xii hsc - board - 2018...rao iit academy/ xii hsc - board exam 2018 / mathematics / qp + solutions 3...

Rao IIT Academy/ XII HSC - Board Exam 2018 / Mathematics / QP + Solutions

1 1

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SECTION - I

Q.1 (A)

(i) If 2 34 1

A

, then adjoint of matrix A is

(A) 1 34 2

(B) 1 34 2

(C) 1 34 2

(D) 1 34 2

Ans. (A)

2 34 1

A

1 34 2

adjA

Topic: Matrix; Sub-topic:Adjoint__L-1__XII-HSC Board Test_Mathematics

(ii) The principal solutions of 2sec3

x are __________.

(A) 11,

3 6

(B) 11,

6 6

(C) 11,

4 4

(D) 11,

6 4

Ans. (B)2sec3

x

3cos cos cos 2 62 6

x

11,6 6

Topic: Trigo. function; Sub-topic:General solution_L-1 __XII-HSC Board Test_Mathematics

XII HSC - BOARD - 2018Date: 03.03.2018 MATHEMATICS (40) - SOLUTIONS


2 2

(iii) The measure of acute angle between the lines whose direction ratios are 3, 2, 6 and –2, 1, 2 is ______.

(A) 1 1cos7

(B) 1 8cos15

(C) 1 1cos3

(D) 1 8cos21

Ans. (D)

2 2 2 2 2 2

3 2 2 1 6 2cos3 2 6 2 1 2

6 2 12 8 8

7 3 2149 9

1 8cos21

Topic: 3D Geometry; Sub-topic:Angle__L-1 __XII-HSC Board Test_Mathematics

Q.1 (B)(i) Wirte the negations of the following statements :

(a) All students of this college live in the hostel.

(b) 6 is an even number or 36 is a perfect square.

Ans. (a) p : All students of this college live in the hostel.Negation :

:p Some students of this college do not live in the hostel. [1 Mark](b) p : 6 is an even number.

q : 36 is a perfect square.Symbolic form : p q

p q p q Negation :6 is not an even number and 36 is not a perfect square. [1 Mark]

Topic: Logic; Sub-topic:Negation__L-1 __XII-HSC Board Test_Mathematics(ii) If a line makes angles , , with the co-ordinates axes, prove that cos 2 cos cos 2 1 0 .

Ans. L.H.S : cos 2 cos 2 cos 2 1 2 2 22cos 1 2cos 1 2cos 1 1 [1 Mark]

2 2 22 cos cos cos 2

2 1 2 2 2 0 [1 Mark]

= R.H.STopic:3D Geometry; Sub-topic: 3D__L-1 __XII-HSC Board Test_Mathematics


3 3

(iii) Find the distance of the point (1, 2, –1) from the plane 2 4 10 0x y z .

Ans. Distance of the point 1 1 1x y z to plane 0ax by cz d is

1 1 12 2 2

ax by cz dDa b c

[1 Mark]

1 1 1 1,2, 1x y z

1, 2, 4a b c

1 2 2 4 1 10 17 17 units1 4 16 21 21

D

[1 Mark]

Topic:Plane; Sub-topic:Distance__L-1 __XII-HSC Board Test_Mathematics

(iv) Find the vector equation of the lines which passes through the point with position vector ˆˆ ˆ4 2i j k and

is in the direction of ˆˆ ˆ2i j k .

Ans. ˆˆ ˆLet 4 2a i j k

ˆˆ ˆ2b i j k

Equation of the line passing through point A a and having direction b is

r a b [1 Mark]

ˆ ˆˆ ˆ ˆ ˆ4 2 2r i j k i j k [1 Mark]

Topic:Line; Sub-topic:Equation__L-1 __XII-HSC Board Test_Mathematics

(v) If ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ3 2 7 , 5 2 and a i j k b i j k c i j k then find ·a b c .

Ans. 1 1 1

2 2 2

3 3 3

a b ca b c a b c a b c

a b c [1 Mark]

3 2 7

5 1 21 1 1

a b c

= 3(–1 + 2) + 2(–5 + 2) +7(5 – 1)= 3 – 6 + 28= 25 [1 Mark]

Topic: Vector; Sub-topic:Triple dot product__L-1 __XII-HSC Board Test_Mathematics


4 4

Q.2 (A)(i) Using vector method prove that the medians of a triangle are concurrent.

