cbse-xii-2018 examination - kopykitabcbse-xii-2018 examination series sgn time : 3 hrs. mathematics...

CBSE-XII-2018 EXAMINATION

Series SGN

Time : 3 Hrs.

MATHEMATICSPaper & Solutions

SET-1

Code : 65/1 Max. M arks : 100

General Instruction :(i) All questions are compulsory.

(ii) The question paper consists of 29 questions divided into four section A, B, C and D. Sections A comprises of 4 questions of one m ark each, Section B comprises of 8 questions of two marks each, Section C comprises of 11 questions of four marks each and Section D comprises of 6 questions of six marks each.

(iii) All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.

(iv) There is no overall choice. However, internal choice has been provided in 3 questions of four marks each and 3 questions of six marks each. You have to attempt only one of the alternatives in all such questions.

(v) Use of calculators is not permitted. You may ask for logarithmic tables, if required.

SECTION - A

Question numbers 1 to 4 carry 1 mark each

1. Find the value of tan-1 (V3) - cot-1 (-V3)

Sol. tan-1 (V3) - cot * 1H /3 )

n 5n 2 n - 5n - 3n -3 6 = 6 = 6

"0 a - 32. If the matrix A = 2 0 -1

b 1 0

"0 a - 3"Sol. A = 2 0 -1

b 1 0

is skew symmetric, find the values of 'a' and 'b'.

For skew symmetric matrix

At = -A

"0 a - 3" T "0 - a 3"2 0 -1 = - 2 0 1b 1 0 - b -1 0

"0 2 b" "0 - a 3"a 0 1 = - 2 0 1

- 3 -1 0 - b -1 0

b = 3

a = -2

(On comparing LHS & RHS)

2

1 / 20


3. Find the magnitude of each of the two vectors a and b , having the same magnitude such that the angle9

between them is 60° and their scalar product is —.

Sol. Magnitude of two vectors a & b are same

a-b = cos 9

a2

cos 60°

2 9= — x 2 = 9

2

a = 3 = b

4. If a * b denotes the larger of 'a' and 'b' and if a o b = (a * b) + 3, then write the value of (5) o (10), where

* and o are binary operations.

Sol. a o b = (a * b ) + 3 5o 10 = (5 * 10 ) + 3

= 10 + 3 = 13

SECTION - BQuestion numbers 5 to 12 carry 2 marks each

2 ’ 2

Sol. 3 sin-1 x = sin-1 (3x - 4x’)Let siiL1 x = 0

5. Prove th a t: 3sin 1 x = sin 1 (3x - 4x’). x e

x = sin 0sin30 = 3sin0 - 4sin’0

sin30 = 3x - 4x’

Case I When —- < x < —2 2

—- < sin0 < —2 2

— < 0 < - 6 6---- < 30 < —

2 2

- 1 1Also — < x < — => -1 < 3x - 4x <1 2 2

sin30 = 3x - 4x’30 = s u f1 (3x - 4x3)3sin 'x = sin 1 (3x- 4x’)

2 / 2 0


6.

Sol.

Given A =2- 4

A =2 - 3

- 4 7

A 1 = - adj AIA | J

-37

, compute A 1 and show that 2A 1 =

114 -12

74

32

27 34 2

To Prove 2A 1 = 9I - A LHS = 2A-1

2 x 1"7 3" "7 3"

2 4 2 4 2

RHS = 9I - A

9 0 0 9

2 - 3- 4 7

= 7 3 = 4 2

LHS = RHS

7. Differentiate tan 1 | 1 + cosx j with respect to x. l sin x )

Sol. -1 f 1 + cosx A tan l 1= y (Let)sinx

y = tan2cos2

______2

A

x x 2sin cosl 2 2

= tan 1 c o t |

= tan 1 ntanl -

l 2x2

ny 2

x1

i = o - idx 2 2

9I - A.

3 / 20


8. The total cost C(x) associated with the production of x units of an item is given by C(x) = 0.005x3 - 0.02x2 + 30x + 5000.Find the marginal cost when 3 units are produced, where by marginal cost we mean the instantaneous rate of change of total cost at any level of output.

Sol. C(x) = 0.005x3 - 0.02x2 + 30x + 5000.

marginal cost (MC) = dC(x) dx

= (0.005)(3x2) - 0.02 (2x) + 30 When x = 3 MC = 0.005 (3 x 9) - 0.02(2 x 3) + 30

= 30.015

Sol.

Evaluate Jcos2x + 2sin2 x

cos2 xdx

cos2x + 2sin2 x cos2 x

2cos2 x -1 + 2(1 - cos2 x)cos2 x

J ~ “~~ " dxcos x

J dx

J ( 2 - d

J f 1 + c o s

2 + ----- 2------1 Idxcos x cos x

+ T~ 1 dxcos x

x + J sec2 xdx

x + tan x + C

9

10.

Sol.

Find the differential equation representing the family of curves y constants.

bx+5y = a eTake log on both sides log y = log (a ebx + 5) logy = loga + log(ebx + 5) Differentiate both sides w.r.t. x

= 0 + — (bx + 5)y dx dx

1dy = by dx

1 d2yy dx2 +

dydx

2

I = 0

a ebx+5, where a and b are arbitrary

d ydx2^ 2 + y

dydx

2I = 0

i.e. the required differential eqn.

4 / 20


11.

Sol.

If 0 is the angle between two vectors i - 2j + 31c and 3i - 2j + 1c, find sin 0.

a • b = |a ||b |cos0

a • b = (i - 2j + 3k) • (3i - 2j + k)= 3 + 4 + 3 = 10

10 = (v 1 + 4 + 9 9 + 4 +1 )cos0

cos0 = 10Vi^Vl4 10 14

sin0 = V1 - cos2 0

cos0 = 10 14

= (1 - :25 = 49 - 25 ' 49 = 49

V24 2yf67 7

12.

Sol.

A black and a red die are rolled together. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.Let F : no. of red die is less than 4.

E : sum of no is 8E = {(2, 6) (3, 5) (4, 4) (5, 6) (6, 2)}

- P(E) = 36

F = {(1, 1) (2, 1) (3, 1 ) ....... (6, 1)(1, 2) (2, 2 ) ................... (6, 2)(1, 3) (2, 3 ) ................... (6, 3)}

P(F) = 16

Also E n F = {(5, 3) (6, 2)} — P(E n F)

P(E/F) = P(E n F) = - ^ 6 P(F) 18/36

= 1 9

2_36

SECTION - CQuestion numbers 13 to 23 carry 4 marks each

13.1

Using properties of determinants, prove that 1 + 3y1

11

1 + 3z

1 + 3x 1 1

9 (3xyz + xy + yz + zx).

1 1 1 + 3xSol. 1 + 3y 1 1

1 1 + 3z 1

5 / 20


1 1 1— b 3

X X X1 1 1— + 3y y y

1 1 „ 1—+ 3

z z z

xyz

Ri — > Ri + R2 + R3

: xyz

--- 1-----1-----h 3 --- 1-----1---- h 3 —+ —+x y z x y z x y

1 " 1 1- + 3y y y

1 1 1—+3

z z z

= (xyz) 1 1 1 .--- 1-----1-----h 3x y z

1 1 11 1 1“ + 3 — —y y y

1 1 1—+3

z z zC2 -> C2 - Ci & C3 -> C3 - Ci

1 1 1 1 .= xyz | — 1----- 1--------h 3x y z

= (yz + zx + xy + 3xyz) 1 (0 + 9 ) = 9(3xyz + xy + yz +zx) = RHS

1 0 0

—+ 3 - 3 - 3y

13 0

z

14. If (x2 + y2)2 = xy, find .dx

OR

If x = a (29 - sin 29) and y = a (1 - cos 29), find — when 9 = —.dx 3

Sol. (x2 + y2)2 = xy

2(x2+ y 2)[2x + 2 y ^ ] = x ^ + y dx dx

4x(x=+y-) + 4 y ( x - + y = ) ^ - x | - + ydx dx

4x(x2+ y 2) - y = ^ [ x - 4 y ( x 2+ y 2)]dx

dy _ 4x(x2 + y2) - y dx x - 4 y (x 2 + y 2)

ORx = a(29 - sin29), y = a (l-co s2 9 )

— = a(2 - 2cos29), — = a(0 + 2sin29)d9 d9

6 / 2 0


dy _ dy/d0 dx dx / d0

_ 2asin20 2a(1 - cos20)

f dy _ sin2n/3

Vdx J ( 0 _ n /3 ) 1 - cos2n/3_ s in (n -n /3)

1 - c o s(n -n /3 )

_ sin n /3 V3/21 + cos n /3 1 +1/2

_ S _ j _ 3 V3

15. d2y dy 2If y _ sin (sin x), prove that 2 + tan x + y cos x _ 0.dx2 dx

Sol. y _ sin(sinx)

— _ cos(sinx) cosx. dx

d ydx2

d ydx2

cos(sinx)(-sinx) + cos2x[-sin(sinx)]

-sinx cos(sinx) - cos2x sin(sinx)

d2y dy 2LHS _ + tanx + y cos x

dx2 dx_ -sinx cos(sinx) - cos2x sin (sinx) + tanx cosx cos(sinx) + cos2x sinx (sinx)

_ -sinx cos(sinx) + sinx cosx cos(sinx) cos x

_ -sinx cos(sinx) + sinx cos(sinx)_ 0 _ RHS

16. Find the equations of the tangent and the normal, to the curve 16x2 + 9y2 _ 145 at the point (xi, yi), where xi _ 2 and yi > 0.

OR

x4 3 2Find the intervals in which the function f(x) _ - x - 5x + 24x + 12 is4

(a) Strictly increasing, (b) Strictly decreasing.Sol. 16x2 + 9y2 _ 145 .... (1)

since (x1, y1) lies on (1)16x2 + 9y2 _ 145

16(2)2 + 9y2 _ 145 ^ y1 _ 3(2, 3 )

16x2 + 9y2 _ 145 Diff. w.r. to x

7 / 20


32x + 18y •d y = 0 dx

dy - 16xdx 9y

dydx ' ( 2 , 3 )

Eqn of tangent(y - 3) = m(x - 2)

- 32(y - 3) = — (x - 2)

27y - 81 = -32x + 64 32x + 27y = 145

-1 6 2 - 32x =9 3 27

-1Eq of normal (y - yi) = (x - xi)

m

27(y - 3) = — (x - 2)

32y - 96 = 27x - 54 27x - 32y + 42 = 0

OR

f(x) = — - x3 - 5x2 + 24x + 12 4

4x3f '(x) = — - 3x2 - 10x + 24.

4f '(x) = 0x3 - 3x2 - 10x + 24 = 0 (x - 2)(x2 - x - 12) = 0 (x - 2)(x - 4)(x + 3) = 0

- x> <- +

-3

increasing in interval (-3, 2) u (4, <»)

decreasing in interval (-<», -3) u (2, 4)

2a>

4

17. An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of material will be least when depth of the tank is half of its width. If the cost is to be borne by nearby settled lower income families, for whom water will be provided, what kind of value is hidden in this question?

Sol. Let the length, width & height of the open tank be x, x & y units Volume = x2 y Total surface area = x2 + 4xy

8 / 20


S - x* 2+4x | A

« = 2x - 1 X . „dx x“

2x" = 4V 2x3 = 4x2y x = 2y

d2Sdx2

= 2 +8V

= 2 + ^ . 2 + v > 08y' y ’

Hence S is minimum when x = 2y ie the depth (height) of the tank is half of the width

18.

Sol.

Find2cosx

( l-s in x )( l + sin2 x)

2cosx

dx

■* ( l-s in x )( l + sin“x)

put sinx = t cos x dx = dt r 2dt J ( l - t ) ( l + t2)

2 A

dx

Bt + C( l - t ) ( l + t2) 1 - t 1 + t2

2 = A(1 + 12) + (Bt + C)(l - 1)put 1 - 1 = 0

t= 12 = A(2)A = 1

Comparing coefficients of t2 & t t2 -> A + (-B) = 0

B = A B = 1

t —> B - C = 0 B = C = 1

1 t + 1 . , + - ---- I dt

1 - t t2 + l

lo g (l- t)-1

f - r -— dt + f -,1 dt J t2 +1 J t2 +1

1 o-log (1 - sinx) + — log (t“ + 1) + tan 11 + C

1 o-log (1 - sinx) + — log (sin“x + 1) + tan (sinx) + C

9 / 2 0


x x 2 7119. Find the particular solution of the differential equation ex tan y dx + (2 - ex) sec“ y dy = 0, given that y = —

when x = 0.OR

dv %Find the particular solution of the differential equation — + 2y tan x = sin x, given that y = 0 when x = — .dx 3

Sol. ex tany dx + (2 - ex) sec2y dy = 0 ex tany dx = (ex - 2) sec2y dydy ex tanydx e sec y -2 s e c y

dx e sec y -2 s e c ydy ex tany

dx _ 2 o 2sec y 2 sec y^_:dy tany tany

dx _ sec2 y | dy tany

I - ' \l> | 1 clxJ tany J l -2 e ~ xtany = t sec2y dy = dt

r * = p L ^ , xJ t J ex - 2

ex - 2 = u ex dx = du

logt = log u + log C log(tany) = log(ex -2)C tany = C(ex - 2)

p u ty = —, x = 0

C = - ltan y = -(ex -2)

dy— + (2tan x)y = sinx dx

IF = eJ2tanx_ 21og sec xe= sec2x.

2 f* 2y.sec“x = sec“ x-sinxdx + C

tan — = C(1 - 2)

Ismx

cos2 Xdx + C

OR

= tanxsecxdx + C

y sec“x = secx + C

1 0 / 20


20.—^ / \ /V /V —^ /V /V /V ---->

Let a = 4i + 5j - k , b = i - 4j + 5k and c/V /V /V ---->

3i + j - k . Find a vector d which is perpendicular to both

c and b and d . a = 21.

Sol. a = 4i + 5 j - k

b = i — 4 j + 5k

c = 3i + j - k

To find vector d , such that

d -c = 0

d-b = 0

& d-a = 21

Let d = xi + yj + zk

(xi + yj + zk) • (3i + j - k) = 0 3x + y - z = 0

(xi + yj + zk) • (i - 4j + 5k) = 0 x - 4y + 5z = 0

(xi + yj + zk)-(4i + 5j - k ) = 214x + 5y - z = 21

eqn (1) x 4 + eqn (2)12x + 4y - 4z = 0 x - 4y + 5z = 0 13x + z = 0eqn (2) x 5 + eqn (3) x 4 5x - 20y + 25z = 0 16x + 20y - 4z = 84 21x + 21z = 84 21(x + z) = 84 x + z = 4 eq. (4) - (5)12x = -4

= ~ 4 ~1 12 “ 3

z = 4 - x„ 1 13z = 4 + — = —

3 3Put x & z in (1)3x + y - z = 0 . 13 _3 x m + y " T ' 0

316

y = Td = xi + yj + zk-r - b 16“ 13“ d = — 1+ — 1+— k

3 3 3

( 1)

(2)

•(3)

• •(4)

(5)

1 1 / 20


21.

Sol.

22.

Sol.

23.

Sol.

Find the shortest distance between the lines

r = (4i - j) + X (i + 2j - 31c) and r

r = (4i - .j) + X(i + 2j - 3k)

Sj = 4j - j bj = j + 2j - 31c

r = (i - j + 2k) + p(2i + 4j - 51.)

S2 = j - .j + 2k , b2 = 2i + 4j - 51.

(a2 - !ij) • (bJ X b2 )S.D. =

bJ x b2 =

lbJ x b2 I

i .j 1.1 2 - 32 4 - 5

(j - j + 21c) + p (2i + 4j - 5k).

S.D.

= i(-10 +12) - j (-5 + 6) + 1c(4 - 4)

= 2 j - j

= (-3 j + 0j + 2k) • (2j - j )4 +1

- 6 = 6

V5 % /5

Suppose a girl throws a die. If she gets 1 or 2, she tosses a coin three times and notes the number of tails. If she gets 3, 4, 5 or 6, she tosses a coin once and notes whether a 'head' or 'tail' is obtained. If she obtained exactly one 'tail', what is the probability that she threw 3, 4, 5 or 6 with the die?Let E1 be the event that the girl. Gets 1 or 2 on the roll

P(E1)2 = 16 = 3

Let E2 be the event that the girl gets 3, 4, 5 or 6 on the roll P(E2)4 = 26 = I

Let A be event that she obtained exactly one tailsIf she tossed a coin 3 times & exactly 1 tail shows then [HTH, HHT, THH} = 3

P(A/E1) = 3/8P(A/E2) = 1/2 (If she tossed a coin only once & exactly 1 shows)

P(E2/A) = P(E2)P (A /E 2)P(E1)P (A /E 1) + P(E2)P (A /E 2)

1 2x2 3

1 2 3 1x + x 2 3 8 3

_8_11

Two numbers are selected at random (without replacement) from the first five positive integers. Let X denote the larger of the two numbers obtained. Find the mean and variance of X.The first five positive integers are 1, 2, 3, 4, 5 we select two positive numbers in 5 x 4 = 20 ways Out of these two no. are selected at random & let X denote larger of the two no.X can be 2, 3, 4 or 5

12 / 20


P(X = 2) = P(larger no. is 2) = {(1, 2) and (2, 1)}= _2 _

304

P(X = 3) =30

P(X = 4) = —308

P(X = 5) =30

2 4 6 8Mean = E(X) = 2 x + 3 x + 4 x + 5 x30 30 30 30

= 4 +12 + 24 + 40= 30

= 80 30

2 2 2 4 2 6 2 8Variance = 22 x + 32 x + 42 x + 52 x30 30 30 30

= 8 + 36 + 96 + 200= 30= 340 _34

30 = 3

SECTION - DQuestion numbers 24 to 29 carry 6 marks each

24. Let A = {x e Z : 0 < x < 12}. Show thatR = {(a, b) : a, b e A, |a - b| is divisible by 4} is an equivalence relation.Find the set of all elements related to 1. Also write the equivalence class [2].

ORx

Show that the function f : R ^ R defined by f(x) = — , V x e R is neither one-one nor onto. Also,x 2 + 1

if g : R ^ R is defined as g(x) = 2x - 1, find fog(x).Sol. R = {(a, b) : a, b e A, |a - b| is divisible by 4}

Reflexivity : for any a e A|a - a| = 0, which is divisible by 4 (a, a) e R.So, R is reflexive.

Symmetry : Let (a, b) e R |a - b| is divisible by 4 ^ |b - a| is also divisible by 4 So R is symmetry

Transitive : Let (a, b) e R & (b, c) e R|a - b| is divisible by 4|a - b | = 4Xa - b = ± 4X ....... (1)|b - c| is divisible by 4 |b - c| = 4qb - c = ±4q ...... (2)

Add (1) & (2)

13 / 20


a - b + b - c = ±4(L + p) a - c = ±4(L + p)^ (a, c) e R

So, TransitiveHence, R is reflexive, Symmetry & Transitive so it is an equivalence relation Let x be an element of A such that (x, 1) e R, then

|x - 1| is divisible by 4 x - 1 = 0, 4, 8, 12.^ x = 1, 5, 9

Hence, the set of all element of A which are related to 1 in {1, 5, 9}OR

f(x) = -T —x +1for one-one f(x) = f(y)

x yx2 +1 _ y2 +1

xy2 + x = yx2 +yxy(y - x) = y - xxy = 1

1x = — y

x ^ ySo not one-one

for onto f(x) = yx

x2 +1= y

2x = yx + y x2y + y - x = 0x cannot be express in y so not onto

As g(x) = 2x - 1

fog(x) = f [g(x)] = f(2x - 1)2x - 1

(2x - 1)2 +12x - 1

4x2 - 4x + 2

25. If A =2 - 3 53 2 - 41 1 - 2

, find A 1. Use it to solve the system of equations

2x - 3y + 5z = 11 3x + 2y - 4z = - 5 x + y - 2z = - 3.

ORUsing elementary row transformations, find the inverse of the matrix

' 1 2 3A = 2 5 7

- 2 - 4 - 5

14 / 20


2 - 3 5Sol. |A| = 3 2 - 4

1 1 - 22(-4 + 4) +3(-6 + 4) +5(3 2)

|A| = 0 - 6 + 5 = -1 * 0 Now An = 0; A12 = 2; A13 = 1A21 = -1; A22 = - 9; A23 = -5 A31 = 2; A32 = 23; A33 = 13

0 2 1-1 - 9 - 52 23 13

A

' 0 2 1 "T "0 -1 2"Adj A = [Aij]T = -1 - 9 - 5 = 2 - 9 23

2 23 13 1 - 5 13

"0 -1 2 " 0 1 - 2= AdjA 1 2 - 9 23 = - 2 9 - 23

1A 11 - 5 13 -1 5 -13

"2 - 3 5 " x " 11"Now 3 2 - 4 y = - 5

1 1 - 2 z - 3

^ A X = B"0 1 - 2" " 11"

^ X = A-1 B = - 2 9 - 23 - 5-1 5 -13 - 3

x " 0 - 5 + 6 " "1"

y - 22 - 45 + 69 = 2z -11 - 25 + 39 3

x = L; y = 2 and z = 3

A = IA (Inverse of matrix)"1 2 3 " "1 0 0"

2 5 7 = 0 1 0- 2 - 4 - 5 0 0 1

R2 — R2 - 2R1 R3 —— R3 + 2R]"1 2 3" "1 0 0"0 1 1 = - 2 1 00 0 1 2 0 1

R1 —— R1 - 3R3R2 — R2 R3"1 2 0" " - 5 0 -3"0 1 0 = - 4 1 -10 0 1 2 0 1

OR

15 / 20


Ri ^ Ri — 2R2"1 0 0" "3 - 2 -1"0 1 0 = - 4 1 -10 0 1 2 0 1

"3 - 2 -1A-1 = - 4 1 -1

2 0 1

A

26.

