ( )2 2 2 2 2 2 2 21 2 3 16r r r r R a b c+ + + = − + +

O. If 13, 14, 15a b c= = = show that 1 265 21

, 4, , 128 2

R r r r= = = = and 3 14r =



` KUKATPALLY CENTREKUKATPALLY CENTREKUKATPALLY CENTREKUKATPALLY CENTRE

ipe 2015- i a final - fiitjee · pdf filefiitjee kukatpally centre: # 22-97, plot no.1, opp....

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