Ans. Let a,b,c,d ,e the position vectors of the vertices A, B, C of ABC and , ,d e f be the position vectorsof the midpoints D, E, F of the sides BC, CA and AB respectively

A

F E

DB C

2

G1

[1 Mark]

Then by the midpoint formula,

, ,2 2 2

b c c a a bd e f

2 ; 2 ;2d b c e c a f a b

2d a a b c

2e b a b c

2 f c a b c [1 Mark]

2 2 2 let2 1 2 1 2 1 3

d a e b f c a b c g

lies on the three medians AD, BE and CF dividing each of them internally in the ratio 2 : 1.Hence, the medians are concurrent at point G . [1 Mark]

Topic: Vector; Sub-topic:Theorem__L-1 __XII-HSC Board Test_Mathematics(ii) Using the truth table, prove the following logical equivalence :

p q p q p q Ans. 1 Mark 1 Mark

p q A B

1 2 3A4

5 6B7

8

By column number 3 and 8

p q p q p q [1 Mark]

Topic: Logic; Sub-topic:Truth table__L- 1__XII-HSC Board Test_Mathematics


5 5

(iii) If the origin is the centroid of the triangle whose vertices are A(2, p, –3), B(q, –2, 5) and R(–5, 1, r), thenfind the values of p, q, r.

Ans. Let , ,a b c be the position vectors of ABC whose vertices are A(2, p, –3), B(q, –2, 5), C(–5, 1, r)ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ 2 3 , 2 5 , 5a i pj k b qi j k c i j rk

Given that origin O is the centroid of ABC

3

a b cO a b c O [1 Mark]

ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ2 3 2 5 5i pj k qj j k i j rk O

ˆ ˆˆ ˆ ˆ ˆ2 5 2 1 3 5 0 0 0q i p j r k i j k [1 Mark]by equality of vectors2 5 0 3q q

2 1 0 1p p

3 5 0 2r r 1, 3 and 2p q r [1 Mark]

Topic:Vector; Sub-topic:Section formula__L-1__XII-HSC Board Test_Mathematics

Q.2 (B)

(i) Show that a homogeneous equation of degree two in x and y, i.e., 2 22 0ax hxy by represents a pair

of lines passing through the origin is 2 0h ab .Ans. Consider a homogenous equation of degree two in x and y

2 2ax 2hxy by 0 ..... i

In this equation at least one of the coefficients a ,b or h is non zero.We consider two casesCase I : If b = 0, then the equation of lines x = 0 and (ax + 2 hy) = 0 [1 Mark]These lines passes through the origin.Case II : b 0 ,Multiplying both the sides of equation (i) by b, we get [1 Mark]

2 2 2abx 2hbxy b y 0 2 2 2b y 2hbxy abx

To make L.H.S a complete square, we add h2x2 on both the sides.2 2 2 2 2 2 2b y 2hbxy h x abx h x

2 2 2by hx h ab x

2

2 2by hx h ab x

2

2 2by hx h ab x 0

2 2by hx h ab x by hx h ab x 0 [1 Mark]


6 6

It is the joint equation of two lines

2 2by hx h ab x 0 and by hx h ab x 0

i.e. 2 2h h ab x by 0 and h h ab x by 0

These lines passes through the origin. [1 Mark]

Topic: Pair of straight line; Sub-topic:Theorem__L-1__XII-HSC Board Test_Mathematics

(ii) In ABC , prove that tan cot2 2

C A c a Bc a

.

Ans. In ABC , by sine Rule,

sin sin sina b c k

A B C

sin , sin , sina k A b k B c k C [1 Mark]Consider,

sin sinsin sin

c a k C k Ac a k C k A

[1 Mark]

sin sinsin sin

C AC A

2cos sin2 2

2sin cos2 2

C A C A

C A C A

[1 Mark]

cot ·tan2 2

C A C A

tan tan2 2B C A

tan cot2 2

C A C a BC a

[1 Mark]

Hence proved.Topic: Trigo. function; Sub-topic:Theorem L-1 __XII-HSC Board Test_Mathematics


7 7

(iii) Find the inverse of the matrix, 1 2 21 3 0

0 2 1A

using elementary row transformations.

Ans.