Sol.

Using integration, find the area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x2 + y2 = 32.x2 + y2 = 32

x2 = 16x = ±4 M(4, 4)Required Area = area of shaded region = area of OMA = area OMP + area MPA

4 4 2= J y A + J y2dx

0 44 4 2

= J xdx + J (4 2 )2 - x 2 dx0 4

( x2 A4

+ N2 2i y0rV(W2) 2 2 , - x + (W 2)2 .sin ■1I x

= 16 (4 /2= 2 + l

2 14V2A

2V (W 2)2 - ( W 2 ) 2 + -3 2 sin-11

2 y

4V2

4

= 8 + (2 ^2 (0 ) + 16 x - ) - (2 x 4 + 16 x - )

= 8 + 8n - 8 - 4n = 4n

16 / 20


27.

Sol.

n / 4 .r sinx + cosXd J 16 + 9sin2x

3

Evaluate J (x2 + 3x + ex ) dx,i

as the limit of the sum.

OR

n / 4sin x + cos x

dx0

n / 4

0

n / 4

16 + 9sin2x

sin x + cos x 16 + 9[1 - (sin x - cos x )2 ]

sinx + cosx

dx

0 25 - 9 (sin x -co sx )2

sin x - cos x = t (cos x + sin x) dx = dt

f dx

dx

25 - 9t2

J1

1 9| — - t 2 9

dt

1 J —

9 J f ’ Y - t ’

2 x 11log

5---+ t3___

3- - t

- 1

1 1X9 10

31

f 5 5 2

log 35 - log I

8V 3 ) 3

30log 1 - log—

1 1 log 4 = log 2

30 15

j

J ( x 2 + 3x + ex) dx

0

1

OR

= Limh[f(1) + f(1 + h) + f(1 + 2h) + ... + f(1 + 2(n - 1)h)]h ^ 0

= Lim h[(1 + 3 + e) + ((1 + h)2 + 3(1 + h) + e1+h) + ((1 + 2h)2 + 3(1 + 2h) + e1 + 2h) + . ]h ^ 0

= Limh[4 + e + (1 + h2 + 2h + 3 + 3h + e1+h) + (1 + 4h2 + 4h + 3 + 6h + e1 + 2h) +...]h ^ 0

17 / 20


= Limh[4 + e + (4 + h2 + 5h + e1+h) + (4 + 4h2 + 10h + e1 + 2h) +...]h — 0

= Lim h[4n + e(1 + eh + e2h + ...) + h2[12 + 22 + ...] + 5h[1 + 2 + )]h — 0

= Lim h 4n + eh — 0

= Lim hh — 0

4 n + e

1(ehn-1 )

eh -1

enh-1

eh -1

D 1 + ^n (n -1 )(2 n -1 )

1 + 5hf °j ^ 6 J l

h2n3 L 1 Y . 1 1 5hn2 (+ 1 - II 2 - | + |

6 l n J l n j 2 l 1 - o j

= Lim

( nx2 12 4n+ e e n -1 4

” 1 (1- 1 1( 2 - 1 15 2 2 f+ x n2 x V 1 1

n e2/n-1l J

n2 6 i nJl n j 2 n V n j j

= Limn ——w n

4n+ e( e2 - 1 1 +i ! f1- 11 (2 - 11 + 4 > - 1 11e2/n- 1 l e 1J 6 f n Jl n j l nJJ

= Lim 2n — w

4 + + 2 f1_ 1 'If 2 - 1 1 + 5 f i- 1n e2/n -1 3 1 n

2 4= 8 + e(e -1) + — + 5

= U l i l i i + e(e2 - 1)

= y + e(e2 - 1)

2

n n

28. Find the distance of the point (- 1, - 5, - 10) from the point of intersection of the line

r = 2i - j + 2k + X (3i + 4j + 2k) and the plane r . (i - j + k) = 5.

Sol. Equation of line is

r = ( 2i - j + 2 k ) + X( 3i + 4j + 2 k )

Equation of plane is r -(i - j + k ) = 5

Now combined equation

[ ( 2 i - j + 2k) + X(3j + 4j + 2 k )](j - j + k ) = 5

[(2 + 3X)i + (- 1 + 4X) j + (2 + 2X)k ]-(i - j + k ) = 5

2 + 3X - (- 1 + 4X) + 2 + 2X = 5 X + 5 = 5 X = 0So, equation of line is

r = (2i - j + 2k) + 0( 3 i + 4j + 2k)

r = 2i - j +2k

18 / 20


Let the point of intersection be (x, y, z)

So, r = xi + yj + 2k

x = 2, y = -1, z = 2

point of intersection is (2, - 1, 2)Distance b/w (2, - 1, 2) & (- 1, - 5, - 10)

= 7(-1 - 2)2 + (-5 + 1)2 + (-10 - 2)2

= V9 +16 +144 = 13

29. A factory manufactures two types of screws A and B, each type requiring the use of two machines, an automatic and a hand-operated. It takes 4 minutes on the automatic and 6 minutes on the hand-operated machine to manufacture a packet of screws 'A'. While it takes 6 minutes on the automatic and 3 minutes on the hand-operated machine to manufacture a packet to screws 'B'. Each machine is available for at most 4 hours on any day. The manufacturer can sell a packet of screws 'A' at a profit of 70 paise and screws 'B' at a profit of ' 1. Assuming that he can sell all the screws he manufactures, how many packets of each type should the factory owner produce in a day in order to maximize his profit? Formulate the above LPP and solve it graphically and find the maximum profit.

Sol. Let the number of package of screw A = x Number of packages of screw B = y

Item Number Machine A Machine B ProfitScrew A x 4 minutes 6 minutes To paise = .7 RsScrew B y 6 minutes 3 minutes Rs. 1Max time Available

4 hours = 240 min

4 hours = 240 minutes

Automated Machine Works for screw A ^ 4 min

Works on screw B ^ 6 min

Hand operated machine

Works on screw A ^ 6 min

Works on screw B ^ 3 min

4x + 6y < 240 2x + 3y < 120

x, y > 0Now max Z = 0.7 x + y s.t. 2x + 3y < 120

2x + y < 80 x, y > 0

2x + 3y < 120 2x + y < 80x 0 60 x 0 40

y 40 0 y 80 0

6x + 3y < 240 2x + y < 80

x, y > 0

19 / 20


y

(0, 80) 80

70

60

50( 0, 40) 4

30

20

10

10 20 30 4^ 50 6 ^ 70(40, 0) (60, 0)

Corner Points Value of Z(0,40) 40

(30,20) 41(40,0) 28

Hence, profit will be maximum, if the company produces, 30 packages of screw A 20 packages of screw B Maximum Profit = Rs 41

80 *x

20 / 20


S e r ie s G B M

Time : 3 Hrs._______General Instruction :

MATHEMATICSP a p e r & S o lu t io n s

SET-1

C ode: 65/1 Max. Marks : 100

(i) All questions are compulsory.(ii) The question paper consists of 29 questions divided into four section A, B, C and D. Sections A comprises of 4

questions of one mark each, Section B comprises of 8 questions of two marks each, Section C comprises of 11 questions of four marks each and Section D comprises of 6 questions of six marks each.

(iii) All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.

(iv) There is no overall choice. However, internal choice has been provided in 3 questions of four marks each and 3 questions of six marks each. You have to attempt only one of the alternatives in all such questions.

(v) Use of calculators is not permitted. You may ask for logarithmic tables, if required.

SECTION - A

Question numbers 1 to 4 carry 1 mark each

1. If for any 2 x 2 square matrix A, A(adj A) =8 0 0 8

, then write the value of |A|.

A(adj A) =0 8

by using propertyA(adj A) = A In

o0©

=> A In =0 8

=> A In = 8"1 0" 0 1

|A| = 8

2. Determine the value of 'k' for which the following function is continuous at x = 3 :

f(x)(x + 3)2 -3 6

x - 3 k

,x?^3

, x = 3

Sol. lim f(x) = lim ---- —x->3 x —>3 X — 3

_ j. (x + 3 - 6)(x + 3 + 6) x—>3 (x - 3)

= 12given that f(x) is continuous at x = 3 .'. lim f(x) = f(3)

x —>3

=> k = 12

1 / 1 9


3.. rsin2x - c o s 2x ,

Find : -------------------dxJ sin x cos x

Sol.

4.

Sol.

' sin2 x - cos2 xdx

J S i l l X • COS X

, 2 |^ c o s 2 x dx J sin ?.x

= - 2 J" cot 2x dx

_ -2 log | sin 2x |2

= - log | sin 2x| + C

C

Find the distance between the planes 2x - y + 2z = 5 and 5x - 2-5y + 5z = 20.

2 x - y + 2z = 5 •••(!)5x - 2-5 y + 5z = 20or 2x - y + 2z = 8 ... (2)Distance between plane (1) & (2)

= 1di - d 2 3

Va2 + b 2 + c 2 £

SECTION - B

Question numbers 5 to 12 carry 2 marks each

5. If A is a skew-symmetric matrix of order 3, then prove that det A = 0.

0 a bSol. Let A = - a 0 c b e a skew symmetric matrix of order 3

- b - c 0

0 a b - a 0 c - b - c 0

|A| =

|A| = - a (0 + be) + b(ac - 0) = - abc + abc = 0 Proved

6. Find the value of c in Rolle's theorem for the function f(x) = x ’ - 3x in \-y[3 , 0],

Sol. f(x) = x3 - 3x

(i) f(x) being a polynomial is continuous on [- y/3 , 0]

(h) f(->/3 ) = f (0) = 0

(iii) f (x) = 3x2 - 3 and this exist uniquely on [-^ 3 , 0]

f(x) is derivable on (--^3 , 0) f(x) satisfies all condition of Rolle's theorem

There exist atleast one c e (-V3 , 0) where f (c) = 0 => 3c2 - 3 = 0 => c = ± 1 => c = — 1

2 / 1 9


7. The volume of a cube is increasing at the rate of 9 cnrVs. How fast is its surface area increasing when the length of an edge is 10 cm ?

Sol. Assumed volume of cube = V

Given that, = 9 cm Vscc dt

dA = ? dt/ = 10 cm

dV dt

d/ dt

i - (/)3 = 9dt

3/2 — = 9 dt

.(1)

dAdt

_ddt

Now — = — (6 /“) = 12 / — = 12 / x" " dt /-

36 36 _ 2,= — = — = 3.6 cn r/ sec

1 10

(form (1))

8. Show that the function f(x) = x ’ - 3x2 + 6x - 100 is increasing on R.

Sol. f(x) = x3 - 3x2 + 6x - 100

f (x) = 3x2 - 6x + 6

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

f(x) = 3 [ ( x - l ) 2 + l]

f (x) > 0 for all x e R So, f(x) is increasing on R.

9. The x-coordinate of a point on the line joining the points P(2, 2, 1) and Q(5, 1, -2) is 4. Find its z-coordinate.

Sol.

( 2 . 2 . 1)

Let R divides PQ in the ratio k : 1

I"5k + 2 k + 2 - 2k + l 'j{ k + 1 ’ k + 1 ’ k + 1 J

given x co-ordinate of R = 45 k + 2 „

.-. -------- = 4k + 1

k = 2

z co-ordinate-2(2)+ 1

2 + 11

Q(5. 1. -2)

3 / 1 9


10.

Sol.

11.

Sol.

A die, whose faces are marked 1, 2, 3 in red and 4, 5, 6 in green, is tossed. Let A be the event "number obtained is even" and B be the event "number obtained is red". Find if A and B are independent events.

A = {2, 4,6}

B = {1, 2, 3}

A n B = {2}

™ - W

p(B) 4 4

P(A n B) = - 6

Here, P(A) P(B) = I x I = I

Since, P ( A n B ) # P(A) P(B), so events A and B are not independent events.

Two tailors, A and B, earn t 300 and f 400 per day respectively. A can stitch 6 shirts and 4 pairs of trousers while B can stitch 10 shirts and 4 pairs of trousers per day. To find how many days should each of them work and if it is desired to produce at least 60 shirts and 32 pairs of trousers at a minimum labour cost, formulate this as an LPP.

Tailor A Tailor B Minimum Total No.No. of shirts 6 10 60No. of trousers 4 4 32Wage Rs 300/day Rs 400/dayLet tailor A and tailor B works for x days and y days respectively

x > 0, y > 0minimum number of shirts = 60

6 x + 10y> 60 3x + 5y > 30

minimum no. of trousers = 324x + 4y > 32

=> x + y > 8Let z be the total labour cost

z = 300x + 400yThe given L.P. Problem reduces to : z = 300 x + 400 yx > 0, y > 0, 3x + 5y > 30 and x + y > 8

12. Find: 1

dx5 - 8x - :

Sol. Jdx

5 - 8 x - x “

-/■

I

dx« x + 4)2 -21}

dx(V2I)2 - ( x + 4)2

2V2Ilog

V 2l + (x + 4) V 2 l- (x + 4)

C

4 / 1 9


SECTION - CQuestion numbers 13 to 23 carry 4 marks each

13. If tan 1 ——— + tan 1 X + = —, then find the value of x.x - 4 x + 4

Sol. tan

x - 3 x + 3+ -x - 4 x + 4

1 -f x2 - 9 A vx2 - 16y

K~4

(x + 4)(x - 3) + (x + 3)(x - 4) (x2 - 1 6 ) - ( x 2 -9 )

2x2 - 24 = - 7 2x2 = - 7 + 24

17

= 1

2X =

X = ± if14. Using properties of detenninants, prove that

a“ + 2a 2a + 1 1 2a + 1 a + 2 1

1= ( a - i r

Find matrix A such thatOR

Sol.

' 2 - U f - 1 -8 "1 0 A = 1 - 2

- 3 4 v , 9 22 y

Use Ri — Ri - R2; R2 — F-2~~ R3; R3 — R3 L.H.S.

a2 -1 a -1 0 2a - 2 a -1 0

3 3 1

(a - l) (a + l) (a — 1) 0 2(a — 1) ( a -1) 0

3 3 1Taking common (a — 1 )2

(a +1) 1 0= (a - l)2 2 1 0

3 3 1= (a - l)2 [(a + 1) (1 - 0) - 1 (2 - 0)]= (a - l)2 [(a + 1) - 2]= (a - l)3 = R.H.S.

5 / 1 9


OR

A =

Let matrix A is a b c d

2 -11 0

- 3 4

-1 -8"a b _ 1 - 2c d

9 22

2a - c 2b - d " -1 -8"a b = 1 - 2

3a + 4c -3 b +4d 9 22Comparing both the sides

2a - c = - 1,2b - d = - 8 And a = 1 , b = - 2

After solving we get c = 3, d = - 4

“1 - 2 3 - 4

So, A =

15.

Sol.

If xy + yx = a , then find — .dx

If ey(x + 1) = 1, then show that

We have xy + yx = a .

ORd2y _ fd y " 2dx2 dx.

Differentiating w.r.t. x, we get — (xy) + — (yx) = 0.dx dx

Let u = xy log u = y log x

u dx

_d_dx

Let v = y

or (xy) = xy 1 ^

dx

i dy- + logx—x dx

dx

. . . d )

1 du 1 , , dy du ( y , dy= > ------- =y. — + lo g x .— ; => — = u —+ logx—dx

...(2)

log v = x log y

1 dv 1 dy= > -------= x .---------+ log y . 1

v dx y dxdvdx

= vX dy ,—— + logyy dx

(or — (y ) = y —— + logy

dxx dyy dx

Using (2) and (3) in (1),

,y I y , , _ d ywe get xy | — + log x —x dx

x dy+ logy

y dx= 0 .

(xy log x + xyx 1) — = - (yx log y + yxy 1) or — dx dx

...(3)

...(4)

yx logy + yxy~' xy logx + xyx~*

6 / 1 9


Let ey (x + 1) = 1

ey(l) + (x + 1) ey =0dx

dy

d ydx2

dydx

(x + 1)

OR

=> (x + 1 ) - i - + 1 = 0 dx

Again differentiating w.r.t. x

••• (x + 1 ) + M . 1 = 0dx2 l dx J

...(1)

d j_ = dy dy [equation (i)]dx

dVdx2

dx dx

16. F ind :

Sol

cos 9

• J

(4 + sin2 9)(5 - 4cos2 9)

cos 9

d9

(4 + sin“ 9)(5 - 4cos“ 9)d9

cos 9 d9

f -J ( o f

(4 + sin2 9)(5 - 4(1 - sin2 9))

cos 9 d9(sin“ 9 + 4)(4sin“ 9 + 1)

Put sin9 = t Cos9 d9 = dt

••• 1= f-------J (4 + t 2)(l + 4 t2)

Consider1 At + B Ct + D

(4 + t 2)(l + 4 t2) _ 4 + t 2 1 + 4 t2

1 = (At + B) (1 + 4t2) + (Ct + D)(4 + 12)= At + B + 4At3 + 4Bt2 + 4Ct + Ct3 + 4D + Dt2 = (4A + C)t3 + (4B + D)t2 + (A + 4C)t + (B + 4D)

4A + C = 0 ^ C = - 4 A 4B + D = 0 ^ D = - 4 B A + 4C = 0 ^ A = - 4 C B + 4D = 1

1 4By solving we get A = 0, B = ----- , C = 0, D = —

7 / 1 9


1 - 1/15.+ . /15

(4 + t2)(l + 4 t2) 4 + t 2 1 + 4t2

1 r 1.-.1= - i r i 4 i r i— ------7<it + — x — - —15 J 4 + t 2 15 4 J 1

dt- + t"

1 1= ---------x — tail

15 2

( \( 0 1 1 -i t— -I---- x — tanU J 15 1 1 /

v /2 )+ C

= — — tan 1+ — tan 1(2t)+C30 U J 15

= — tan-1 (2sin0)-^~ tan-1 ( S*n 1 + C

17.

Sol.

Evaluate71

lxtanx

secx + tanx

OR

-dx

4Evaluate : j " { |x - l | + | x - 2 | + | x - 4 |}dx

1 =l

71 xtanx o secx +tanx

dx . . . d )

I= r Q t-x X -tan x ) Jo

■-J,o - se c x - ta n x

11 (7 1-x )ta n x ,----------------dx

o secx + tanx

Adding (1) & (2)

...(2)

21 =

Jo secx + tanxdx

f 11 7itanx : + tanI.ti/2 tanx

2I = 2ti fJo-dx

o sec x + tanx

.' j" f (x)dx = 2j" f (x)dx wheneverf (2a - x) = f (x)

dxI = 71l

71/2 tanx:CX + tc

in x(se(0 sec2 x - tan2 x

o sec x +tanx

71/2 tanx(secx - tanx)PT* - tan a. — tan a. ; 1I = 71 ------------------ ------ dx

Jo - - 2 - — 2

<•71/2 0I = 7i (sec xtan x - tan“ x) dx

Jo

fTl/2 0(sec x tan x - sec“ x +1) dx

Jo= 71

8 / 1 9


I = n [sec x - tail x + x] n/ 20

71= n lim (sec x - tan x) h------sec 0

n~ 2x —>—

,. 1 - sin x n2= n lu n ------------1------- n

ti“ cosx 2X —>—

2

= n lim 1 - sin“ xx > 'j cosx(l + sinx) 2

- 71

Let f (x) = | x - 11 +| x - 2 | + |x - 4|We have three critical points x = 1, 2, 4(i) when x < 1(ii) when 1 < x < 2(iii) when 2 < x < 4(iv) when x > 4

OR

f(x) - (x - 1) - (x - 2) - (x - 4) if x < 1(X — 1) — (X — 2) — (x — 4) if 1 < x < 2(x - 1) + (x - 2) - (x - 4) if 2 < x < 4( x - 1) + (x - 2) + (x - 4) if x > 4

.\f(x) = -3x + 7 = -x + 5= x + l = 3x -7

if x < 1if 1 < x < 2if 2 < x < 4if x > 4

4

1= Jf(x)dx 12 4

" I = J"f(x)d x + j f (x)dx 1 2

2 4

I = j" (- x + 5)dx + j" (x + l)dx1 2

9 / 1 9


18. Solve the differential equation (tan 1 x -y )d x = (1 + x2) dy.

Sol. We havedy tan 1x - ydx 1 + x 2

dy + y

X71

dx 1 + x 2 1 + x 2f ^ d x

I.F = eJ 1+x" = etan x

_tany.e" = J

tan 1 +

" f x e ^ d xx 2

Put t = tan 1X

1 +2

X

- J < e* d t

I II

= t . e ‘- J

l . e ‘ d t

; = t e ‘ - e * + c

= ( t a n ■’ x - - 1 )tan- 1 3

e

= t a n 1' x - 1 + s t a n c e

19. Show that the points A, B, C with position vectors 2 i - j + k , i - 3 j - 5 k and 3 i - 4j - 4k respectively, are the vertices of a right-angled triangle, Hence find the area of the triangle.

Sol. AB = - i - 2j - 6k

BC = 2i - j + k

CA = l - 3j - 5k

BC-CA = 0

BC 1 CAAABC is a right angled triangle

A = i |B C | |A C |

A = — V4 + 1 + 1 Vl + 9 + 25 2

= - V6 a/352

= - V2102

20. Find the value of X, if four points with position vectors 3i + 6j + 9k, i + 2 j + 3k, 2i + 3 j + k and 4i + 6j + /.k, are coplanar.

Sol. We haveP.V. of A = 3i + 6j + 9k

P.V. of B = i + 2j + 3k

1 0 / 1 9


AB = -2 i - 4 j - 6k

AC = —i — 3 j — 8k

AD = i + (A - 9)k

Now __„ __ „A B .(A C xA D )^

-2-11

- 4- 30

-6- 8

-9 )

= 0

=> -2 (-3A, + 27) + 4(-L + 9 +8) -6(0 + 3) = 0

=> 6A, - 54 - 4A, + 68 - 18 = 0

2A, - 4 = 0

L = 2

v AB, AC, AD are coplanar and so the points A, B, C and D are coplanar.

21. There are 4 cards numbered 1, 3, 5 and 7, one number on one card. Two cards are drawn at random without

replacement. Let X denote the sum of the numbers on the two drawn cards. Find the mean and variance of X.

Sol. X denote sum of the numbers so, X can be 4, 6, 8, 10, 12

X Number on card P(x) X P(x) X2 P(x)

4 (1,3) 1 1 . 14 3 6

2/3 8/3

6 (1,6) 1 x 1 x 2 - 14 3 6

1 6

8 (3, 5) or (1,7) 1 1 . 1 1 . 14 3 4 3 3

8/3 64/3

10 (3,7) 1 1 . 14 3 6

5/3 50/3

12 (5,7) 1 1 . 14 3 6

2 24

Mean = 2 X P(x) = 8

717 70Variance = I X2 P(x) - ( I X P(x))2 = — - 64 = —

22. Of the students in a school, it is known that 30% have 100% attendance and 70% students are irregular.

Previous year results report that 70% of all students who have 100% attendance attain A grade and 10%

irregular students attain A grade in their animal examination. At the end of the year, one student is chosen at

random from the school and he was found to have an A grade. What is the probability that the student has

100% attendance? Is regularity required only in school? Justify your answer.

1 1 / 1 9


Sol. Let E, be students having 100% attendance E2 be students having irregular attendance E be students having A grade

P(Ei) =30

P(E2)70

Too ” Too

r e i 70 30= 21%p

FvEw 100 100

f E AFVn 2 y

10 70100 100

By Baye's theorem,

= 7%

So, PI ^ | =P ( E j ) P

( E ^FVn i 7

f E TPCEJ P — + P (E 2) P

vh i )( _j l "VE2 y

30 21100X100

30 21 70 7Tm xTm + T(K)xToo

6363 + 49

63112

23. Maximize Z = x + 2y Subject to the constraints

x + 2y > 100 2x - y < 0 2 x + y <200 x ,y > 0

Solve the above LPP graphically.Sol. x + 2y = 100

2x - y = 0 .......(1)2x + y = 200 .......(2)x = 0 ,y = 0 .......(3)

Comer points are A (100, 0), B(50, 100), C(20, 40)

minimum

maximum minimum

Maximum at point B and maximum value 250

Comer points Z = x + 2yA(100, 0) 100B(50, 100) 250C(20, 40) 100

1 2 / 1 9


SECTION - DQuestion numbers 24 to 29 carry 6 marks each

II__ " 1 - 1 1 '- 7 1 3 1 - 2 - 25 - 3 -1 2 1 3

24. Determine the product

x - y + z = 4, x - 2y - 2z = 9, 2x + y +3z = 1.

Sol. Product of the matrices

and use it to solve the system of equations

11__ " 1 - 1 1 '- 7 1 3 1 - 2 - 25 - 3 - 1 2 1 3

- 4 +4 +8 - 7 +1 +6

5 - 3 - 2

8 0 0

0 8 0

0 0 8

"1 - 1 i r

4 - 8 +4 7 - 2 +3

- 5 +6 - 1

= 81,

- 4 - 8 +12 - 7 - 2 +9

5 +6 - 3

Hence 1 - 2 - 2 2 1 3

- 4- 75

41

- 3 -1

Now, given system of equations can be written in matrix form, as follows

i -1 1 ' X "4"i - 2 - 2 y = 92 1 3 z 1

X "1 - 1 1 ' -i "4"

y = 1 - 2 --2 = 9z 2 1 3 1

X1

y = —8

z _X

1y = —

8z

' - 4 4 4 "4"- 7 1 3 = 95 - 3 -1 1

-16 + 36 + 4"-2 8 + 9 + 320 -2 7 -1

24

24-16- 8

-16x = — , y =

8x = 3,

8y = -2,

-8 z = —

8z = -1

1 3 / 1 9


f 4 ] (A ] 4 x + 325. Consider f : R < — > —> R - < — > given by f (x) = —----- . Show that f is bijective. Find the inverse of f

[ 3J [3 J 3 x + 4and hence find f '(()) and x such that f '(x) = 2.

ORLet A = Q x Q and let * be a binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for (a, b), (c, d) e A. Determine, whether * is commutative and associative.Then, with respect to * on A(i) Find the identity element in A.(ii) Find the invertible elements of A.

Sol. f ( x ) = — , x e R - f - - 3x + 4 1 3

f is one - one ->

Let xi, x2 e R - j--^ -j and f (xi) = f (x2)

4 x j + 3 _ 4x2 + 3 3x j + 4 3x2 + 4

=> 12xi x2 + 16 xi + 9 x2 + 12 = 12xi x2 + 9 x\ + 16 x2 + 12 => 7x] = 7x2 =} X! = x2

f is one - one f is onto ->

Let k e R - j - ^ - 1 be any number

f(x) = k4x + 3

=> x —

3x + 4 4x + 3 = 3kx + 4k

4k - 3

Also

4 -3 k 4 k - 34 -3 k 3

implies - 9 = - 16 (which is impossible)

f (— —- 1 = k i.e. f is ontoU - 3kJ

The function f is invertible i.e. f exist inverse of f Let r 1 (x) = k

f (k) = x 4 k + 33 k + 4

, 4x - 3 k =

••• f ' ( x ) =

4 - 3x 4x + 34 - 3 x

3

x s R - -

r ( 0 )= -

and whenr 1 (x) = 2

4 x - 3 ^=> -------- = 2

4 - 3 x4x - 3 = 8 - 6x

=> lOx =11 11

=> x —10

1 4 / 1 9


OR(i) Let (e, f) be the identify element for *

for (a, b) e Q x Q, we have(a, b) * (e, f) = (a, b) = (e, f) * (a, b)

=> (ae, af + b) = (a, b) = (ea, eb + f)=> ae = a, af + b = b, a = ea, b = eb + f => e = l , a f= 0 , e = l, b = ( l)b + f

( v a need not be '0')=> e = l , f = 0, e = l , f = 0

(e, f) = (1, 0) e Q x Q (1, 0) is the identify element of A

(ii) Let (a, b) e Q x Q Let (c, d) e Q x Q such that(a, b) * (c, d) = (1, 0) = (c, d) * (a, b)=> (ae, ad + b) = (1, 0) = (ea, eb + d)=> ae = 1, ad + b = 0, ea = 1, eb + d = 0

=> c = —, d = ——, ( —1 b + d = 0 (a ^ 0) a a ^ a )

(c,d) = (a ^ ° )

for a ^ 0, (a, b)_1 = ( —, —- l a a

26. Show that die surface area of a closed cuboid with square base and given volume is minimum, when it is a cube.

Sol. If each side of square base is x and height is h then volumeT VV = x-h => h = A r

S is surface area then

S = 4hx + 2x2 = 4 1 X r I x ■

4VS = ------r 2x“

■ 2x“

+ 4x and

Diff w. r. to x dS _ _ 4 Vdx x2

Now — = 0 4x = dx

1/3 d“Sat x = V , — - > 0dx2

d2Sdx24V

■ = +-8V + 4

S is minimum when x = V1/3

and h =V V

2 7-2/3 = V1/3

x- V"• x = h means it is a cube

: = h

1 5 / 1 9


27. Using the method of integration, find the area of the triangle ABC, coordinates of whose vertices A (4, 1), B (6, 6) and C (8, 4).

ORFind the area enclosed between the parabola 4y = 3x2 and the straight line 3x - 2y + 12 = 0.

Sol.

5x=> y = -------92

Equation of BC is, 4 - 6 , ^=> y - 6 = ------ (x - 6)

8 - 6=} y = - x + 12 Equation of AC is

=> y - 1 = (x - 4)8 - 4

=> 4y - 4 = 3x - 123x^ y= T ~2

Area of AABC = area ABED + area BEFC - area ADFC

' Ilf" 91dx +1’(_ x+12) dx" /It _ 21dx-A 2

r 5x2 Y-9 x

f 2- X Y+ 12x

4 y 4

r 3x 2 Y- 2x = 7 sq units

OR

Parabola 4y = 3x2 •••(1)line 3x - 2y + 12 = 0 •••(2)

from (2) y = 3 x ^ 12

putting this value of y is (1) we get6x + 24 = 3x2 => x = 4, - 2when x = 4 then y = 12

x = - 2 then y = 3 Required area

= J4 (y of line) dx - (y of parabola) dx

are

1 6 / 1 9


rf 3x + 12 3x

J-2 ^ 2 4

§ (8 + 2x - x 24 -2

3 3Q | 2 X8x + x ------

4 3

2 Adx

27 sq. imits

28.

Sol.

Find the particular solution of the differential equation (x - y) dydx

dy dxx + 2y

(x -y ) = (x + 2y)dx

dy dx x - y

Let y = Vx

L - V + x ^dx

V + x

V + x

dVdxdV

dxX + 2(Vx)

x - V x 1 + 2V

dx 1 -VdVdx

1 + 2 V -V + V-

1 -V1 + V + y -

1 -V

-dV =rdx

J X

(2 V +1) - 3 1 + V + V 2

• 2V + 1 1 + V + V2

v =•dx

d V -3 dV1 + V + V2_

dV

2 f v + i ) 2 +l 2 J l2J

r dx J x

— -=log |x | + C

j !°g |1 + V + V2| + | -j= tan 1V + -____ 2

V32

= log |x| + C

(x + 2y), given that y = 0 when x = 1.

1 7 / 1 9


“ i 108

xr xr2v

2—+ 11 _i_ A _j_ A + a/3 tan 1 X

X X sv y

= log |x| + C

we have y = 0 when x = 1

=> 0 + V3 tan-1 = 0 + c

C = VJ tan -1 _ L

Solution

- - ! o g2

v2 — + 1

1+ y + y + a/3 tan 1 X

X X

v y

= log |x| + V3 tan 11

1 *

29.

Sol.

Find the coordinates of the point where the line through the points (3, -4, -5) and (2, -3, 1), crosses the plane determined by the points (1, 2, 3), (4, 2, -3) and (0, 4, 3).

ORA variable plane which remains at a constant distance 3p from the origin cuts the coordinate axes at A, B, C.

Show that the locus of the centroid of triangle ABC is — ^ .2 2 2 2 x y z p

. . . d )

Equation of line passing through ( 3 , - 4 , - 5 ) and (2 ,-3 ,1 ) x - 3 _ y + 4 _ z + 5~ i 6Equation of plane passing through(1,2, 3) (4, 2 , - 3 ) and (0, 4,3) x — 1 y - 2 z - 3

3 0 - 6 = 0- 1 2 0

(x - 1) (12) - (y - 2) (- 6) + (z - 3) (6) = 0 =0 2x + y + z - 7 = 0 • • • (2)Let any point on line (1)

is P (- k + 3, k - 4, 6k - 5) it lies on plane.-. 2 ( -k + 3 ) + k - 4 + 6 k - 5 - 7 = 0

5k = 10 => k = 2

••• P (1 ,-2 , 7)

1 8 / 1 9


ORLet the equation of planex y z- + - + - = 1 ...(a b c

It cut the co-ordinate axes at A, B and C .'.A (a, 0,0), B (0, b, 0), C (0, 0, c)Let the centroid of AABC be (x, y, z)

given that distance of plane (1) from origin is 3p

...(2)

11 1 1 - 3 P— H---— H---—

_L _L J_ = J_2 i 2 2 a 2a b c 9p

from (2)1 _ 1

9z2 9p2

Proved

1 9 / 1 9


Time : 3 Hrs.

MATHEMATICSPaper & Solutions Code : 65/2/N

Max. M arks : 100General Instruction :(i) All questions are compulsory.(ii) Please check that this question paper contains 26 questions.(iii) Questions 1-6 in Section-A are very short-answer type questions carrying 1 mark each.(iv) Questions 7-19 in Section-B are long answer I type questions carrying 4 marks each.(v) Questions 20-26 in Section-C are long answer II type questions carrying 6 marks each.(vi) Please write down the serial number of the question before attempting it.

SECTION - A

Question numbers 1 to 6 carry 1 m ark each

—— —— —— ——

If a = 4i - j + k and b = 2i - 2j + lc, then find a unit vector parallel to the vector a + b— —

a + b = 4j - j + k + 2i - 2j + k

= 6j - 3j + 2k

(6i - 3j + 2k) (6j - 3j + 2k)

1.

Sol.

2.

Sol.

3.

Sol.

unit vector parallel to a + b = -V36 + 9 + 4 7

Find X and p if

(i + 3j + 9k) x (3i - Xj + pk) = 0 .——

(i + 3j + 91c) x (3i - Xj + pic) = 0 .

i j k^ 1 3 9 = 0

3 -X p

^ i(3p + 9X) - j ( p - 27) + k ( - X - 9) = 0j + oj + 0k ^ 3p + 9X = 0 ^ p —27 = 0 ^ |p = 27(i) to 3x 27 + 9X = 0

^ X = -9

Write the sum of intercepts cut off by the plane r (2i + j - 1c) - 5 = 0 on the three axes.

Plane r (2i + j - k) - 5 = 0

^ (xi + yj + zk) • (2i + j - k) - 5 = 0 ^ 2x + y - z = 5

x y z -+ + = 1(5/2) (5) (-5)

sum of intercepts = 5 + 5 -5 =

1 / 20


Sol.

4.

5.Sol.

6.

Sol.

For what values of k, the system of linear equationsx + y + z = 2 2x + y - z = 3 3x + 2y + kz = 4 has a unique solution ? x + y + z = 2 (i)2x + y - z = 3 (ii)3x + 2y + kz = 4 (iii)The system of linear equation has unique solution then

1 1 12 1 -1 * 03 2 k

^ 1 (k + 2) -1 (2k +3) +1 (4 - 3) * 0 ^ k + 2 - 2 k - 3 + 1 * 0 ^ k * 0So values of k = R - {0}

If A is a 3 x 3 matrix and |3A| = k|A|, then write the value of k. A is a matrix of 3 x3 say

A =

3B =

a11 a 12 a 13

a 21 a 22 a 23

a31 a 32 a33

3a11 3a 12

3a21 3a 22

3a31 3a 32

3 a 13

3 a 23

3 a 33

|3A| = 3 x 3 x 3 |A| = 27 |A|Which is given as k |A| so k = 27

If A =f cos a sin a ^v- sin a cos a j

, find a satisfying 0 < a < n when A + AT = V2I2; where AT is transpose of A.

A =f cos a sin a

r,a t =

cos a - sin a

^ A + A t =r,cos a + cos a

00 f 2

2cos acos a 0 ^

2cos a

v a + a t = V2"i 2

f 2cosa 0 \ ( J l 0 ^So =

0 2c o sa j I 0 2

by comparing : - 2 cos a = 4 21

cos a =4 2

n I nso a = el 0,

4 I 2

v sin cos j

0

2 / 20

7.

Sol.


SECTION - BQuestion numbers 7 to 19 carry 4 marks each

A bag X contains 4 white balls and 2 black balls, while another bag Y contains 3 white balls and 3 black balls. Two balls are drawn (without replacement) at random from one of the bags and were found to be one white and one black. Find the probability that the balls were drawn from bag Y.

ORA and B throw a pair of dice alternately, till one of them gets a total of 10 and wins the game. Find their respective probabilities of winning, if A starts first.

X Y4 White 2 Black 3White 3Black

Event Ei ^ Bag X is selected Event E2 ^ Bag Y is selected Event A ^ 1 white 1 black is taken out

1 1P(E1) = - , P (E ) = - , P (A /E 1) =

• 2C15C,

3C • 3CP(A / E2) = - 4 ^

P(A) = P(E1) P(A/E1) + P(E2) P(A/E2)

1 4C1 ■ 2C1 + 1 3- ■ 3C12 6C 2 6C

probability balls are drawn from bag 1 : P(E2/A) =

1 3c1' 3C1

P(E2PPA ;/E2) (Using Baye's Theorem)

2 6,C2 9_ = ______= _9_1 . V 2- . 1 V 3- = 4-2 + 9 = 17- + —

2 62 6C2

total of 10 :

A starts first1 11 11 1 11 11 11 11 1P(A wins) = + • •■ + • • • • +.

12 12 12 12 12 12 12 12 12

C 2

(6, 4) (4, 6) (5, 5)

P = — = ±36 12•/ x x V

A A B A

OR

x x x x VA B A B A

12 f 1+f 11r +(11 j 4 + ■ :(

2

P(A wins)

121 - i 11

i 12 j j

12

1223

1 14412 (144 -121)

P(B wins) = 1 - P(A wins) = 1 -1223

11 23 .

3 / 20


Sol.

8.

9.

Sol.

Find the coordinates of the foot of perpendicular drawn from the point A(-1, 8, 4) to the line joining the points B(0, -1, 3) and C(2, -3, -1). Hence find the image of the point A in the line BC.Equation of BC x - 0 y +1 z - 32 - 0 - 3 +1 -1 - 3

x y +1 z - 3 ^ = =

2 - 2 - 4

x y +1 z - 3 ^ = = = X (say)

1 -1 - 2

^ D(X, - X - 1, -2X + 3)DR's of AD^ X + 1, - X - 9, - 2X + 3 - 4

^ X + 1, - X - 9, - 2X - 1

v AD 1 BC

^ (X + 1) (1) + (-X - 9) (-1) + (-2X - 1) (-2) = 0

^ X + 1 + X + 9 + 4X + 2 = 0

^ 6X + 12 = 0 ^ X = - 2 foot of perpendicular to D(-2, 1, 7) let image of A w.r.to line Bc is E(x1, y1, z1)

f x, -1 y, + 8 z, + 4 'jmid point of AE l 1 , 1 , 1 I = (-2, 1, 7)

f A ( - 1 , 8, 4)

x1 - 1 = - 4 (x1 = - 3)

2 2

y1 + 8 = 2 (y1 = - 6)

2

z1 + 4= 14

(z1 = 10)E (-3, - 6, 10) Ans.Show that the four points A(4, 5, 1), B(0, -1 , -1), C(3, 9, 4) and D(-4, 4, 4) are coplanar.

AB = (0 - 4)i + (-1 - 5)j + (-1 - 1)k

^ AB = -4 i - 6j - 2)2

^ AC = (3 - 4)i + (9 - 5)j + (4 - 1)1

= - i + 4j + 3k

^ AD = (-4 - 4)i + (4 - 5)j + (4 - 1)k

= - 8i - j + 3k

We know that three vectors a,b and c are coplanar if [a b c] = 0

- 4 - 6 - 2^ [A BA C AD] = - 1 4 3 = - 4 (12 + 3) + 6 (-3 + 24) - 2 (1 + 32)

- 8 -1 3

= - 60 + 126 - 166 = 0

A,B,C,D are coplanar.

4 / 20


10.

Sol.

Find the particular solution of the differential equation2y ex/y dx + (y - 2x ex/y)dy = 0given that x = 0 when y = 1.2yex/ydx + (y - 2xex/y)dy = 0

^ 2ex/y = -(1 - x ex / y dy

v y j-dx

xput — = t

yx = yt

dy dt1 = t + y

dx dx

S L = i f 1 - y iLdx t v dx

^ 2e' ■ - (1 - 2,el) Tf1 - j

^ 2e' - - (1 - 2te,) 7 f1 - l | -

^ 2tet = (2tet - 1) ( 1 - y — | = 2tet - 1 - y2tet— + V dx j dx dx

^ 1 = y(1 - 2tet) — = x (1 - 2tet)— dx t dx

I IT = IV 1 - * >^ Inx = Int - 2et + c

^ Inx = 1n (—j - 2ex/y + c

^ Iny = - 2ex/y + c At x = 0, y = 1 ^ 0 = - 2e0 + c ^ c = 2

Iny = 2 - 2ex/y

11.

Sol.

dyFind the particular solution of differential equation:

dxGiven differential equation dy = x + ycosx dx 1 + sin x

dy cos x -x^ + y =

dx 1 + sin x 1 + sin x

I| |dxIntegrating factor = e v 1+sinx j

= e i°g (1 + sinx) = 1 + sin x

x + ycosx 1 + sinx

given that y = 1 when x = 0.

5 / 20


• (1 + sin x) = J - x(1 + sin x)

(1 + sin x) dx + c

^ y • (1 + sin x) = - — + c

At x = 0 , y = 11- (1 + 0) = c ^ c = 1

So y (1 + sin x) = - — +1

y

2

2

12.

Sol.

Find : J (x + 3)^3 - 4x - x2dx.

J (x + 3)V3 - 4x - x2dx

d 2Set ^ x + 3 = p • (3 - 4x - x2) + qdx

^ x + 3 = p • (-4 - 2x) + q

so - 2p = 1 ^ p = - 2

- 4p + q = 3 ^ q = 1 so given integral

j { - "2 (-4 - 2x) + 1 V3 - 4x - x 2 dx

^ J- -2 ( -4 - 2x)V3 - 4x - x 2dx + J ' l 3 - 4x - x2 dx

2 tMake perfect sq.

put 3 - 4x - x2 = t ^ (- 4 - 2x)dx = dt

- 1 J-y/tdt + Jyj 7 - (x + 2)2dx

^ - H ^ 2 + a/7 - (x + 2)2 + 2 sin-x + 2

T +c

^ - 3 ( 3 - 4x - x 2)3/2 + f x +2 y 3 - 4x - 2 7 . - x + sin ( x + 2 '

V 7 ,2

13. Find: J(2x - 5)e2x (2x - 3)3

dx

OR

Find: Jx 2 + x +1

(x2 + 1) (x + 2)dx

+c

6 / 20


Sol. (2x - 5)e2x (2x - 3)3

dx put 2x = t

1 (t - 5)etr j

dt2 1 (t - 3)3

t - 3(t - 3)3I ' J et (

2 J e 'f 7T1(t - 3)2

2

(t - 3)3

2

(t - 3)3

Adt

Adt

yf(t) f '(t)

Q J ex(f(x) + f'(x))dx

= ex f(x) +c

1 _t 1so e2 (t - 3)2

- + c

1 2x 1-e2 --------- - + c2 (2x - 3)2

OR+ x +1

(x2 + 1) (x + 2)dx

' x 2 + x + 1 (x2 + 1)(x + 2) dx =J

CAx + B(x2 + 1) (x + 2)

dx

^ x2 + x + 1 = (Ax + B) (x + 2) + C(x2 + 1)^ x2 + x + 1 = Ax2 + 2Ax + Bx + 2B + Cx2 + C

x

A + C = 1 ....(i)2A + B = 1 ....(ii)2B + C = 1 _(iii)by equation (i), (ii) and (iii) we getC = 3/5

B = 6 -1 ^ 5

B = 1 5

3 2 A = 1 - - = -

5 5

••• J

Adx1 (2x +1) 3 1

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

1 J ^ dx + - f ^ dx + - J 15 x2 +1 5 x 2 +1 5 x +5 x + 2

dx

1 1 3= 5 *log(x2 + 1) + -5-tan-1 x + 5 .log | x + 2 1 +c

7 / 20


Sol.