1 2 2| | 1 3 0

0 2 1A

1 3 2 1 2 2 13 2 4 1 0 exists.A [1 Mark]

We know,1AA I

1

1 2 2 1 0 01 3 0 0 1 0

0 2 1 0 0 1A

[1 Mark]

2 2 1R R R

1

1 2 2 1 0 00 5 2 1 1 00 2 1 0 0 1

A

2 2 32R R R

1

1 2 2 1 0 00 1 0 1 1 20 2 1 0 0 1

A

1 1 2 3 3 22 and 2R R R R R R

1

1 0 2 1 2 40 1 0 1 1 20 0 1 2 2 5

A

[1 Mark]

1 1 32R R R

1

1 0 0 3 2 60 1 0 1 1 20 0 1 2 2 5

A

1

3 2 6 1 1 2

2 2 5A

[1 Mark]

Topic: Matrix; Sub-topic:Inverse__L-1 __XII-HSC Board Test_Mathematics


8 8

Q.3 (A)(i) Find the joing equation of the pair of lines passing through the origin, which are perpendicular to the lines

represented 2 25 2 3 0x xy y .Ans. Given homogeneous equation is

2 25 2 3 0x xy y Which is factorisable

2 25 5 3 3 0x xy xy y

5 3 0x x y y x y

5 3 0x y x y

0 and 5 3 0x y x y are the two lines represented by the given equation.

Their slopes are –1 and 53 [1 Mark]

Required two lines are respectively perpendicular to these lines.

Slopes of required lines are 1 and 35

and the lines pass through origin.

Their individual equations are31· and 5

y x y x

i.e., 0 and 3 5 0x y x y [1 Mark]Their joint equation is

3 5 0x y x y

2 23 3 5 5 0x xy xy y 2 23 2 5 0x xy y [1 Mark]

Topic: Pair of straight line_Sub-topic:Formulation of equation__L-1 __XII-HSC Board Test_Mathematics

(ii) Find the angle between the lines 1 3

4 1 8x y z

and 2 1 4

2 2 1x y z

.

Ans. Let a and b be the vectors in the direction of the lines 1 3

4 1 8x y z

and 2 1 4

2 2 1x y z

respectively.

4 8a i j k and ˆ2 2b i j k

4 2 1 2 8 1 8 2 8 18a b [1 Mark]

and 16 1 64 81 9a

4 1 4 9 3b

Let be the acute angle between the two given lines


9 9

18 2cos9 3 3

a ba b

[1 Mark]

1 2cos3

[1 Mark]

Topic: Line; Sub-topic:Line__L-1 __XII-HSC Board Test_Mathematics(iii) Write converse, inverse and contrapositive of the following conditional statement :

If an angle is a right angle then its measure is 90°.Ans. Converse : If the measure of an angle is 90° then it is a right angle. [1 Mark]

Inverse : If an angle is not a right angle then its measure is not 90°. [1 Mark]Contra positive : If the measure of an angle is not 90° then it is not a right angle. [1 Mark]

Topic:Logic; Sub-topic:Condictional__L-1 __XII-HSC Board Test_MathematicsQ.3 (B)

(i) Prove that : 1 1 13 12 56sin cos sin5 13 65

Ans. 1 12Let cos13

x

12cos13

x

5sin

13x [1 Mark]

and let 1 3sin5

y

3sin5

y

4cos5

y [1 Mark]

using sin sin cos cos sinx y x y x y [1 Mark]

5 4 12 313 5 13 5

20 3613 5

5665

1 56sin

65x y

1 1 112 3 56cos sin sin13 5 65

[1 Mark]

Hence proved.Topic: Trigo. function; Sub-topic:ITF__L-1 __XII-HSC Board Test_Mathematics


1010

(ii) Find the vector equation of the plane passing through the points A(1, 0, 1), B(1, –1, 1) and C(4, –3, 2).Ans. Let the p.v. of points A(1, 0, 1), B(1, –1, 1) and C(4, –3, 2) be

, and 4 3 2a i k b i j k c i j k

, 3 3b a j c a i j k [1 Mark]

0 1 0 33 3 1

i j kb a c a i k

[1 Mark]

Equation of plane through A, B, C in vector form is

· 0r a b a c a [1 Mark]

· 3 0r a i j

· 3 · 3 1 3 2r i j i k i j

· 3 2r i j [1 Mark]

Topic: Plane; Sub-topic:Equation of plane__L-1 __XII-HSC Board Test_Mathematics(iii) Minimize Z = 7x + y subject to

5 5, 3, 0, 0x y x y x y

Ans. Z 7x y Subject to

5x y 5

x y 3

x 0, y 0

Line Inequation Points on x Points on y Feasible region

AB 5 5x y 1,0A 0,5B Non - origin side

CD 3x y 3,0C 0,3D Non - origin side

[1 Mark]