14. Find the equation of tangents to the curve y = x3 + 2x - 4, which are perpendicular to line x + 14y + 3 = 0.

x + 14y + 3 = 0

m = -14

slope of perpendicular line = 14.

curve y = x3 + 2x - 4

dydx

| = 3x2 + 2 = 14iX.yO

3x2 = 12

x? = 4

xi = ± 2

x1 = 2

x1 = -2

tangent at (2, 8)

and at (-2, -16)

y 1 = 8 + 4 - 4 = 8

y 1 = - 8 - 4 - 4 = - 16

y - 8 = 14(x - 2) = 14x- 28,

y + 16 = 14(x + 2)

y + 16 = 14x + 28,

15. If x cos(a + y) = cosy then prove thatdydx

cos2(a + y) sin a

d2yHence show that sina + sin2(a + y) = 0 .dx2

dydx

6x - W T -Find — if y = sin 1 dx

Sol. xcos(a + y) = cosy

cosyx =cos(a + y)

Differentiate with respect to 'x'

4x2

1 =

dycos(a + y)| - sin y — j - cos y|^- sin(a + y) —

cos2(a + y)

OR

dy

5

point (2, 8)

point (-2, -16)

y = 14x - 20

y = 14x + 12

(sin(a + y)cos y - cos(a + y) sin y)1 =

dydx

cos2(a + y)

8 / 20

16.

Sol.


cos2(a + y) = sin(a + y - y) —dx

dy _ cos2 (a + y) dx sin a

and

dy 1 + cos2(a + y) sina _

dx 2

Again, Differentiate with respect to 'x'

sina d2y _ 0 _ sin2(a + y) 2 dy dx2 2 dx

d2y dysina 2 + sin2(a + y) _ 0dx dx

y _ sin i 6x _ W l _ 4x2 5

= sin 3 2x _ - V l _ 4x2 5 5

= sin 2xi _ ( i )

• - 1 -1 y = sin 2x - sin45

-dy _ 1 2 _ 0 _ 2dx 1 _ (2x)2 1 _ 4x2

OR

2 2 xEvaluate : I dx .

_2 1 + 5x

2 2 2 tx . + (_x)

2 2i _ r- ^ d x _ r

_21 + 5x 0 ^ 1 + 5x 1 + 5_x jdx

z

I2 x A 2

x x 5 ' dx _ r x0 v 1 + 5x 5x +1 j

2 (1 + 5x) 5x +1

dx

3

I x ! f a _ ^ _ 8 _ 0 = 83 3

2

0

9 / 20


17. If f(x) =

sin(a + 1)x + 2 sinxx2

V1 + bx -1

, x < 0

j x = 0

, x > 0

Sol.

is continuous at x = 0, then find the values of a and b.

" sin(a + 1)x + 2 sinx

f(x) =x2

V i+bx -1

, x < 0

j x = 0

, x > 0

value at x = 0 is 2

LHL : lim f(0 - h)h 0

= iim - sin(a + 1)h - 2 sinhh ^ 0 - h

= lim I sin(a + l)h + ^ 1 IQ i ,m iin i = 1h ^ 0 e^0 0

= (a + 1) + 2 = a + 3

RHL : lim f(0 +h)h ^ 0

- lim - ‘) + 1limh ^ 0 h ( 1 + bh + 1) 2

= (rationalize)

x

h h

since it is continuous

LHL = RHL = f(0)

a + 3 = - = 2 2

.'. a = -1, b = 4

18. A typist charges 7 145 for typing 10 English and 3 Hindi pages, while charges for typing 3 English and

10 Hindi pages are ? 180. Using matrices, find the charges of typing one English and one Hindi page

separately. However typist charged only 7 2 per page from a poor student Shyam for 5 Hindi pages. How

much less was charged from this poor boy ? Which values are reflected in this problem ?

10 / 20


Sol. Charges of typing one English page = Rs x Charges of typing one Hindi page = Rs y 10x + 3y = 145 3x + 10y = 180"10 3 “ x "145“

3 10 _ y_ 180

Ax = BA-1 Ax = A-1B

Ix = -adjAB |A| = 100 - 9 = 91|A | ' '

1 " 10 - 3“ "145“91 - 3 10 180

x“ 1 " 1450 - 540 “ " 910 “

y_ = 91 - 435 +1800 1365

1015

x = 10 y = 15Poor student he charged = 2* 5 = Rs 10 actual cost = 15*5 = Rs 75 less charged = 75 - 10 = Rs 65 ?This problem reflect on human values "Kindness" etc.

19. Solve for x : tan 1 (x - 1) + tan 1 x + tan '(x + 1) = tan '3x

(Prove that : tan

6x - 8x 3

v 1 - 12x 2 J- tan

OR4x _i . , „ , 1

1 _ 4x2J 3

Sol. tan 1(x-1) + tan 1 (x +1) = tan :3x - tan :x

^ tanx _ 1 + x +1

1 - (x - 1)(x + 1)tan

3x - x 1 + 3x2

2x 2x1 - (x2 -1) 1 + 3x2

^ 2x(1+ 3x2) = 2x (2-x2) ^ 2x[(1 + 3x2) - 2 + x2 ] = 0 ^ x(4x2 - 1) = 0

^ x = 0 x 2 = 1

^ x = ±2

^ x = 0, + —, - 1 2 2

4

11 / 20


tan 3(2x) - (2x) 1 - 3(2x)2

3 (- tan

let 2x = tan0 |2x| <

(tan

33tan 0 - tan3 0- tan

2(2x) v1 - (2x)2

-11 2 tan 0

A

1 - 3tan2 0

^ tan-1(tan30) - tan-1(tan20)

^ 30 - 20 = 0 = tan-1 2x H.P.

1 - tan2 0

OR

SECTION - C

Question number 20 to 26 carry 6 mark each

20.

Sol.

Using the method of integration, find the area of the triangular region whose vertices are (2, -2), (4, 3)

and (1, 2).

equation of line AB :

y - 2 - ( H ) (x -1)

y - 2 - 1 (x - 1)

^ x = 3y - 5

equation of line AC :

y - 2 -- 2 - 2

2 -1(x - 1)

^ y - 2 = -4x + 4

x - 6 - y 4

(a) (line x 1)

(b) (line x2)

12 / 20


21.

Sol.

equation of line BC :

'3 + 2y + 2 = ,

' 4 - 22y + 4 = 5x - 10

2y +14

(x - 2)

x =5

Area of AABC

(c) (line x3)

J (x3 - x 2)dy + J(x3 - x i )dy- 2

^ 2 , i i l l i - ^ ]dy + J f i n i i - ( 3 y -5 ) Idy

13^ sq. units

2

Using properties of determinants, prove that

(x + y)2 zx zyzx (z + y)2 xy zy xy (z + x)2

= 2xyz(x + y + z)3

OR

If A =( 1 0 2

0 2 1

v2 0 3I

and A3 - 6A2 + 7A + kI3 = O find k.

(x + y)2 zx zyzx (z + y)2 xy zy xy (z + x)2

R1 ^ zR1, R2 ^ xR2, R3, ^ yR3 we get

1xyz

2 2z xz(x + y)2 2 2

2z yzx x(z + y) x 2yzy2 xy2 y(z + x)2

Taking z, x, y common from, C1, C2, C3 respectively, we get

xyzxyz

(x + y)2 2 2z zx 2 (z + y) x2 2 x

2 2(z + x)

C1 ^ C1- C3 , C2 ^ C2 - C3

(x + y)2 - z20 2 2 2 (z + y) - x x

y 2 - (z + x)2 y 2 - (z + x)2 (z + x)2

y y

20 z

13 / 20


(x + y + z)(x + y - z) 0 2Z

=> 0 (z + y - x) (z + y + x) 2X

(y - z - x)(y + z + x) (y - z - x) (y + z + x) (z + x )2

x + y - z 0 2Z

=> (x + y + z)2 0 z + y - x X2

y - z - x y - z - x (z + x )2 |

R 3 —> R3 — R2 —Ri

x + y - z 0 z 2

=0 (x + y + z)2 0 z + y - x x 2

- 2 x - 2z 2zx

C i-> Ci + - C 3, c 2^ c 2 + - c 3Z X

=> (x + y + z)2

x + y2Z 2z

Xx 2

z + y X2z0 0 2zx

Expanding along R3

=>2x z ( x + y + z)2

^>2x z ( x + y + z)2 (xz + x y + yz + y2 - xz) => 2xyz (x + y + z f H.P.

A =1 0 2 0 2 12 0 3

=> A“ = A. A =

2X 2 '\ zZ

2 xz)

0 212 10 3J

OR

1 + 0 + 4 0 + 0 + 0 2 + 0 + 6 0 + 0 + 2 0 + 4 + 0 0 + 2 + 3 2 + 0 + 6 0 + 0 + 0 4 + 0 + 9

5 0 82 4 58 0 13

"5 0 8 " "1 0 2"2 4 5 0 2 18 0 13 2 0 3

"5 + 0 + 16 0 + 0 + 0 10 + 0 + 24" "21 0 34"2 + 0 + 10 0 + 8 + 0 4 + 4 + 15 = 12 8 238 + 0 + 26 0 + 0 + 0 16 + 0 + 39 34 0 55

14/20


Taking A3 - 6A2 + 7A + kI3 = 0

"21 0 34" "5 0 8 " "1 0 2"12 8 23 - 6 2 4 5 +7 0 2 1 + kI334 0 55 8 0 13 2 0 3

= 0

"21 0 34" "30 0 48" "7 0 14"Or 12 8 23 - 12 24 30 + 0 14 7

34 0 55 48 0 78 14 0 21

"- 2 0 0 " "k 0 0" "0 0 0"Or 0 - 2 0 + 0 k 0 = 0 0 0

0 0 - 2 0 0 k 0 0 0

+ kI3 =0

^ k - 2 = 0

^ k = 2

22. A retired person wants to invest an amount of 7 50,000. His broker recommends investing in two type of bonds 'A' and 'B' yielding 10% and 9% return respectively on the invested amount . He decides to invest at least T 20,000 in bond 'A' and at least ? 10,000 in bond 'B'. He also wants to invest at least as much in bond 'A' as in bond 'B'. Solve this linear programming problem graphically to maximise his returns.

Sol. Let Rs 'x' invest in bond A and Rs 'y' invest in bond B. then A.T.P.

10 9Maxmise z = x + y

100 100...(1)

subject to constraintsx + y < 50,000 ...(a)

x > 20,000 ...(b)

y > 10,000 ...(c)

and x > y or x - y > 0

and x > 0, y > 0

...(d)

Now change inequality into equationsx + y = 50,000, x = 20,000, y = 10,000 and x = y

x 0 50,000

y 50,000 0

Region: put (0, 0) in (a), (b), (c), (d)0 < 50,000 (towards origin)

0 > 20,000 (away from origin)

0 > 10,000 (away from origin)

15 / 20


Points Z 10 + 9 100 100

A(20,000 10,000) Z = Rs, 2900B(40,000, 10,000) Rs 4900 <------C(25000, 25000) Rs 4750

D(20,000, 20,000) Rs 3800

Maximize

So he has to invest Rs 40,000 in 'A' and Rs 10,000 in bond 'B' to get maximum return Rs 49,00

23. Find the equation of the plane which contains the line of intersection of the planes.

r • (i - 2j + 3k) - 4 = 0 and r • (-2i + j + It) + 5 = 0

and whose intercept on x-axis is equal to that of on y-axis.

Sol. Planes are r • (i - 2j + 3k) - 4 = 0

r • (-2 j + j + k) + 5 = 0

^ x - 2y + 3z - 4 = 0}Two planes

& -2x + y + z + 5 = 0

16 / 20


24.

Any plane passing through the line of intersect

(x - 2y + 3z - 4) + X (-2x + y + z + 5) = 0

x (1 — 2X) + y ( — 2 + X) + z (3 + X) + (5X — 4) = 0

Intercepts are equal on axes

4 - 5X 4 - 5Xso =

1 - 2X - 2 + X

^ — 2 + X = 1 — 2 X

^ 3X = 3 ^ X = 1

required plane

—x — y + 4z + 1 = 0

x + y — 4z — 1 = 0 Ans.

Prove that y4sin 0

2 + cos 00 is an increasing function of 0 on

0 f

OR

Show that semi-vertical angle of a cone of maximum volume and given slant height is

Sol. y = 24sin00 — 0, 0 e [0, n/2]2 + cos 0

Diff. w. r. to 0, we get

dy = 4 I (2 + cos 0) cos 0 — sin 0(— sin 0) d0 [ (2 + cos 0)2

= 4 I_2 cos0 + ^ I — 1 [ (2 + cos0)2 J

= 8cos 0 + 4 — (2 + cos 0)2 (2 + cos 0)2

= 4 cos0 — cos2 0 (2 + cos 0)2

= cos 0(4 — cos 0)(2 + cos 0)2

[q 0 < cos 0 < 1 V 0 e [0, n/2]]

cos 0(4 — cos 0) (2 + cos 0)2

> 0 V 0 e [0, n/2]

> 0 V 0 e [0, n/2] d0

^ y is increasing in [0, n/2]

- / 1 1 cos .W 3 J

17 / 20


1 2Volume V = nr h

3

V = —n(12 - h 2)h

For maxima and minima

dvdh

1

■ = 0

3n(12 - 3h2) = 0

^ l 2 = 3h2

h = lh = 3

and -^-2 = — n(0 - 6h) dh2 3

= - 2nh

= - 2n( l ^o / 3 j

< 0

lso at h = volume of cone is maximum

3

and semivertical angle a as :

hcos a =

l

cos a =-l /V3

cos a =13

f — 1cos 1 3 J Ans.

OR

18 / 20


25. Let A = R x R and * be a binary operation on A defined by (a, b) * (c, d) = (a + c, b + d)

Show that * is commutative and associative. Find the identity element for * on A. Also find the inverse of

every element (a, b) e A.

Sol. A = R x R

(a, b)*(c, d) = (a + c, b +d)

Commutative : let (a., b), (c, d) eA

(a, b)*(c, d) = (a + c, b + d)

= (c + a, d + b)

= (c, d) (a, b) V (a, b) (c, d) eA

* is commutative

Associative : Let (a, b), (c, d), (e, f) e A

((a, b) * (c, d)) * (e, f) = ((a + c, b + d)) * (e, f)

= (a + c + e, b + d + f)

= (a + (c + e), b + (d + f))

= (a, b) * (c + e, d + f)

= (a, b) * ((c, d) * (e, f))

V (a, b), (c, d) (e, f) e A

* is associative

Identity element:

Let (ei, e2) e A

is identify element for * operation by definition

^ (A b) * (e^ e2) = (a, b)

^ (a + ei, b + e2) = (a, b)

a + e1 = a, b + e2 = b

^ ei = 0, e2 = 0

^ (0, 0) eA

^ (0, 0) is identify element for *

Inverse : let (b1, b2) eA is inverse of element (a, b)eA then by definition.

(a, b) * (bi, b2) = (0, 0)

(a + bi, b + b2) = (0, 0)

^ a + bi = 0, b + b2 = 0

^ (- a, -b) eA is inverse of every element (a, b) eA.

26. Three numbers are selected at random (without replacement) from first six positive integers. Let X denote the

largest of the three numbers obtained. Find the probability distribution of X. Also, find the mean and

variance of the distribution.

19 / 20


Sol. First six positive integrs are 1, 2, 3, 4, 5, 6 Three numbers are selected at random

without replacement so, total no. of ways 6C3 = 20 let, x denote the larger of three numbers so x can take values 3, 4, 5, 6

p(x = 3)

p(x = 4)

p(x = 5)

p(x = 6)

20

20_6_201020

x 3 4 5 6p(x) 1 3 6 10

20 20 20 20xp(x) 3 12 30 60

20 20 20 20x2p(x) 9 48 150 360

20 20 20 20^ mean 2 x p (x)

3 12 30 6020 20 20 10520

= 5.25

20

Varience = 2 x 2p(x) - ( 2 x p (x ) f

= ( f r H 28.35 - 27.56 = 0.79

20 / 20

II | Mathematics

Paper 2015 – All India Set – 1

1

CBSE Board

Class XII Mathematics

Board Paper – 2015

All India Set – 1 Time: 3 hrs Total Marks: 100

General Instructions:

1. All questions are compulsory.

2. Please check that this question paper contains 26 questions.

3. Question 1 – 6 in Section A are very short – answer type questions carrying 1 mark

each.

4. Questions 7 – 19 in Section B are long – answer I type question carrying 4 marks each.

5. Questions 20 – 26 in Section B are long – answer II type question carrying 6 marks

each.

6. Please write down the serial number of the question before attempting it.

SECTION – A

Question numbers 1 to 6 carry 1 mark each.

1. If a 2i j 3k and b 3i 5j 2k, thenfind a b .

2. Find the angle between the vectors i jand j k.

3. Find the distance of a point (2, 5, −3) from the plane r. 6i 3j 2k 4.

4. Write the element a12 of the matrix A = [aij]2 × 2, whose elements aij are given by aij = e2ix

sin jx.

5. Find the differential equation of the family of lines passing through the origin.

6. Find the integrating factor for the following differential equation: dy

x log x y 2log xdx

II | Mathematics


2

SECTION – B

Question numbers 7 to 19 carry 4 marks each.

7. If

1 2 2

A 2 1 2 ,

2 2 1

then show that A2 – 4A – 5I = O, and hence find A-1

OR

If

2 0 1

A 5 1 0 ,

0 1 3 then find A-1 using elementary row operations.

8. Using the properties of determinants, solve the following for x:

x 2 x 6 x 1

x 6 x 1 x 2 0

x 1 x 2 x 6

9. Evaluate:2/2

0

sin xdx

sinx cosx.

OR

Evaluate 2

3x

1

e 7x 5 dx as a limit of sums.

10. Evaluate:

2

4 2

xdx

x x 2

11.In a set of 10 coins, 2 coins are with heads on both the sides. A coin is selected at

random from this set and tossed five times. If all the five times, the result was heads,

find the probability that the selected coin had heads on both the sides.

OR

How many times must a fair coin be tossed so that the probability of getting at least one

head is more than 80%?

12. Find x such that the four points A(4, 1, 2), B(5, x, 6) , C(5, 1, -1) and D(7, 4, 0) are

coplanar.

II | Mathematics


3

13. A line passing through the point A with position vector a 4i 2j 2k is parallel to the

vector b 2i 3j 6k . Find the length of the perpendicular drawn on this line from a

point P with vector 1r i 2j 3k .

14. Solve the following for x:

sin-1 (1 - x) – 2 sin-1 x =

2

OR

Show that:

1 13 172sin tan

5 31 4

15. If y = eax. cos bx, then prove that

2

2 2

2

d y dy2a a b y 0

dxdx

16. If xx + xy + yx = ab , then find dy

.dx

17. If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t) then find dy

dx at t .

4

18. Evaluate:

x

3

x 3 edx

x 5

19. Three schools X, Y, and Z organized a fete (mela) for collecting funds for flood victims

in which they sold hand-helds fans, mats and toys made from recycled material, the sale

price of each being Rs. 25, Rs. 100 andRs. 50 respectively. The following table shows the

number of articles of each type sold:

School SchoolX SchoolY SchoolZArticle

Hand - held fans 30 40 35

Mats 12 15 20

Toys 70 55 75

Using matrices, find the funds collected by each school by selling the above articles and

the total funds collected. Also write any one value generated by the above situation.

II | Mathematics


4

SECTION – C

Question numbers 20 to 26 carry 6 marks each.

20. Let A = Q × Q, where Q is the set of all rational numbers, and * be a binary opearation

on A defined by (a, b) * (c, d) = (ac, b + ad) for (a. b), (c, d) A. Then find

(i) The identify element of * in A.

(ii) Invertible elements of A, and hence write the inverse of elements (5, 3) and

1,4

2.

OR

Let f : W → W be defined as

n 1, if n is oddf n

n 1, if n is even

Show that f is invertible and find the inverse of f. Here, W is the set of all whole

numbers.

21. Sketch the region bounded by the curves 2y 5 x and y x 1 and find its area

using intergration .

22. Find the particular solution of the differential equation x2dy = (2xy + y2) dx, given that y

= 1 when x = 1.

OR

Find the particular solution of the differential equation

12 m tan xdy1 x e y ,

dx

given that y =1 when x = 0.

23. Find the absolute maximum and absolute minimum values of the function f given by f(x)

= sin2x – cos x, x (0, π)

24. Show that lines:

r i j k i j k

r 4j 2 k 2j j 3k are coplanar.

Also, find the equation of the plane containing these lines.

II | Mathematics


5

25. Minimum and maximum z = 5x + 2y subject to the following constraints:

x – 2y ≤ 2

3x + 2y ≤ 12

−3x + 2y ≤ 3

x ≥ 0, y ≥ 0

26. Two the numbers are selected at random (without replacement) from first six positive

integers. Let X denote the larger of the two numbers obtained. Find the probability

distribution of X. Find the mean and variance of this distribution.

II | Mathematics

Paper 2015 – All India Set – 1 Solution

1

CBSE Board


Board Paper – 2015 Solution

All India Set – 1

SECTION – A

1. Given that a 2i j 3k and b 3i 5j 2k

2 2 2

We need to find a b

i j k

a b 2 1 3

3 5 2

i 2 15 j 4 9 k 10 3

17i 13j 7k

Hence, a b 17 13 7

a b 507

2. Let a i j; b j k

22 2

22 2

a b i j j k 1 0 1 1 0 1 1

a 1 1 0 2

b 0 1 1 2

We know that a b a b cos

a b 1 1Thus, cos =

22 2a b

cos cos120

120

II | Mathematics


2

3. Consider the vector equation of the plane.

1 1 1

r 6i 3j 2k 4

xi yj zk 6i 3j 2k 4

6x 3y 2z 4

6x 3y 2z 4 0

Thus the Cartesian equation of the plane is

6x 3y 2z 4 0

Let d be the distance between the point 2, 5, 3

to the plane.

ax by cz dThus, d=

a

2 2 2

22 2

b c

6 2 3 5 2 3 4d

6 3 2

12 15 6 4d

36 9 4

13d

49

13d units

7

4. Given that of aij = e2ixsin(jx)

2 1 x 2x12

Substitute i = 1 and j = 2

Thus, a e sin 2 x e sin 2x

II | Mathematics


3

5. Consider the equation, y = mx, where m is the parameter.

Thus, the above equation represents the family of lines which pass through the origin.

y mx....(1)

ym....(2)

x

Differentiating the above equation (1) with respect to x,

y mx

dym 1

dx

dym

dx

dy yfrom equation (2)

dx x

dy y0

dx x

Thus we have eliminated the constant, m.

The required differential equation is

dy y0

dx x

6. Consider the given differential equation:

dyxlog x y 2log x

dx

Dividing the above equation by xlogx, we have,

x log x dy y 2log x

xlog x dx xlog x xlog x

dy y 2....(1)

dx xlog x x

Consider the general linear differential equation,

dyPy Q,where P and Q are funct

dx

ions of x

Pdx

dxPdx x log x

Comparing equation (1) and the general equation, we have,

1 2P x and Q x

xlog x x

The integrating factor is given by the formula e

Thus,I.F. e e

II | Mathematics


4

dx

log log xx log x

dxConsider I=

xlog x

dxSubstituting logx=t; dt

x

dtThus I= log t log log x

t

Hence,I.F. e e log x

SECTION – B

7.

2

1 2 2

A 2 1 2

2 2 1

1 2 2 1 2 2

A 2 1 2 2 1 2

2 2 1 2 2 1

1 1 2 2 2 2 1 2 2 1 2 2 1 2 2 2 2 1

2 1 1 2 2 2 2 2 1 1 2 2 2 2 1 2 2 1

2 1 2 2 1 2 2 2 2 1 1 2 2 2 2 2 1 1

1 4 4 2 2 4 2 4 2

2 2 4 4 1 4 4 2 2

2 4 2 4 2 2 4 4 1

9 8 8

8 9 8

8 8 9

2Consider A 4A 5I

9 8 8 1 2 2 1 0 0

8 9 8 4 2 1 2 5 0 1 0

8 8 9 2 2 1 0 0 1

9 8 8 4 8 8 5 0 0

8 9 8 8 4 8 0 5 0

8 8 9 8 8 4 0 0 5

9 9 8 8 8 8

8 8 9 9 8 8

8 8 8 8 9 9

II | Mathematics


5

0 0 0

0 0 0

0 0 0

2

2

2 1 1 1 1

1

1

Now

A 4A 5I 0

A 4A 5I

A A 4AA 5IA Postmultiply by A

A 4I 5A

1 2 2 4 0 0

2 1 2 0 4 0 5A

2 2 1 0 0 4

1

1

3 2 2

2 3 2 5A

2 2 3

3 2 2

5 5 5

2 3 2A

5 5 5

2 2 3

5 5 5

OR

1

1

1

2 0 1

A 5 1 0

0 1 3

2 3 0 0 15 0 1 5 0

6 0 5

1

0

Hence A exists.

A A I

2 0 1 1 0 0

A 5 1 0 0 1 0

0 1 3 0 0 1

II | Mathematics


6

1 1

1

2 2 1

1

1Applying R R

2

1 11 0 0 0

2 2

A 5 1 0 0 1 0

0 1 3 0 0 1

Applying R R 5 R

1 11 0 0 0

2 2

5 5A 0 1 1 0

2 2

0 1 3 0 0 1

3 3 2

1

3 3

1

Applying R R 1 R

1 11 0 0 0

2 2

5 5A 0 1 1 0

2 2

1 50 0 1 1

2 2

Applying R 2 R

1 11 0 0 0

2 2

5 5A 0 1 1 0

2 2

0 0 1 5 2 2

Applying R1 1 3

2 2 3

1

1

1R R

2

5 R R R

2

1 0 0 3 1 1

A 0 1 0 15 6 5

0 0 1 5 2 2

3 1 1

A 15 6 5

5 2 2

II | Mathematics


7

8.

2 2 1 3 3 1

2 2 1 3 3 1

x 2 x 6 x 1

Let x 6 x 1 x 2

x 1 x 2 x 6

Applying C C C andC C C

x 2 4 3

x 6 7 4

x 1 3 7

Applying R R R andR R R

x 2 4 3

4 11 1

3 1 10

2 2 3

3 3 2

1

Applying R R R

x 2 4 3

1 12 9

3 1 10

Applying R R 3 R

x 2 4 3

1 12 9

0 37 37

Expanding along C

12 9 4 3x 2 1

37 37 37 37

x 2 444 333 1 148 111

x 2 111 1 37

0 111x 259

259 7x

111 3

II | Mathematics


8

9.

22

0

2a a2

0 0 0

22

0

2

sin xLet I dx.....(i)

sin x cosx

sin x2

I dx Using Property, f(x)dx f(a x)dx

sin x cos x2 2

cos xI dx........(ii)

sin x cosx

Adding,(i) and (ii),

sin x cos2I=

22

0

xdx

sin x cosx

2

0

2

0

2

0

dx2I=

sin x cosx

1 dx2I

1 12 sin x. cosx2 2

1 dx2I

2 sin x.cos cosx.sin4 4

2

0

2

0

2

0

1 dx2I

2 sin x4

12I cosec x dx

42

12I ln cosec x cot x

4 42

12I ln cosec cot ln cosec 0 cot 0

4 2 4 2 4 42

12I ln 2 ( 1) ln 2 1

2

1 2 1I ln

2 2 2 1

II | Mathematics


9

OR

23x

1

3x

2 n3x

nr 11

n n3 1 rh

n nr 1 r 1

3 3h 3h 6h 3nh 2

n

e 7x 5 dx

Here f(x)=e 7x 5

b-a 3a 1, b 2, h=

n n

By definition e 7x 5 dx lim h.f(a rh)

lim h.f( 1 rh) lim h. e 7 1 rh 5

lim h.e .e e e .... e 7h 1 2 3 ... n

3h 3nh2

3 3hn

33 3n 3n

n3 3h 2n

9

3

9

3

12nh

n n 1he e 1lim 7h 12nh

2e e 1

n n 13e 3h n 63lim e 1 12 3

3 3 2ne e 1 n

Now applying the limit we get

e 1 6336

23e

e 1 9

23e

II | Mathematics


10

10. 2

4 2

2

2 2

2

2

2

2 2

2 22

2 2

xdx

x x 2

xdx

x 1 x 2

xdx

x 1 x 1 x 2

Using partial fraction,

x A B Cx D

x 1 x 1x 1 x 1 x 2 x 2

A x 1 x 2 B x 2 x 1 Cx D x 1 x 1x

x 1 x 1 x 2 x 1 x 1 x 2

Equating the coefficients from both the numerators we

2

2 2

get,

A B C 0........(1)

A B D 1........(2)

2A 2B C 0........(3)

2A 2B D 0........(4)

Solving the above equations we get,

1 1 2A ,B ,C 0,D

6 6 3

Our Integral becomes,

x 1 1 2dx dx

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

1log x

61

1

1 2 1 x1 log x 1 tan C

6 3 2 2

1 xlog x 1 log x 1 2 2 tan C

6 2

II | Mathematics


11

11. 1 1Let E , E and A be the events defined as follows:

1

1

1

E Selecting a coin having head on both the sides

E Selecting a coin not having head on both the sides

A Getting all heads when a coin is tossed five times

We have to find P E / A .

There are 2 coins having

21

1 101

81

2 101

51

5

2

1 11

1 1 2 2

heads on both the sides.

C 2P E

10C

There are 8 coins not having heads on both the sides.

C 8P E

10C

P A /E 1 1

1P A /E

2

By Baye's Theorem, we have

P E P A /EP E / A

P E P A /E P E P A /E

2

105

1

2 8 11

10 10 2

2

82

32

8

9

II | Mathematics


12

OR

Let p denotes the probability of getting heads.

Let q denotes the probability of getting tails.

1p

2

1 1q 1

2 2

Suppose the coin is tossed n times.

Let X denote the number of times of getting heads in n tr

r n r nn r n r n n

r r r

nn

0

ials.

1 1 1P X r C p q C C ,r 0,1,2,......,n

2 2 2

80P X 1

100

80P X 1 +P X 2 + +P X n

100

80P X 1 +P X 2 + +P X n P X 0 P X 0

100

801 P X 0

100

1P X 0

5

1 1C

2 5

1

2

n1

5

n 3,4,5.......

So the fair coin should be tossed for 3 or more times for getting the required probability.

II | Mathematics


13

12. Position vector of OA 4i j 2k

Position vector of OB 5i xj 6k

Position vector of OC 5i j k

Position vector of OD 7i 4j 0k

AB OB OA

5i xj 6k 4i j 2k i (x 1)j 4k

AC OC OA

5i j k 4i j 2k i 3k

AD OD OA

7i 4j 0k 4i j 2k 3i 3j 2k

The above three vectors are coplanar

AB. AC AD 0

1 x 1 4

1 0 3 0

3 3 2

1(0 9) (x 1)( 2 9) 4(3 0) 0

9 7(x 1) 12 0

7(x 1) 21

x 1 3

x 4

II | Mathematics


14

13. Let the equation of the line be r a b

Here, a 4i 2j 2k

b 2i 3j 6k

Equation of the line is r 4i 2j 2k 2i 3j 6k

Let L be the foot of the perpendicular and P be the required point from

which we have to find the length of the perpen

dicular

P( ) i 2j 3k

PL position vector of L position vector of P

=4i 2j 2k 2i 3j 6k i 2j 3k

3i k 2i 3j 6k ........(i)

Now,PL.b 0[Since PL is perpendicular to b]

3i k 2i 3j 6k . 2i 3j 6k 0

i(3 2 ) j(3 ) k( 1 6 ) . 2i 3j 6k 0

(3 2 )2 (3 )3 ( 1 6 )6 0

2 2

6 4 9 6 36 0

49 0

0

PL 3i k [from(ii)]

PL 3 ( 1)

PL 10

Length of the perpendicular drawn on the line from P = 10

II | Mathematics


15

14. sin-1 (1 – x) – 2sin-1 x = 2

⇒ sin-1 (1 – x) = 2

+ 2sin-1 x

⇒ (1 – x) = sin (2

+ 2sin-1 x )

⇒ (1 – x) = cos(2sin-1 x)

⇒ (1 – x) = cos(cos-1 (1 – 2x2))

⇒ (1 – x) = (1 – 2x2)

⇒ 1 – x = 1 – 2x2

⇒ 2x2 – x = 0

x = 0, x = 1

2

OR

2 sin-1 3

5

− tan-1 17

31

= 4

L.H.S.,

= cos -1 9

1 225

− tan-1

17

31

= cos-1 7

25

− tan-1 17

31

= tan-1 24

7

− tan-1 17

31

= tan-1 24

7

− tan-1 17

31

= tan-1

24 17

7 3124 17

17 31

= tan-1 24 31 17 7

31 7 24 17

= tan-1 625

625

= tan-1 1

= 4

= R.H.S.

Hence Proved

II | Mathematics


16

15. axy e .cosbx

ax ax

ax

2ax ax

2

2ax 2 ax

2

22 ax

2

2

2

dyae .cosbx be .sinbx.....(i)

dx

dyay be .sinbx

dx

d y dya b ae .sinbx be .cosbx

dxdx

d y dya abe .sinbx b e .cosbx

dxdx

d y dy dya a ay b y [Substituting be sinbx from (i)]

dx dxdx

d y da

dx

2 2

22 2

2

y dya y a b y

dx dx

d y dy2a (a b )y 0

dxdx

Hence Proved

II | Mathematics


17

16. xx + xy + yx = ab ……….(i)

Let u = xx

log u = x log x

x

1 du 1. x. log x

u dx x

dux 1 log x

dx

Let v = xy

log v = y log x

y

1 dv y dy. log x

v dx x dx

dv y dyx log x

dx x dx

Let w = yx

Log w = x log y

x

1 dw x dy. . log y

w dx y dx

dw x dyy log y .

dx y dx

(i) can be written as

u + v + w = ab

du dv dw0

dx dx dx

x y x

x x y y x x

y x x x y x

y x 1 x x y 1 x

x

y dy x dyx 1 log x x log x y log y 0

x dx y dx

y dy x dyx x log x x . x .log x. y .log y y . . 0

x dx y dx

dy x yx .log x y . x x log x x . y .log y

dx y x

dyx .log x xy x x log x yx y .log y

dx

xdy

dx

x y 1 x

y x 1

x log x yx y .log y

x .log x xy

II | Mathematics


18

17. x = a sin 2t(1 + cos 2t),

y = b cos 2t(1 – cos 2t)

2 2

dx2acos2t(1 cos2t) asin2t( 2sin2t)

dt

2acos2t 2acos 2t 2asin 2t

2acos2t 2acos4t

dy2bsin2t(1 cos2t) bcos2t(2sin2t)

dt

2bsin2t 2bsin2tcos2t 2bcos2t sin2t

2bsin2t 4bsin2tcos2t

2bsin2t 2bsin4t

dy

dtdx

dt

2bsin2t 2bsin4t

2acos2t 2acos4t

t4

t4

t4

t4

dy 2bsin2t 2bsin4t

dx 2acos2t 2acos4t

2 42bsin 2bsin

dy 4 42 4dx 2acos 2acos4 4

2bsin 2bsindy 2dx 2acos 2acos

2

dy 2b b

dx 2a a

dy b

dx a

II | Mathematics


19

18. x

3

x

3

x

3 3

x

2 3

(x 3)edx

(x 5)

(x 5 2)edx

(x 5)

(x 5) 2e .dx

(x 5) (x 5)

1 2e .dx

(x 5) (x 5)

x x

x

2 3

x

2

This is of the form

e [f(x) f '(x)]dx e f(x) C

1 2e .dx

(x 5) (x 5)

eC

(x 5)

19.

30 12 70 25

40 15 55 100

35 20 75 50

30 25 12 100 70 50

40 25 15 100 55 50

35 25 20 100 75 50

5450 X

5250 Y

6625 Z

The funds collected by X = Rs. 5450, Y = Rs. 5250, Z = Rs. 6625

Total funds collected = Rs. 17325

Value generated: team work

II | Mathematics


20

SECTION – C

20. Let A = Q × Q, where Q is the set of rational numbers.

Given that * is the binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for

(a, b), (c, d) A.

(i)

We need to find the identity element of the operation * in A.

Let (x, y) be the identity element in A.

Thus,

(a, b) * (x, y) = (x, y) * (a, b) = (a, b), for all (a, b) A

(ax, b + ay) = (a, b)

ax = a and b + ay =b

y = 0 and x = 1

Therefore, (1, 0) A is the identity element in A with respect to the operation *.

(ii)

We need to find the invertible elements of A.

Let (p, q) be the inverse of the element (a, b)

Thus,

(a, b) * (p, q) = (1, 0)

(ap, b + aq) = (1, 0)

ap = 1 and b + aq = 0

1 bp and q =

a a

1 bThus the inverse elements of (a, b) is ,

a a

1Now let us find the inverse of (5, 3) and ,4

2

1 3Hence, inverse of (5, 3) is ,

5 5

1 4And inverse of ,4 is 2,

12

2

2, 8

II | Mathematics


21

OR

Let f: WW be defined as

n 1, if n is odd

f nn + 1, if n is even

We need to prove that 'f' is invertible.

In order to prove that 'f' is invertible it is sufficient to prove that f is a bijection.

A function f: AB is a one-one function or an injection, if

f(x) = f(y) x = y for all x, y A.

Case i:

If x and y are odd.

Let f(x) = f(y)

x − 1 = y − 1

x = y

Case ii:

If x and y are even,

Let f(x) = f(y)

x + 1 = y + 1

x = y

Thus, in both the cases, we have,

f(x) = f(y) x = y for all x, y W.

Hence f is an injection.

Let n be an arbitrary element of W.

If n is an odd whole number, there exists an even whole number n − 1 W such that

f(n − 1) = n − 1 + 1 = n.

If n is an even whole number, then there exists an odd whole number n + 1 W such

that f(n + 1) = n + 1 − 1 = n.

Also, f(1) = 0 and f(0) = 1

Thus, every element of W (co-domain) has its pre-image in W (domain).

So f is an onto function.

Thus, it is proved that f is an invertible function.

Thus, a function g: BA which associates each element y B to a unique element x A

such that f(x) = y is called the inverse of f.

That is, f(x) = y g(y) = x

The inverse of f is generally denoted by f -1.

Now let us find the inverse of f.

Let x, y W such that f(x) = y

x + 1 = y, if x is even

II | Mathematics


22

And

x − 1 = y, if x is odd

1

1

1

y 1, if y is oddx

y+1, if y is even

y 1, if y is oddf y

y +1, if y is even

Interchange, x and y, we have,

x 1, if x is oddf x

x+1, if x is even

Rewriting the above we have,

x+1, if x isf x

-1

even

x 1, if x is odd

Thus, f x f x

21. Consider the given equation

2y= 5 x

This equation represents a semicircle with centre at

the origin and radius = 5 units

Given that the region is bounded by the above

semicircle and the line y = x 1

2

Let us find the point of intersection of the

given curve meets the line y= x 1

5 x x 1

22

2 2

2

2

2

2

Squaring both the sides, we have,

5 x x 1

5 x x 1 2x

2x 2x 5 1 0

2x 2x 4 0

x x 2 0

x 2x x 2 0

x x 2 1 x 2 0

x 1 x 2 0

x 1,x 2

When x = 1,y 2

When x = 2,y 1

Consider the following figure.

II | Mathematics


23

Thus the intersection points are 1,2 and 2,1

Consider the following sketch of the bounded region.

2

2 1

1

1 22 2

1 1

1 1 1 2 2 22 2

1 1 1 1 1 1

1 1212 1

1

11

22

2 1

1

Required Area, A= y y dx

5 x x 1 dx 5 x x 1 dx

5 x dx xdx dx 5 x dx xdx dx

x 5 x x5 x sin x

2 2 25

x 5 x x5 x sin

2 2 25

2

2

1

1

1 1

1 1

x

5 1 5 2 1sin sin

2 2 25 5

5 1 5 2 1Required Area = sin sin sq.units

2 2 25 5

II | Mathematics


24

22. x2dy = (2xy + y2)dx 2

2

dy 2xy y.......(i)

dx x

Let y vx,

dy dvv x

dx dx

2 2 2

2

2

2

2

Substituting in (i), we get

dv 2vx v xv x

dx x

dvv x 2v v

dx

dvx v v

dx

dv dx

xv v

2

Integrating both sides,

dv dx

xv v

v 1 v dx.dv

v(v 1) x

dv dv dx

v v 1 x

log v log | v 1| log x logC

vlog log Cx

v 1

yxlog log Cx

y 1x

2

yCx [Removing logarithm in both sides]

y x

y Cxy Cx ,which is the general solution.

2

2

Putting y=1 and x=1,

1 C C

2C 1

1C

2

xy xy

2 2

2y xy x ,which is the particular solution.

II | Mathematics


25

OR

1

1

1

1

2 mtan x

mtan x

2 2

mtan x

2 2

mtan x

2 2

dy(1 x ) e y

dx

dy e y

dx (1 x ) (1 x )

dy y e

dx (1 x ) (1 x )

1 eP ,Q

(1 x ) (1 x )

2

1

1

1 1

1

1

Pdx

dx

(1 x )

tan x

Pdx Pdx

mtan xtan x tan x

2

(m 1)tan xtan x

2

I.F. e

e

e

Thus the solution is

ye Qe dx

ey e .e .dx

(1 x )

ey e dx........(i)

(1 x )

1(m 1)tan x

2

-1

2

edx..............(ii)

(1 x )

Let (m+1)tan x z

(m 1)dx dz

(1 x )

1

2

z

z

(m 1)tan x

dx dz

(m 1)(1 x )

Substituting in (ii),

1e dz

(m 1)

e

(m 1)

e

(m 1)

II | Mathematics


26

1

1(m 1)tan x

tan x

Substituting in (i),

ey e C.....(iii)

(m 1)

1

1

1

1

(m 1)tan 1tan 1

(m 1)4

4

(m 1)4

4

(m 1)(m 1)tan x 4

tan x 4

Putting y=1 and x=1, in the above equation,

ey e C

(m 1)

e1 e C

(m 1)

eC e

(m 1)

e eParticular solution of the D.E. is y e e

(m 1) (m 1)

II | Mathematics


27

23. f(x) = sin2x – cos x,

f’(x) = 2sin x.cos x + sin x

= sin x(2cos x + 1)

Equating f’(x) to zero.

f’(x) = 0

sin x(2cos x + 1) = 0

sin x = 0

x = 0,

2cos x + 1 = 0

⇒cos x = 1

2

x = 5

6

f(0) = sin20 – cos 0 = − 1

2

2

5 5 5f sin cos

6 6 6

sin cos6 6

1 3

4 2

1 2 3

4

f( ) = sin2 – cos = 1

Of these values, the maximum value is 1, and the minimum value is −1.

Thus, the absolute maximum and absolute minimum values of f(x) are 1 and −1, which

it attains at x = 0 and x = .

II | Mathematics


28

24. r i j k (i j k)........(i)

1 1 1

1 1 1

Convert ing into cartesian form,

x 1 y 1 z 1

1 1 1

(x ,y ,z ) (1,1,1)

a 1,b 1,c 1

r 4j 2k (2i j 3k)........(ii)

2 2 2

2 2 2

Convert ing into cartesian form,

x 0 y 4 z 2

2 1 3

(x ,y ,z ) (0,4,2)

a 2,b 1,c 3

1 1 1

Condition for the lines to be coplanar is

0 1 4 1 2 1 1 3 1

1 1 1 1 1 1 0

2 1 3 2 1 3

the lines are coplanar.

Intersection of the two lines is

Let the equation be a(x x ) b(y y ) c(z z ) 0.....(iii)

Direction ratio

s of the plane is

a b c 0

2a b 3c 0

Solving by cross-multiplication,

a b c

3 1 2 3 1 2

a 2 ,b ,c

Since the plane passes through (0,4,2) from line (ii)

a(x 0) b(y 4) c(z 2) 0

2 x (y 4) (z 2) 0

2x y 4

z 2 0

2x y z 2

2x y z 2

II | Mathematics


29

25. x – 2y ≤ 2

3x + 2y ≤ 12

−3x + 2y ≤ 3

x 0, y 0

Converting the inequations into equations, we obtain the lines

x – 2y = 2…..(i)

3x + 2y = 12……(ii)

−3x + 2y = 3……(iii)

x = 0, y = 0

From the graph, we get the corner points as

A(0, 5), B(3.5, 0.75), C(2, 0), D(1.5, 3.75), O(0, 0)

II | Mathematics


30

The values of the objective function are:

Point (x, y) Values of the objective function

Z = 5x + 2y

A(0, 5) 5 × 0 + 2 × 5 = 10

B(3.5, 0.75) 5 × 3.5 + 2 × 0.75 = 19 (Maximum)

C(2, 0) 5 × 2 + 2 × 0= 10

D(1.5, 3.75) 5 × 1.5 + 2 × 3.75 = 15

O(0, 0) 5 × 0 + 2 × 0 = 0 (Minimum)

The maximum value of Z is 19 and its minimum value is 0.

26. First six positive integers are {1, 2, 3, 4, 5, 6}

No. of ways of selecting 2 numbers from 6 numbers without replacement = 6C2 = 15

X denotes the larger of the two numbers, so X can take the values 2, 3, 4, 5, 6.

Probability distribution of X:

X 2 3 4 5 6

p(x) 1

15

2

15

3

15

4

15

5

15

Computation of Mean and Variance:

xi P(X = xi) pixi pixi2

2 1

15

2

15

4

15

3 2

15

6

15

18

15

4 3

15

12

15

48

15

5 4

15

20

15

100

15

6 5

15

30

15

180

15

pixi =

70

15=

14

3 pixi2 =

350

15=

70

3

Mean = pixi = 70

15= 4.67

Variance = pixi2 – ( pixi)2 = 70 196 210 196 14

3 9 9 9

II | Mathematics

Paper 2014 – Set 3

1

CBSE Board


Board Paper 2014

Set – 3 Time: 3 hrs Total Marks: 100

Note:

Please check that this question paper contains 5 printed pages.

Code number given on the right hand side of the question paper should be written on the

title page of the answer-book by the candidate.

Please check that this question paper contains 29 questions.

Please write down the Serial Number of the question before attempting it.

15 minutes time has been allotted to read this question paper. The question paper will be

distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m., the students will read the question

paper only and will not write any answer on the answer-book during this period.



2. The question paper consists of 29 questions divided into three sections A, B and C. Section

A comprises of 10 questions of one mark each, Section B comprises of 12 questions of four

marks each and Section C comprises of 7 questions of six marks each.

3. All questions in Section A. are to be answered in one word, one sentence or as per the exact

requirement of the question.

4. There is no overall choice. However, internal choice has been provided in

4 questions of four marks each and 2 questions of six marks each. You have to attempt only

one of the alternatives in all such questions.

5. Use of calculators is not permitted. You may ask for logarithmic tables, if required.

SECTION – A

1. If A is a square matrix such that A2 =A, then write the value of 7A − (I + A)3, where I is an

identity matrix.

2. x-y z -1 4

If =2x-y w 0 5

, find the value of x + y.

3. 1 1 πIf tan x+tan y= ,xy<1

4, then write the value of x + y + xy.

4. If3x 7 8 7

=-2 4 6 4

, find the value of x.

II | Mathematics


2

5. If f(x) = x

0 t sin t dt, write the value of f’(x).

6. Find the value of 'p' for which the vectors ˆˆ ˆ3i+2j+9k and ˆˆ ˆi 2p j+3k are parallel.

7. If R = {(x, y): x + 2y = 8} is a relation on N, write the range of R.

8. If the cartesian equations of a line are 3-x y+4 2z-6

5 7 4, write the vector equation for the

line.

9. a

2

0

1 πIf dx=

4+x 8 find the value of a.

10. If a

and b

are perpendicular vectors, a b 13

and a 5

and find the value of

b .

SECTION – B

11. Solve the differential equation -12 tan xdy

(1 x ) y =e .dx

12. Show that the four points A, B, C and D with position vectors

4i 5 j k , j k ,3i 9 j 4k

and 4( i j k)

respectively are coplanar.

OR

The scalar product of the vector a i j k

with a unit vector along the sum of vectors

b 2i 4 j 5k

and c λ i 2 j 3k

is equal to one. Find the value of and hence find the

unit vector along b c .

13. Evaluate: π

2

0

4x sin xdx

1+ cos x

OR

Evaluate:

2

x+2dx

x +5x+6

II | Mathematics


3

14. Find the value(s) of x for which y = [x (x − 2)]2 is an increasing function.

OR

Find the equations of the tangent and normal to the curve 2 2

2 2

x y1

a bat the point 2 a,b .

15. If the function f : R R be given by f (x) = x2 + 2 and g: R R be given by

x

g x ,x 1,x-1

find fog and gof and hence find fog (2) and gof ( −3).

16. Prove that

1 11 x 1 x π 1 -1

tan cos x, x 14 21+x 1 x 2

OR

-1 -1x-2 x+2 πIf tan +tan

x-4 x+4 4

, find the value of x.

17. An experiment succeeds thrice as often as it fails. Find the probability that in the next five

trials, there will be at least 3 successes.

18. If y = Peax + Q ehx , show that 2

2

d y dy(a+b) + aby=0

dx dx

19. Using properties of determinants, prove that:

1 a 1 1

1 1+b 1 abc + bc + ca + ab

1 1 1 c

20. If x = cost (3 − 2 cos2 t) and y = sin t (3 − 2 sin2 t), find the value of dy π

at t = .dx 4

21. Find the particular solution of the differential equation log

dy3x + 4y,

dx given that y =

0 when x = 0.

II | Mathematics


4

22. Find the value of p, so that the lines

1

1 x 7y 14 3:

3 p 2

zl and 2

7 7x y 5 6:

3p 1 5

zl

are perpendicular to each other. Also find the equations of a line passing through a point (3,

2, − 4) and parallel to line l1.

SECTION – C

23. Find the equation of the plane through the line of intersection of the planes x + y + z = 1

and 2x + 3y + 4z = 5 which is perpendicular to the plane x- y + z = 0. Also find the

distance of the plane obtained above, from the origin.

OR

Find the distance of the point (2, 12, 5) from the point of intersection of the line

r 2i 4 j 2k λ 3i 4 j 2k

and the plane r . i 2 j k 0.

24. Using integration, find the area of the region bounded by the triangle whose

vertices are (-1, 2), (1, 5) and (3, 4).

25. A manufacturing company makes two types of teaching aids A and B of Mathematics for

class XII. Each type of A requires 9 labour hours of fabricating and 1 labour hour for

finishing. Each type of B requires 12 labour hours for fabricating and 3 labour hours

for finishing. For fabricating and finishing, the maximum labour hours available per week

are 180 and 30 respectively. The company makes a profit of 80 on each piece of type A and

120 on each piece of type B. How many pieces of type A and type B should be manufactured

per week to get a maximum profit? Make it as an LPP and solve graphically. What is the

maximum profit per week?

26. There are three coins. One is a two-headed coin (having head on both faces), another is a

biased coin that comes up heads 75% of the times and third is also a biased coin that

comes up tails 40% of the times. One of The three coins is chosen at random and tossed,

and it shows heads. What is the probability that it was the two-headed coin?

OR

Two numbers are selected at random (without replacement) from the first six positive

integers. Let X denote the larger of the two numbers obtained. Find the probability

distribution of the random variable X, and hence find the mean of the distribution.

II | Mathematics


5

27. Two schools A and B want to award their selected students on the values of sincerity,

truthfulness and helpfulness. The school A wants to award x each, y each and z each for

the three respective values to 3, 2 and 1 students respectively with a total award money

of 1,600. School B wants to spend 2,300 to award its 4, 1 and 3 students on the

respective values (by giving the same award money to the three values as before). If the

total amount for one prize on each value is 900, using matrices, find the award money for

each value. Apart from these three values, suggest one more value which should be

considered for award.

28. If the sum of the lengths of the hypotenuse and a side of a right triangle is given, show that

the area of the triangle is maximum, when the angle between them is 60°.

29. Evaluate:

4 2 2 4

1dx

sin x+sin xcos x+cos x

II | Mathematics

Paper 2014 – Delhi Set 3 Solution

1

CBSE Board


Board Paper 2014 Solution

Delhi Set – 3

SECTION – A

1. Given that 2A A .

We need to find the value of 3

7A I A , where I is the identity matrix.

Thus,

3 3 2 2 3

3 3 2 2 3 2 2 2

3 2

3

3

3

7A I A 7A I 3I A 3IA A

7A I A 7A I 3A 3A A A I I,I A A,IA A

7A I A 7A I 3A 3A A A A

7A I A 7A I 3A 3A A

7A I A 7A I 7A

7A I A I

2. Given that x y z 1 4

2x y w 0 5

We need to find the value of x + y.

ij ij

x y z 1 4

2x y w 0 5

Two matrices A and B are equal to each other, if they have the same dimensions

and the same elements a b , for i = 1,2,...,n and j = 1,2,...,m.

x y 1...(1)

2x y 0...(2)

Equa

tion (2) (1) is x = 1

Substituting the value of x = 1 in equation (1), we have

1 y 1

y 2

Therefore, x + y = 1 + 2 = 3

II | Mathematics


2

3. 1 1Given that tan x tan y and xy<1.4

1 1

1

1

We need to find the value of x+y+xy.

tan x tan y4

x ytan xy 1

1 xy 4

x ytan tan tan

1 xy 4

x y1

1 xy

x y 1 xy

x y xy 1

4. Given that 3x 7 8 7

2 4 6 4

.

We need to find the value of x

3x 7 8 7

2 4 6 4

12x 14 32 42

12x 14 10

12x 10 14

12x 24

x 2

II | Mathematics


3

5. Since differentiation operation is the inverse operation of integration, we have

sin f x x x

Let 0

sin x

f x t tdt

Let us do this by integration by parts.

Therefore assume u = t; du = dt

sin

cos

tdt dv

t v

00

Therefore,

= t cos cos

cos sin

Differentiating the above function with respect to x,

f x sin cos cos sin

x

xf x t t dt

f x x x x C

x x x x x x

6. Since the vectors are parallel, we have

a b

3i 2j 9k i 2pj 3k

3i 2j 9k i 2 pj 3 k

Comparing the respective coefficients, we have

3;

2 p 2

2 3 p 2

1p

3

7. The set of natural numbers, N = 1, 2, 3, 4, 5, 6....

The relation is given as

R = x, y : 2 8

Thus, R = 6, 1 , 4, 2 , 2, 3

Domain = 6, 4, 2

Range = 1, 2, 3

x y

II | Mathematics


4

8. Given that the cartesian equation of the line as

3 4 2 6

5 7 4

That is,

3 4 2 3

5 7 4

43 3

5 7 2

x y z

x y z

yx z

Any point on the line is of the form:

5 3,7 4,2 3

Thus, the vector equation is of the form:

r , where is the position vector of any

point on the line and b is the vector parallel to the lin

a b a

e.

Therefore, the vector equation is

r 5 3 7 4 2 3

r 5 7 2 3 4 3

r 3 4 3 5 7 2

i j k

i j k i j k

i j k i j k

9.

a

2

0

dxGiven that

4+x 8

a

2

0

-1

0

We need to find the value of a.

dxLet I=

4+x 8

1Thus, I= tan

2 2 8

a

x

1

1

1

1tan

2 2 8

tan 22 8

tan2 4

12

2

a

a

a

a

a

II | Mathematics


5

10. Given that a and b are two perpendicular vectors.

Thus, a b 0

Also given that, a b 13 and a =5.

We need to find the value of b.

2

2 2 2

22 2

2

2

2

Consider a b :

a b = a 2 a b b

13 5 2 0 b

169 25 b

b 169 25

b 144

b 12

II | Mathematics


6

SECTION – B

11. Given differential equation is:

12 tan1+x

xdyy e

dx

1tan

2 21+x 1+x

This a linear differential equation of the form

xdy y e

dx

dyPy Q

dx

1

1

1

1 1

tan

2 2

Pdx tan

tantan tan

2

1where P= and Q

1+x 1+x

Therefore,

I.F.=e

Thus the solution is,

. .

1+x

x

x

xx x

e

e

y I F Q I F dx

ey e e dx

1

1

tan

tan

2

Substitute ;

1

1+x

x

x

e t

e dx dt

1

1

1

1

tan

2tan

2tan

tan

Thus,

2

2

x

x

x

x

y e tdt

ty e C

ey e C

II | Mathematics


7

12. Given position vectors of four points A,B,C and D are:

OA 4 5

OB

OC 3 9 4

OD 4

These points are coplanar, if the vectors, AB,AC and AD are

coplanar.

i j k

j k

i j k

i j k

AB

4 5 4 6 2

OB OA

j k i j k i j k

AC

3 9 4 4 5 4 3

OC OA

i j k i j k i j k

AD

4 4 5 8 3

These vectors are coplanar if and only if, they can be expressed

as a linear combination of other two.

So let

AB AC+yAD

4 6 2 4

OD OA

i j k i j k i j k

x

i j k x i j

3 8 3

4 6 2 8 4 3 3

k y i j k

i j k x y i x y j x y k

Comparing the coefficients, we have,

8 4;4 6;3 3 2

Thus, solving the first two equations, we get

4 2x= and y=

3 3

These values of x and y satisfy the equation 3 3 2.

Hence the vectors are copl

x y x y x y

x y

anar.

II | Mathematics


8

OR

Given that

b 2 4 5

c 2 3

Now consider the sum of the vectors b c :

b c 2 4 5 2 3

b c 2 6 2

i j k

i j k

i j k i j k

i j k

2 2 2

ˆLet n be the unit vector along the sum of vectors b c :

2 6 2n

2 6 2

The scalar product of a and n is 1. Thus,

i j k

2 2 2

2 2 2

2 6 2ˆa n

2 6 2

1 2 1 6 1 21

2 6 2

i j ki j k

2 2 2

2 2 2

2 2

2 2

2 6 2 2 6 2

2 6 2 6

2 40 6

4 4 40 12 36

4 44 12 36

8 8

1

II | Mathematics


9

2 2 2

2 2 2

Thus, n is :

2 1 6 2n

2 1 6 2

3 6 2n

3 6 2

3 6 2n

49

3 6 2n

7

3 6 2n

7 7 7

i j k

i j k

i j k

i j k

i j k

II | Mathematics


10

13. We need to evaluate the integral

2

0

2

0

2 2

0 0

4xsin xI dx....(1)

1 cos x

Using the property f a x dx f x dx, we have

4 x sin xI= dx

1 cos x

4 sin x 4xsin xI= dx dx....(2)

1 cos x 1 cos x

2 2 2

0 0 0

2

0

2

0

Adding equations (1) and (2), we have,

4xsin x 4 sin x 4xsin x2I= dx dx dx

1 cos x 1 cos x 1 cos x

4 sin x2I= dx

1 cos x

sin x2I=4 dx

1 cos x

1

2

1

1

2

1

Substitute t = cosx; dt = sin xdx

when x = 0,t = 1

when x = , t = 1

1 dt2I = 4

1 t

dtI = 2

1 t

1

2

0

11

0

1

2

dtI=2 2

1 t

I=2 2 tan t

I 4 tan 1

I 44

II | Mathematics


11

OR

2

2

We need to evaluate the integral

x+2

x 5 6

x+2 I=

x 5 6

Consider the integrand as follows:

dxx

Let dxx

2

2 2

x 5 6x+2

x 5 6 x 5 6

2 2 5

2 2 5

dA x Bdx

x x

x A x B

x A x A B

2

2

Comparing the coefficients, we have

2A=1;5A B 2

Solving the above equations, we have

1 1A= and B=

2 2

Thus,

x+2I dx

x 5x 6

2x+5 1

2 2 dxx 5x 6

2 2

1 2x 5 1 1dx dx

2 2x 5x 6 x 5x 6

1 2

1 1I I I ,

2 2

II | Mathematics


12

1 2

2 2

1

1 2

2

1

2

2 5where I d

x 5 6

1and I d

x 5 6

Now consider I :

2 5I d

x 5 6

Substitute

x 5 6 t; 2 5 dx dt

dtI

t

2 t

2 x 5 6

xx

x

xx

xx

x

x x

x

2

2 2

2 2

2

Now consider I :

1

x 5 6

1

5 5x 5 6

2 2

I dxx

dx

x

2

2

2 2

1

5 25x 6

2 4

1

5 1x

2 4

1

5 1x

2 2

dx

dx

dx

22

1 2

2 2

5log x 5 6

2

1 1Thus, I=

2 2

1 55 6 log x 5 6

2 2

I x x C

I I

I x x x x C

II | Mathematics


13

14. Given function is

2

22

f x x x 2

f x x 2 x 2 x 2 2x

f x 2x x 2 x x 2

f x 2x x 2 2x 2

f x 2x x 2 2 x 1

f x 4x x 1 x 2

Since f x is an increasing function, f x 0.

f x 4x x 1 x 2 0

x x 1 x 2 0

0<x<1 or x>2

x 0,1 2,

II | Mathematics


14

OR

2 2

2 2

2 2

2 2

22 2 2

2

xLet 1 be the equation of the curve.

Rewriting the above equation as,

x1

Differentiating the above function w.r.t. x, we get,

y

a b

y

b a

by x b

a

2

2

2

2

dy2y 2

dx

dy

dx

bx

a

b x

a y

2

22 ,

1 1

dy 2 2

dx

2Slope of the tangent is m=

Equation of the tangent is

y

a b

b a b

a b a

b

a

y m x x

2y 2

y 2 2

2 2 0

2 0

1Slope of the normal is

2

bb x a

a

a b b x a

bx ay ab ab

bx ay ab

b

a

1 1

2 2

2 2

Equation of the normal is

y

y 22

2 y 2

2 2 2 0

2 2 0

y m x x

ab x a

b

b b a x a

ax by b a

ax by a b

II | Mathematics


15

15. 2 xGiven that f x x 2 and g x

x 1

2

2

22

2

Let us find f g:

f g=f g x

f g= g x 2

xf g= 2

x 1

x 2 x 1f g=

x 1

2 2

2

2

2

2

2

x 2 x 2x 1f g=

x 2x 1

3x 4x 2f g=

x 2x 1

3 2 4 2 2Therefore, f g 2

2 2 2 1

12 8 2f g 2 6

4 4 1

Now let us find g f:

g f=g f x

f xg f=

f x 1

2

2

2

2

2

2

x 2g f=

x 2 1

x 2g f=

x 1

3 2 9 2 11Therefore, g f 3

9 1 103 1

II | Mathematics


16

16. We need to prove that

-1 11 1 1 1tan cos , 1

4 21 1 2

Consider x=cos2t;

x xx x

x x

-1

-1

-1

1 cos2 1 cos2L.H.S=tan

1 cos2 1 cos2

2cos 2sintan

2cos 2sin

1 tantan

1 tan

t t

t t

t t

t t

t

t

-1

-1

tan tan4tan

1 tan tan4

tan tan4

t

t

t

1

4

1cos

4 2

R.H.S

t

x

II | Mathematics


17

OR

1 1

1 1

2 2Given that tan tan

4 4 4

We need to find the value of x.

2 2tan tan

4 4 4

x x

x x

x x

x x

1

2 2

4 4tan2 2 4

14 4

2 2

4 4 tan2 2 4

14 4

x x

x xx x

x x

x x

x xx x

x x

2 2

2 2

2

2

2

2

2 4 2 41

4 4 2 2

2 8 2 81

16 4

2 161

12

2 16 12

2 4

2

2

x x x x

x x x x

x x x x

x x

x

x

x

x

x

II | Mathematics


18

17. An experiment succeeds thrice as often as it fails.

Therefore, there are 3 successes and 1 failure.

3Thus the probability of success =

4

1And the probability of failure =

4

5 3 2 5 4 1 5 5 0

3 4 5

3 2 4 1 5

5 5 53 4 5

We need to find the probability of atleast 3 successes

in the next five trials.

Required Probability=P X=3 4 5

3 1 3 1 3 1

4 4 4 4 4 4

P X P X

C p q C p q C p q

C C C

0

3 2 4 1 5 03 1 3 1 3 1

10 54 4 4 4 4 4

918

1024

459

512

II | Mathematics


19

18. Given that

axy=Pe

Differentiating the above function w.r.t. x,

dy

dx

bx

ax bx

Qe

Pae Qbe

22 2

2

2 2

2 2

Differentiating once again, we have,

d y

dx

dyLet us now find a+b :

dx

dya+b a+b

dx

dya+b

dx

dya+b

dx

ax bx

ax bx

ax bx bx bx

ax bx bx

Pa e Qb e

Pae Qbe

Pa e Qabe Pabe Qb e

Pa e P Q abe Qb e

ax

2

2

2 2 2 2 ax

we have,

aby=ab Pe

d y dyThus, a+b

dx dx

+abPe

0

bx

ax bx ax bx bx bx

Also

Qe

aby

Pa e Qb e Pa e P Q abe Qb e abQe

II | Mathematics


20

19. Consider the detrminant

1+a 1 1

1 1+b 1

1 1 1+c

Taking abc common outside, we have

1 1 1+1

a b c

1 1 1+1

a b c

1 1 1+1

a b c

abc

1 1 2 3Apply the transformation, C ,

1 1 1 1 11 +

a b c b c

1 1 1 1 11 + +1

a b c b c

1 1 1 1 11 + +1

a b c b c

C C C

abc

1 11

b c

1 1 1 1 11 + 1 +1

a b c b c

1 11 +1

b c

abc

2 2 1 3 3 1Apply the transformations, R and R

1 11

b c1 1 1

1 + 0 1 0a b c

0 0 1

R R R R

abc

1Expanding along C , we have

1 01 1 1= abc 1 + 1

a b c 0 1

1 1 1= abc 1 +

a b c

abc ab bc ca

II | Mathematics


21

20. 2cos 3 2cos x t t

2

and

y=sin 3 2sint t

dyWe need to find :

dx

dy

dx

dy

dtdx

dt

2

2

2 2

Let us find :

cos 3 2cos

cos 4cos sin 3 2cos sin

3sin 4cos sin 2cos sin

dx

dt

x t t

dxt t t t t

dt

dxt t t t t

dt

2

2

2 2

Let us find :

sin 3 2sin

sin 4sin cos 3 2sin cos

3cos 4sin cos 2sin cos

dy

dt

y t t

dyt t t t t

dt

dyt t t t t

dt

2 2

2 2

2

2

2

2

3cos 4sin cos 2sin cosThus,

3sin 4cos sin 2cos sin

3cos 6sin cos

3sin 6cos sin

3cos 1 2sin

3sin 1 2cos

dy t t t t t

dx t t t t t

dy t t t

dx t t t

t tdy

dx t t

2

2

2 2

3cos 1 2sin

3sin 2cos 1

cos 2cos 1 1 2sin

sin

t tdy

dx t t

dy tt t

dx t

II | Mathematics


22

4

cot

cot 14

t

dyt

dx

dy

dx

21. Consider the differential equation,

dy

log 3 4dx

x y

dylog

3 4dx

3 4

3 4

3

4

Taking exponent on both the sides, we have

e

dy

dx

dy

dx

dy

x y

x y

x y

x

y

e

e

e e

e dxe

3

4

4 3

3 4

Integration in both the sides, we have

dy

4 3

We need to find the particular solution.

We have, y=0, when x=0

1 1

4 3

1 1

4 3

3 4 7

12 12

7Thus, the solution is

3 4 12

x

y

y x

x y

e dxe

e eC

C

C

C

e e

II | Mathematics


23

22. 1The equation of line L :

2

1 2

1 7 14 3

3 2

1 2 3....(1)

3 2

7

The equation of line L :

7 7 5 6

3 1 5

1 5 6....(2)

3 1 5

7

Since line L and L are perpendicular to each other, we have

x y z

p

x y z

p

x y z

p

x y z

p

3

3 1 2 5 07 7

910

7 7

10 70

7

p p

p p

p

p

1 2

1

Thus equations of lines L and L are:

1 2 3

3 1 2

1 5 6

3 1 5

Thus the equation of the line passing through the point 3, 2, 4

and parallel to the line L is:

3 2 4

3 1 2

x y z

x y z

x y z

II | Mathematics


24

SECTION – C

23. Equation of the plane passing through the intersection

of the planes x+y+z=1 and 2x+3y+4z=5 is :

1 2 3 4 5 0

1 2 1 3 1 4 1 5 0

This plane has to be perpendicular to the plane x-y+z=0.

x y z x y z

x y z

Thus,

1 2 1 1 3 1 1 4 1 0

1 2 1 3 1 4 0

1 3 0

1

3

Thus, the equation of the plane is :

1 1 1 11 2 1 3 1 4 1 5 0

3 3 3 3

2 4 51 1 1 1 1 0

3 3 3

20

3 3 3

2

Thus, the distance of this

x y z

x y z

x z

x z

2 2 2

plane form the origin is :

2 22

21 0 1

II | Mathematics


25

OR

Any point in the line is

2+3 , 4 4 ,2 2

The vector equation of the plane is given as

r 2 0

Thus the cartesian equation of the plane is x 2 0

i j k

y z

Since the point lies in the plane

2+3 1 4 4 2 2 2 1 0

2 8 2 3 8 2 0

12 3 0

12 3

4

2 2 2

Thus, the point of intersection of the line and the

plane is:2+3 4, 4 4 4,2 2 4

14,12,10

Distance between 2, 12, 5 and 14, 12, 10 is:

d= 14 2 12 12 10 5

144 25

169

13

d

d

d units

24. Consider the vertices, A 1, 2 ,B 1, 5 and C 3, 4 .

Let us find the equation of the sides of ABC.

Thus, the equation of AB is:

y 2 x 1

5 2 1 1

3x 2y 7 0

Similarly, the equation of BC is:

y 5 x 1

4 5 3 1

x 2y 11 0

Also,the equation of CA is:

y 4 x 3

2 4 1 3

x 2y

5 0

II | Mathematics


26

x

y

(-1,2)

(1,5)

(3,4)

A

B

C

D

1

1

3

1

1

1

Now the area of ABC=Area of ADB + Area of BDC

3x 7 x 5Area of ADB= dx

2 2

11 x x 5Similarly, Area of BDC= dx

2 2

Thus, Area of ADB+Area of BDC

3x 7 x 5 11 x x 5= dx

2 2 2 2

3

1

1 3

1 1

1 3

1 1

1 32 2

1 1

dx

2x 2 6 2xdx dx

2 2

x 1 dx 3 x dx

x xx 3x

2 2

9 12 9 3

2 2

9 52

2 2

4 square units

II | Mathematics


27

25. Let x be the number of pieces manufactured of type A

and y be the number of pieces manufactured of type B.

Let us summarise the data given in the problem as follows:

Time for Time for Maximumlabour

Product Fabricating Finishinghours available

(in hours) (in hours)

Type A 9 1 180

Type B 12 3 30

Maximum80 120

Profit(in Rupees)

Thus,the mathematical form of above L.P.P. is

Maximise Z = 80x + 120y

subject to

9x + 12y 180

x + 3y 30

Also, we have x 0, y 0

Let us now find the feasible region, which is the set of all

points whose coordinates satisfy all constraints.

Consider the following figure.

II | Mathematics


28

Thus, the feasible region consists of the points A, B and C.

The values of the objective function at the corner points are

given below in the following table:

Points Value of Z

A 12, 6 Z = 80 12+120 6 Rs. 1680

B 0, 10 Z = 80 0+120 10 Rs. 1200

C 20, 0 Z = 80 20+120 0 Rs. 1600

Clearly,Z is maximum at x = 12 and y = 6 and the maximum profit is Rs. 1680.

II | Mathematics


www.topperlearning.com 29

26. 1 2 3Let E ,E ,E and A be the events defined as follows:

1

2

3

1 2 3

1 2 3

E Choosing 2 headed coin

E Choosing coin with 75% chance of getting heads

E Choosing coin with 40% chance of getting heads

A= Getting heads

1Then P E P E P E

3

75 3 40 2Also,P A /E 1,P A /E ,P A /E

100 4 100 5

1

1 1

1 1 2 2 3 3

Required probability

=P E / A

P E P A /E

P E P A /E P E P A /E P E P A /E

11

31 1 3 1 2

13 3 4 3 5

1

31 1 2

3 4 15

1203

43 43

60

II | Mathematics


30

OR

If 1 is the smallest number,

the other numbers are:2,3,4,5,6


the other numbers are:3,4,5,6


the other numbers are:4,5,6


the other numbers are:5,6


the other number is:6

12,13,14,15,16

23,24,25,26

Thus, the sample space is S= 34,35,36

45,46

56

i i

Thus, there are 15 set of numbers in the sample space.

Let X be

X :2 3 4 5 6

1 2 3 4 5P X :

15 15 15 15 15

We know that,

E X X P X

1 2 3 4 52 3 4 5 6

15 15 15 15 15

2 6 12 20 30

15

70

15

4.66

II | Mathematics


31

27. From the given data, we write the following equations:

3

x y z 2 1600

1

4

x y z 1 2300

3

1

x y z 1 900

1

From above system, we get:

3x+2y+z=1600

4x+y+3z=2300

x+y+z=900

Thus we get:

3 2 1 x 1600

4 1 3 y 2300

1 1 1 z 900

-1

11 12 13

21 22 23

31 3

This is of the form

3 2 1 x 1600

AX=B, where A= 4 1 3 ;X y and B= 2300

1 1 1 z 900

3 2 1

A 4 1 3 3 1 3 2 4 3 1 4 1 6 2 3 5 0

1 1 1

We need to find A :

C 2;C 1;C 3

C 1;C 2;C 1

C 5;C

2 335;C 5

T

-1

2 1 3 2 1 5

Therefore, adj A= 1 2 1 1 2 5

5 5 5 3 1 5

2 1 5adjA 1

Thus, A 1 2 5A 5

3 1 5

II | Mathematics


32

-1Therefore,X A B

x 2 1 5 16001

y 1 2 5 23005

z 3 1 5 900

x 2 1600 1 2300 5 9001

y 1 1600 2 2300 5 9005

z 3 1600 1 2300 5 900

x 10001

y 15005

z 2000

x 200

y 300

z 400

Awards can be given for discipline.

28. Let ABC be the right angled triangle with base b and hypotenuse h.

2 2

2 2 2 2

222 2

22 2 2 2

22 2

2 2 32

Given that b + h = k

Let A be the area of the right triangle.

1A= b h b

2

1A b h b

4

bA k b b h k b

4

bA k b 2kb b

4

bA k 2kb

4

b k 2kbA

4

2 2

Differentiating the above function w.r.t. x, we have

dA 2bk 6kb2A ....(1)

db 4

II | Mathematics


33

2 2dA bk 3kb

db 2A

For the area to be maximum, we have

dA0

db

2 2

2

2 2 2

2

bk 3kb 0

bk 3b

kb

3

Again differentiating the function in equation (1), with

respect to b, we have

dA d A 2k 12kb2 2A ....(2)

db db 4

22

2

2 2 2

2

dA kNow substituting 0 and b in equation (2), we have

db 3

k2k 12k

d A 32A

db 4

d A 6k 12k2A

db 12

2 2

2

2 2

2

1

d A k2A

db 2

d A k0

db 4A

kThus area is maximum at b= .

3

k 2kNow, h=k

3 3

Let be the angle between the base of the triangle

and the hypotenuse of the right triangle.

kb 13Thus, cos =

2kh 2

3

1cos

2

3

II | Mathematics


34

29. 4 2 2 4

dxWe need to evaluate

sin x sin xcos x cos x

4 2 2 4

4

dxLet I=

sin x sin xcos x cos x

Multiply the numerator and the

denominator by sec x, we have

4

4 2

2 2

4 2

2 2

2 2

4 2

2

2

2 4

sec xdxI=

tan x tan x 1

sec x sec xdxI=

tan x tan x 1

We know that sec x 1 tan x

Thus,

1+tan x sec xdxI=

tan x tan x 1

Now substitute t=tanx;dt=sec xdx

Therefore,

1+t dtI=

1 t t

2 2

2 4 2 2

Let us rewrite the integrand as

1+t 1+t

1 t t t t 1 t t 1

2

2 4 2 2

2 2 2

2 4 2 2

Using partial fractions, we have

1+t At+B Ct+D

1 t t t t 1 t t 1

1+t At+B t t 1 Ct+D t t 1

1 t t t t 1 t t 1

2

2 4

3 2 2 3 2 2

2 2

1+t

1 t t

At At At Bt Bt B Ct Ct Ct Dt Dt D

t t 1 t t 1

II | Mathematics


35

2

2 4

3 2

2 2

1+t

1 t t

t A C t A B C D t A B C D B D

t t 1 t t 1

So we have,

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

Solving the above equations, we have

1A=C=0 and B=D=

2

2

2 4

2 2

2 2

2 2

1 2

1 22 2

1+t dtI

1 t t

1 1dt

2 t t 1 2 t t 1

dt dt

2 t t 1 2 t t 1

1 dt 1 dt

2 t t 1 2 t t 1

I I

1 dt 1 dtwhere, I and I

2 t t 1 2 t t 1

1

1 2

2

Consider I :

1 dtI

2 t t 1

1 dt

1 12 t t 14 4

2

1

1

1

1 dt

2 1 3t

2 4

1t

1 1 2tan2 3 3

4 4

1 2t 1tan

3 3

1 2tan x 1tan

3 3

II | Mathematics


36

2

2 2

2

2

Similarly,

Consider I :

1 dtI

2 t t 1

1 dt

1 12 t t 14 4

1 dt

2 1 3t

2 4

1

1

1

1 2

1 1

1 1

1t

1 1 2tan2 3 3

4 4

1 2t 1tan

3 3

1 2tan x 1tan

3 3

Thus,I = I I

1 2tan x 1 1 2tan x 1I tan tan

3 3 3 3

1 2tan x 1 2tan x 1I tan tan C

3 3 3

II | Mathematics

Paper 2013 – Delhi Set 1

1

CBSE Board


Board Paper 2013

Delhi Set – 1 Time: 3 hrs Total Marks: 100



2. The question paper consists of 29 questions divided into three Section A, B and C.

Section A comprises of 10 questions of one mark each, Section B comprises of 12

questions of four marks each and Section C comprises of 7 questions of six marks each.

3. All questions in section A are to be answered in one word, one sentence or as per the

exact requirement of the question.

4. There is no overall choice. However, internal choice has been provided in 4 questions

of four marks each and 2 questions of six marks each. You have to attempt only one of

the alternatives in all such questions.

5. Use of calculators is not permitted. You may ask for logarithmic tables, if required.

SECTION – A

1. Write the principal value of 1-1 -1tan 1 + cos .2

2. Write the value of tan 1-12tan5

3. Find the value of a if a b 2a + c 1 5

=2a b 3c + d 0 13

4. If x 1 x 1 4 1

x 3 x 2 1 3

, then write the value of x.

5. If 9 1 4 1 2 1

A2 1 3 0 4 9

, then find the matrix A.

6. Write the degree of the differential equation

2 423

2

d y dyx 0

dxdx

7. If a xi 2j zk and b 3i yj k are two equal vectors, then write the value

of x + y + z.

II | Mathematics


2

8. If a unit vector a makes angle 3

with i , with j

4

and acute angle with k , then find

the value of .

9. Find the Cartesian equation of the line which passes through the point (-2, 4, -5) and is

parallel to the line x 3 4 y z 8

3 5 6.

10. The amount of pollution content added in air in a city due to x-diesel vehicles is given

by P(x) = 0.005 x3 + 0.02 x2 + 30x. Find the marginal increase in pollution content

when 3 diesel vehicles are added and write which value is indicated in the above

questions.

SECTION – B

11. Show that the function f in A = R - 2

3 defined as f(x) =

4x 3

6x 4 is one-one and onto.

Hence find f-1.

12. Find the value of the following: 2

1 1

2 2

1 2x 1 ytan sin cos , x 1,y 0 and xy 1.

2 1 x 1 y

OR

Prove that 1 1 11 1 1tan tan tan

2 5 8 4

13. Using properties of determinants prove the following:

21 x x2

2 3x 1 x 1 x

2x x 1

14. Differentiate the following function with respect to x:

(log x)x + xlog x

15. If y = log 2 2x x a , show that (x2 + a2) 2

2

d y dyx 0.

dxdx

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3

16. Show that the function f x x 3 ,x ,R is continuous but not differentiable at x = 3.

OR

If x = a sin t and y = a t

cost log tan2

find 2

2

d y

dx

17. Evaluate: sin x a

dxsin x a

OR

Evaluate: 2

5x 2dx

1 2x 3x

18. Evaluate: 2

2 2

xdx

x 4 x 9

19. Evaluate: 4

0

| x| | x 2| | x 4| dx

20. If a and b are two vectors such that a b a , then prove that vector 2a b is

perpendicular to vector b .

21. Find the coordinates of the point, where the line x 2 y 1 z 2

3 4 2 intersects the plane

x – y + z – 5 = 0. Also find the angle between the line and the plane.

OR

Find the vector equation of the plane which contains the line of intersection of the

planes r. i 2j 3k 4 0 and r. 2i j k 5 0 and which is perpendicular to the

plane r. 5i 3j 6k 8 0.

22. A speaks truth in 60% of the cases, while B in 90% of the cases. In what percent of

cases are they likely to contradict each other in stating the same fact? In the cases of

contradiction do you think, the statement of B will carry more weight as he speaks

truth in more number of cases than A?

II | Mathematics


4

SECTION – C

23. A school wants to award its students for the values of Honesty, Regularity and Hard

work with a total cash award of Rs 6,000. Three times the award money for Hard work

added to that given for honesty amounts to Rs 11,000. The award money given for

Honesty and Hard work together is double the one given for Regularity. Represent the

above situation algebraically and find the award money for each value, using matrix

method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest

one more value which the school must include for awards.

24. Show that the height of the cylinder of maximum volume, which can be inscribed in a

sphere of radius R is 2R.3

Also find the maximum volume.

OR

Find the equation of the normal at a point on the curve x2 = 4y which passes through

the point (1, 2). Also, find the equation of the corresponding tangent.

25. Using integration, find the area bounded by the curve x2 = 4y and the line x = 4y – 2.

OR

Using integration, find the area of the region enclosed between the two circles

x2 + y2 = 4 and (x – 2)2 + y2 = 4.

26. Show that the differential equation 2yex/y dx + (y – 2x ex/y) dy = 0 is homogeneous.

Find the particular solution of this differential equation, given that x = 0 when y = 1.

27. Find the vector equation of the plane passing through three points with position vectors

i j 2k,2i j k and i 2j k. Also, find the coordinates of the point of intersection of

this plane and the line r 3i j k 2i 2j k .

28. A cooperative society of farmers has 50 hectares of land to grow two crops A and B.

The profits from crops A and B per hectare are estimated as Rs 10,500 and Rs 9,000

respectively. To control weeds, a liquid herbicide has to be used for crops A and B at

the rate of 20 litres and 10 litres per hectare, respectively. Further not more than 800

litres of herbicide should be used in order to protect fish and wildlife using a pond

which collects drainage from this land. Keeping in mind that the protection of fish and

other wildlife is more important than earning profit, how much land should be

allocated to each crop so as to maximize the total profit? Form an LPP from the above

and solve it graphically. Do you agree with the message that the protection of wildlife is

utmost necessary to preserve the balance in environment?

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5

29. Assume that the chances of a patient having a heart attack is 40%. Assuming that a

meditation and yoga course reduces the risk of heart attack by 30% and prescription of

certain drug reduces its chance by 25%. At a time a patient can choose any one of the

two options with equal probabilities. It is given that after going through one of the two

options, the patient selected at random suffers a heart attack. Find the probability that

the patient followed a course of meditation

II | Mathematics


1

CBSE Board


Board Paper 2013 Solution

Delhi Set – 1

SECTION – A

1. Let 1tan 1 y

tan y = 1 = tan π

4

π

y =4

π-1tan 1 =4

1-1cos = z2

cos z = 1 π

= cos =2 3

π 2π

cos π = cos3 3

2

z3

1 2π-1cos =2 3

tan-1 (1) + cos-1 1 π 2π 11π

= + =2 4 3 12

2. We know: 2tan-1 x = tan-1 2

2x

1 - x

1 221 55-1 -1 -1 -152tan = tan = tan = tan

2 245 1211- 25

5

1 5 5-1 -1tan 2tan = tan tan =5 12 12

3. a b 2a +c 1 5

=2a b 3c +d 0 13

Equating the corresponding elements, we get,

a - b = -1, 2a + c = 5, 2a - b = 0, 3c + d = 13

Now, consider the equations:

a - b = -1 and 2a - b = 0

Subtracting first equation from second, we get: a = 1

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2

4. x 1 x 1 4 1

x 3 x 2 1 3

(x + 1) (x + 2) - (x - 1) (x - 3) = 12 + 1

x2 + 3x + 2 - x2 + 4x - 3 = 13

7x - 1 = 13

7x = 14

x = 2

5. 9 1 4 1 2 1

A2 1 3 0 4 9

9 1 4 1 2 1A

2 1 3 0 4 9

9 1 1 2 4 1A

2 0 1 4 3 9

8 3 5A

2 3 6

6.

2 423

2

d y dyx 0

dxdx

We know that the degree of a differential equation is the highest power (exponent) of

the highest order derivative in it.

The highest order derivative present in the given differential equation is 2

2

d y

dx. Its

power is 2. So, the degree of the given differential equation is 2.

7. Given, a xi 2j zk and b 3i yj k are equal vectors.

xi 2j z 3i yj

x 3, y 2, z 1

x y z 3 2 1 0

k k

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3

8. Let a xi yj zk be the unit vector.

1x cos

3 2

1y cos

4 2

2 2 2

222

2

Now, x y z 1

1 1z 1

2 2

1 1 1z 1

4 2 4

1z

2

1cos

2 3

9. The equation of the given line is:

x 3 4 y z 8

3 5 6

x 3 y 4 z 8i.e.,

3 5 6

The required line is parallel to the given line. Therefore, direction ratios of the

required line are same as the direction ratio of the given line. So, the direction ratios of

the required line are 3, -5, and 6.

Thus, the equation of the straight line passing through (-2, 4, -5) and having direction

ratios 3, -5, 6 is

x 2 z 5y 4

3 5 6

x 2 4 y z 5i.e.,

3 5 6

10. P(x) = 0.005x3 + 0.02x2 + 30x

Differentiating w.r.t. x,

Marginal increase in pollution content = 2dP x0.015x 0.04x 30

dx

Putting x = 3 in (1),

x 3

dp x0.015 9 0.04 3 30 30.255

dx

Therefore, the value of marginal increase pollution content is 30.255.

Increase in number of diesel vehicles increases the pollution. We should aim at saving

the environment by reducing the pollution by decreasing the vehicle density on road.

II | Mathematics


4

SECTION – B

11. f(x) = 4x 3

6x 4

Let f(x1) = f(x2)

1 2

1 2

4x 3 4x 3

6x 4 6x 4

24x1x2 – 16x1 + 18x2 – 12 = 24x1x2 + 18x1 – 16x2 – 12

18x2 + 16x2 = 18x1 + 16x1

34x2 = 34x1

x1 = x2

Since, 4x 3

6x 4 is a real number, therefore, for every y in the co-domain of f, there exists

a number x in R - 2

3 such that f(x) =

4x 3y

6x 4

Therefore, f(x) is onto.

Hence, f-1 exists.

Now, let y = 4x 3

6x 4

6xy – 4y = 4x + 3

1 1

6xy 4x 4y 3

x 6y 4 4y 3

4y 3x

6y 4

4x 3y int erchanging the variables x and y

6x 4

4x 3f x put y=f x

6x 4

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5

12.

We know that:

1 1

2

2xsin 2tan x for x 1 ... 1

1 x

21 1

2

1 ycos 2tan y for y 0 ... 2

1 y

21 1 1 1

2 2

2x 1 ysin cos 2tan x 2tan y.

1 x 1 y

2

1 1

2 2

1 2x 1 ytan sin cos

2 1 x 1 y 1

tan2

(2 tan-1x + 2 tan-1y)

= tan (tan-1 x + tan-1 y)

1 1 1 1x y x ytan tan tan x + tan y =tan ,for xy < 1

1 xy 1 xy

x y

1 xy

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6

OR We know that:

tan-1 x + tan-1 y = tan-1 x y

,xy 11 xy

We have:

1 1 11 1 1tan tan tan

2 5 8

1 1 1

1 1

1 1 1tan tan tan

2 5 8

1 11 1 12 5tan tan 1

1 1 8 2 512 5

1 1

1

1

1 1

1 1 1

7 1tan tan

9 8

7 1

9 8tan7 1

19 8

56 9 7 1tan 1

72 7 9 8

65tan tan 1

65 4

1 1 1Hence, tan tan tan

2 5 8 4

II | Mathematics


7

13.

2

2

2

1 x x

x 1 x

x x 1

Applying R1 → R1 + R2 + R3, we have:

2 2 2

2

2

2 2

2

1 x x 1 x x 1 x x

x 1 x

x x 1

1 1 1

1 x x x 1 x

x x 1

Applying C2 → C2 − C1 and C3 → C3 − C1, we have:

2 2 2 2

2

1 0 0

1 x x x 1 x x x

x x x 1 x

2 2

3 2

1 0 0

1 x x 1 x 1 x x 1 x x

x x 1

1 0 0

1 x 1 x x 1 x x

x x 1

Expanding along R1, we have:

3

3 2

3 3

23

1 x x1 x 1 x 1

x 1

1 x 1 x 1 x x

1 x 1 x

1 x

Hence proved.

II | Mathematics


8

14. Let y = (log x)x + xlog x ...(1)

Now let y1 = (log x)x and y2 = xlog x

y = y1 + y2 ...(2)

Differentiating (2) w.r.t. x,

1 2dy dydy

dx dx dx ...(3)

Now consider y1 = (log x)x

Taking log on both sides,

log y1 = x log (log x)

Differentiating w.r.t. x, we get

1

1

11

x1

dy1 1 1x 1 log log x

y dx log x x

dy 1y log log x

dx log x

dy 1log x log log x ...(4)

dx log x

Now, consider y2 = xlog x

log y2 = (log x) (log x) = (log x)2


2

2

logx22

dy1 12 log x

y dx x

dy 2log x 2log xy x ....(5)

dx x x

Using equations (3), (4) and (5), we get:

x logxdy 1 2log xlog x log log x x

dx log x x

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9

15. y = log 2 2x x a ...(1)

y = log 2 2x x a ...(1)

Differentiating (1) w.r.t. x, we get

2 2

2 2

2 2

x1

dy x a

dx x x a

dy 1...(2)

dx x a

2 2

dy xx

dx x a ...(3)

Again, differentiating (2) w.r.t. x, we get

12 2 22

2 2 2

2

2 32 2 2

22 2

2 2 2

2x

2 x ad y

dx x a

d y x

dxx a

d y xx a ...(4)

dx x a

Adding equation (3) and (4), we get 2

2 2

2 2 2 2 2

22 2

2

d y dy x xx a x 0

dxdx x a x a

d y dyx a x 0

dxdx

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10

16. f(x) = 3 x, x 3

x 3x 3, x 3

Let c be a real number.

Case I: c < 3. Then f(c) = 3 – c.

limx c

f(x) = limx c

(3 – x) = 3 – c

Since, limx c

f(x) = f(c), f is continuous at all negative real numbers.

Case II: c = 3. Then f(c) = 3 – 3 = 0

limx c

f(x) = limx c

(x – 3) = 3 – 3 = 0

Since, limx c

f(x) = f(3), f is continuous at x = 3.

Case III: c > 3. Then f(c) = c – 3.

limx c

f(x) = limx c

(x – 3) = c – 3.

Since, limx c

f(x) = f(c), f is continuous at all positive real numbers.

Therefore, f is continuous function.

Now, we need to show that f(x) = x 3 , x R is not differentiable at x = 3.

Consider the left hand limit of f at x = 3

h 0 h 0 h 0 h 0

f 3 h f 3 3 h 3 3 3 h 0 hlim lim lim lim 1

h h h h

h 0 |h| h

Consider the right hand limit of f at x = 3

h 0 h 0 h 0 h 0

f 3 h f 3 |3 h 3| |3 3| |h| 0 hlim lim lim lim 1

h h h h

h 0 |h| h

Since the left and right hand limits are not equal, f is not differentiable at x = 3.

II | Mathematics


11

OR

y = a t

cost log tan2

2

dy d d ta cost log tan

dt dt dt 2

t t 1a sint cot sec

2 2 2

1a sint

t t2sin cos

2 2

2 21 sin t 1 cos ta sint a a

sint sint sint

x = a sin t

2

22 2

2 2

dx da sin t acost

dt dt

cos tdy asin tdy costdt

cot tdxdx acost sin t

dt

d y dt 1 1cosec t cosec t

dx acostdx a sin t cost

II | Mathematics


12

17. sin x a

I dxsin x a

Let (x + a) = t dx = dt

sin t 2aI dt

sin t

sin t cos2a cost sin2adt

sin t

cos2a cot t sin2a dt

cos2a t sin2a log sin t C

cos2a x a sin2a log sin x a C

OR

2

2

2

2

2

2 2

2

2

5x 2dx

1 2x 3x

2x

55 dx1 2x 3x

126x

5 5 dx6 1 2x 3x

126x 2 2

5 5 dx6 1 2x 3x

226x 2

5 5 dx6 1 2x 3x

5 6x 2 5 22 1dx dx

6 6 51 2x 3x 1 23 x

3 9

5 11 1log 1 2x 3x

6 9 1 2x

3

2 1

2 1

dx

9

1x

5 11 3 3log 1 2x 3x tan C

6 9 2 2

3

5 11 3x 1log 1 2x 3x tan C

6 3 2 2

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13

18. 2Let x y

2

2 2

x y A B

y 4 y 9 y 4 y 9x 4 x 9

y A(y 9) B(y 4)

Comparing both sides,

A B 1 and 9A 4B 0

4 9Solving, we get A and B

5 5

2 2

1 1

1 1

4 9I dx

5 x 4 5 x 9

4 1 x 9 1 xtan tan C

5 2 2 5 3 3

2 x 3 xtan tan C

5 2 5 3

19. 4

0

| x| | x 2| | x 4|dx

2 4

0 2

f x dx f x dx

2 4

0 2

x x 2 x 4 dx x x 2 x 4 dx

2 4

0 2

2 42 2

0 2

6 x dx x 2 dx

x x6x 2x

2 2

12 2 8 8 2 4 20

II | Mathematics


14

20.

a b a

2 2

2 2 2

2

a b a

a 2a b b a

2a b b 0 ... 1

2

Now, 2a b b 2a b b b 2a b b 0 [Using (1)]

We know that if the dot product of two vectors is zero, then either of the two vectors is zero or the vectors are perpendicular to each other.

Thus, 2a b is perpendicular to b .

21. The equation of the given line is x 2 y 1 z 2

3 4 2 ....(1)

Any point on the given line is (3 + 2, 4 - 1, 2 + 2). If this point lies on the given plane x – y + z – 5 = 0, then 3 + 2 – (4 - 1) + 2 + 2 – 5 = 0

= 0 Putting = 0 in (3 + 2, 4 - 1, 2 + 2), we get the point of intersection of the given line and the plane is (2, -1, 2). Let be the angle between the given line and the plane.

2 2 2 2 2 2

1

a.b (3i 4j 2k).(i j k) 3 4 2 1sin

29 3 87|a||b| 3 4 2 1 1 1

1sin

87

Thus, the angle between the given line and the given plane is 1 1sin

87.

II | Mathematics


15

OR The equation of the given planes are

r i 2j 3k 4 0 ... 1

r 2i j k 5 0 ... 2

The equation of the plane passing through the intersection of the planes (1) and (2) is

r i 2j 3k 4 r 2i j k 5 0

r 1 2 i 2 j 3 k 4 5 ... 3

Given that plane (3) is perpendicular to the plane r 5i 3j 6k 8 0.

1 2 5 2 3 3 6 0

19 7 0

7

19

Putting 7

19 in (3), we get

14 7 7 35r 1 i 2 j 3 k 4

19 19 19 19

33 45 50 41r i j k

19 19 19 19

r. 33i 45j 50k 41 …..This is the equation of the required plane.

22. Let the probability that A and B speak truth be P(A) and P(B) respectively.

Therefore, 60 3 90 9

P A andP B .100 5 100 10

A and B can contradict in stating a fact when one is speaking the truth and other is not

speaking the truth.

Case 1: A is speaking the truth and B is not speaking the truth.

Required probability = 3 9 3

P A 1 P B 1 .5 10 50

Case 2: A is not speaking the truth and B is speaking the truth.

Required probability = 3 9 9

1 P A P B 1 .5 10 25

Percentage of cases in which they are likely to contradict in stating the same fact =

3 9 3+18100% 100% 42%

50 25 50

From case 1, it is clear that it is not necessary that the statement of B will carry more

weight as he speaks truth in more number of cases than A.

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16

SECTION – C

23. Let the award money given for honesty, regularity and hard work be Rs. x, Rs. y and

Rs. z respectively.

Since total cash award is Rs. 6,000.

x + y + z = 6,000 ...(1)

Three times the award money for hard work and honesty amounts to Rs. 11,000.

x + 3z = 11,000

x + 0×y + 3 z = 11,000 ...(2)

Award money for honesty and hard work is double that given for regularity.

x + z = 2y

x – 2y + z = 0 ...(3)

The above system of equations can be written in matrix form AX = B as:

1 1 1 x 6000

1 0 3 y 11000

1 2 1 z 0

Here,

1 1 1 x 6000

A 1 0 3 , X y and B 11000

1 2 1 z 0

A 1(0 6) 1(1 3) 1( 2 0) 6 0

Thus, A is non-singular. Hence, it is invertible.

6 3 3

Adj A 2 0 2

2 3 1

1

1

6 3 31 1

A adjA 2 0 2A 6

2 3 1

6 3 3 6000 36000 33000 0 30001 1 1

X A B 2 0 2 11000 12000 0 0 120006 6 6

2 3 1 0 12000 33000 0 21000

x 500

y 2000

z 3500

Hence, x = 500, y = 2000, and z = 3500.

Thus, award money given for honesty, regularity and hard work is Rs. 500, Rs. 2000

and Rs. 3500 respectively.

The school can include awards for obedience.

II | Mathematics


17

24. Given, radius of the sphere is R.

Let r and h be the radius and the height of the inscribed cylinder respectively.

We have:

h = 2 22 R r

Let Volume of cylinder = V 2

2 2 2

2 2 2

V r h

r 2 R r

2 r R r

Differentiating the above function w.r.t. r, we have,

2 2 2

32 2

2 2

2 2 3

2 2

2 3 3

2 2

2 3

2 2

V 2 r R r

dV 4 r4 r R r

dr 2 R r

4 r R r 4 r

2 R r

dV 4 rR 4 r 2 r

dr 2 R r

4 rR 6 r

2 R r

For maxima or minima, 2 3dV0 4 rR 6 r 0

dr 3 2

22

6 r 4 rR

2Rr

3 2 3

2 2

dV 4 rR 6 r

dr 2 R r

Now,

2 2 2 2 2 3

2 2 2

2 2 2

2rR r 4 R 18 r 4 rR 6 r

d V 1 2 R r2dr R r

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18

2 2 2 2 2 3

32 2 2

4 2 2 4 2 2

32 2 2

R r 4 R 18 r r 4 rR 6 r1

2R r

1 4 R 22 r R 12 r 4 r R

2R r

Now, when r2 = 2 2

2

2R d V, 0.

3 dr

Volume is the maximum when r2 = 22R

.3

When 2

2 2Rr ,

3 h =

2 22 2R R 2R

2 R 2 .3 3 3

Hence, the volume of the cylinder is the maximum when the height of the cylinder is

2R.

3

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19

OR

The equation of the given curve is x2 = 4y.


dy x

dx 2

Let (h, k) be the co-ordinates of the point of contact of the normal to the curve x2 = 4y.

Now, slope of the tangent at (h, k) is given by

h,k

dy h

dx 2

Hence, slope of the normal at 2

h,kh

Therefore, the equation of normal at (h, k) is

2y k x h

h ...(1)

Since, it passes through the point (1, 2) we have

2 22 k 1 h or k 2 1 h ... 2

h h

Now, (h, k) lies on the curve x2 = 4y, so, we have:

h2 = 4k ...(3)

Solving (2) and (3), we get,

h = 2 and k = 1.

From (1), the required equation of the normal is:

2y 1 x 2 or x y 3

2

Also, slope of the tangent = 1

Equation of tangent at (1, 2) is:

y – 2 = 1(x – 1) or y = x + 1

II | Mathematics


20

25. The shaded area OBAO represents the area bounded by the curve x2 = 4y and line x = 4y – 2.

Let A and B be the points of intersection of the line and parabola.

Co-ordinates of point A are 1

1, .4

Co-ordinates of point B are (2, 1).

Area OBAO = Area OBCO + Area OACO … (1) Area OBCO =

22 2

0 0

2 22 3

0 0

x 2 xdx dx

4 4

1 x 1 x2x

4 2 4 3

1 1 82 4

4 4 3

3 2 5

2 3 6

Area OACO =

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21

20 0

1 1

0 02 3

1 1

2 3

x 2 xdx dx

4 4

1 x 1 x2x

4 2 4 3

1 11 12 1

4 2 4 3

1 1 1 12

4 2 4 3

3 1 7

8 12 24

Therefore, required area = 5 7 9

sq. units6 24 8

OR

Given equations of the circles are 2 2x y 4 ... 1

2 2x 2 y 4 ... 2

Equation (1) is a circle with centre O at the origin and radius 2. Equation (2) is a circle

with centre C (2, 0) and radius 2.

Solving (1) and (2), we have:

(x – 2)2 + y2 = x2 + y2

x2 – 4x + 4 + y2 = x2 + y2

x = 1

This gives y = 3

Thus, the points of intersection of the given circles are A 1, 3 and A' 1, 3 as

shown in the figure.

Required area

= Area of the region OACA'O

= 2 [area of the region ODCAO]

= 2 [area of the region ODAO + area of the region DCAD]

II | Mathematics


22

1 2

0 1

1 22 2

0 1

1 22 1 2 1

10

1 22 1 2 1

10

2 ydx ydx

2 4 x 2 dx 4 x dx

1 1 x 2 1 1 x2 x 2 4 x 2 4sin 2 x 4 x 4sin

2 2 2 2 2 2

x 2 xx 2 4 x 2 4sin x 4 x 4sin

2 2

1 1 1 11 13 4sin 4sin 1 4sin 1 3 4sin

2 2

3 4 4 4 3 46 2 2 6

82 3

3

II | Mathematics


23

26. 2yex/y dx + (y – 2x ex/y) dy = 0

x

y

x

y

dx 2xe y... 1

dy2ye

Let

x

y

x

y

2xe yF x,y

2ye

Then,

x

y

o

x

y

2xe y

F x, y F x,y

2ye

Thus, F(x, y) is a homogeneous function of degree zero. Therefore, the given differential equation is a homogeneous differential equation. Let x = vy Differentiating w.r.t. y, we get dx dv

v ydy dy

Substituting the value of x and dx

dy in equation (1), we get

v v

v v

dv 2vye y 2ve 1v y

dy 2ye 2e

or v

v

dv 2ve 1y v

dy 2e

or v

dv 1y

dy 2e

or v dy2e dv

y

or v dy2e .dv

y

or 2ev = - log |y| + C

Substituting the value of v, we get x

y2e log|y | C ... 2

Substituting x = 0 and y = 1 in equation (2), we get o2e log|1| C C 2

Substituting the value of C in equation (2), we get x

y2e log| y | 2, which is the particular solution of the given differential equation.

II | Mathematics


24

27. Let the position vectors of the three points be,

a i j 2k,b 2i j k and c i 2j k .

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

r a . b c c a 0

r i j 2k . i 3j j 3k 0

r i j 2k . k 3j 9i 0

r. 9i 3j k 14 0

r. 9i 3j k 14 ... 1

So, the vector equation of the required plane is r. 9i 3j k 14.

The equation of the given line is r 3i j k 2i 2j k .

Position vector of any point on the given line is

r 3 2 i 1 2 j 1 k ... 2

The point (2) lies on plane (1) if,

3 2 i 1 2 j 1 k . 9i 3j k 14

9 3 2 3 1 2 1 14

11 25 14

1

Putting 1 in (2), we have

r 3 2 i 1 2 j 1 k

3 2 1 i 1 2 1 j 1 1 k

i j 2k

Thus, the position vector of the point of intersection of the given line and plane (1) is

i j 2k and its co-ordinates are 1, 1, 2 .

II | Mathematics


25

28. Let the land allocated for crop A be x hectares and crop B be y hectares. Maximum area of the land available for two crops is 50 hectares.

x y 50

Liquid herbicide to be used for crops A and B are at the rate of 20 litres and 10 litres per hectare respectively. Maximum amount of herbicide to be used is 800 litres.

20x 10y 800

2x y 80

The profits from crops A and B per hectare are Rs 10,500 and Rs 9,000 respectively. Thus, total profit = Rs (10,500x + 9,000y) = Rs 1500 (7x + 6y) Thus, the linear programming problem is: Maximize Z = 1500 (7x + 6y) subject to the constraints

x y 50 ... 1

2x y 80 ... 2

x 0 ... 3

y 0 ... 4

The feasible region determined by constraints is represented by the shaded region in the following graph:

The corner points of the feasible region are O (0, 0), A (40, 0), B (30, 20) and C (0, 50). The value of Z at these corner points are

Corner point Z = 1500 (7x + 6y) O (0, 0) 0

A (40, 0) 420000 B (30, 20) 495000 Maximum C (0, 50) 450000

The maximum profit is at point B (30, 20). Thus, 30 hectares of land should be allocated for crop A and 20 hectares of land should be allocated for crop B. The maximum profit is Rs 495000. Yes, the protection of wildlife is utmost necessary to preserve the balance in environment.

II | Mathematics


26

29. Let A, E1, and E2, respectively denote the events that a person has a heart attack, the

selected person followed the course of yoga and meditation, and the person adopted

the drug prescription.

P A 0.40

1 2

1

2

1P E P E

2

P A|E 0.40 0.70 0.28

P A|E 0.40 0.75 0.30

Probability that the patient suffering a heart attack followed a course of meditation

and yoga =

1

1 1

1 1 2 2

P E | A

P E P A |E

P E P A |E P E P A |E

10.28

21 1

0.28 0.302 2

28

28 30

28

58

14

29

Now, calculate P(E2|A).

2 22

1 1 2 2

P E P A|EP E | A

P E P A|E P E P A|E

10.30

2 1 1

0.28 0.302 2

30

28 30

30

58

15

29

Since 1 2P E | A P E | A , the course of yoga and meditation is more beneficial for a

patient.

cbse-xii-2018 examination - kopykitabcbse-xii-2018 examination series sgn time : 3 hrs. mathematics...

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