1 unit = 1 cm both axis


1111

B(0, 5)

1 5P ,2 2

P

C(3, 0)A(1, 0)

AB

CD

D(0, 3)

[1 Mark]

common feasible region BPC

Points Minimize 7z x y

0,5B 7 0 5 5Z B

1 5,5 2

P

1 57 62 2

Z P

3,0C 7 3 0 21Z C [1 Mark]

Z is minimum at x = 0, y = 5 and min (z) = 5 [1 Mark]

Topic: LPP; Sub-topic:Graphical solution__L-1 __XII-HSC Board Test_Mathematics

SECTION - IIQ.4 (A) Select and write the appropriate answer from the given alternatives in each of the following

sub-questions : [6](i) Let the p. m. f. of a random variable X be __

3( )10

xP x for 1,0,1,2x

0 otherwiseThen ( )E X is ________.

(A) 1 (B) 2 (C) 0 (D) – 1Ans. (C)

1 0 1 24 3 2 1

10 10 10 104 2 2. 0

10 10 10

x

p x

x p x

. 0x P x Topic: Probability distribution; Sub-topic:Expected value__L-1 __XII-HSC Board Test_Mathematics


1212

(ii) If 20

1 ,2 8 16

k

dxx

then the value of k is ________.

(A) 12 (B)

13 (C)

14 (D)

15

Ans. (A)

20

1162 1 2

k

I dxx

1

0

1 1 tan 22 2 16

kx

1 1tan 2 tan 04

k

2 1k

12

k

Topic:Definite integral; Sub-topic:Definite integral__L-1 __XII-HSC Board Test_Mathematics

(iii) Integrating factor of linear differential equation 22 logdyx y x xdx

is _______.

(A) 2

1x (B)

1x (C) x (D) 2x

Ans. (D)

2 logdy y x xdx x

2Px

22log 2.

dx xxI F e e x

Topic: Diferential equation; Sub-topic:LDE__L-2 __XII-HSC Board Test_Mathematics


1313

Q.4 (B) Attempt any THREE of the following : [6]

(i) Evaluate : 2

cos sinsin

x x xe dxx

Ans. 2 2

cos sinsin sin

x x xI e dxx x

cot cosec cosec'( ) ( )

x x x xe dx

f x f x

[1 Mark]

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

cosxI e ec x C [1 Mark]

Topic: Integration; Sub-topic:Integration__L-1 __XII-HSC Board Test_Mathematics

(ii) If 2 3tan (log ),y x find .dydx

Ans. 2[tan (3log )]y xdifferentiate w.r.t. x both side

2 32[tan (3log )] sec (3log )dy x xdx x

[1 Mark]

3 2 36 tan (log ) sec (log )dy x xdx x

[1 Mark]

Topic: Differentiation; Sub-topic:Composite function__L-1 __XII-HSC Board Test_Mathematics

(iii) Find the area of ellipse 2 2

1.1 4x y

Ans. Required area = 4 Area ( )OAPB

1

0

ydx [1 Mark]

2 2

11 4x y

22 1y x

12

0

Required area 4 2 1 x dx A

BP

O

(0,2)

(0,1)

12 1

0

18 1 sin2 2 1x xx


1414

118 0 sin (1) 02

18 2 sq.units2 2 [1 Mark]

Topic: Definite integral; Sub-topic:AOI__L-2__XII-HSC Board Test_Mathematics

(iv) Obtain the differential equation by eliminating the arbitrary constants from the following equation :2 2

1 2x xy c e c e

Ans. 2 21 2

x xy c e c e

differentiate w.r.t. .x

2 21 22 2x xdy c e c e

dx [1 Mark]

Again diff. w.r.t. x.2

2 21 22 4 4x xd y c e c e

dx

2 21 24( )x xc e c e

4 y

2

2 4 0d y ydx

[1 Mark]

Topic: Differential equation; Sub-topic:Formulation__L-1 __XII-HSC Board Test_Mathematics

(v) Given ~ ( , )X B n pIf 10n and 0.4,p find ( )E X and Var. . ( ).X

Ans. 10, 0.4 1 0.6n p q p

( )E X np

10 0.4 4 [1 Mark]( ) 4 0.6 2.4V x npq [1 Mark]

Topic:Binomial distribution; Sub-topic:Mean__L-1 __XII-HSC Board Test_Mathematics


1515

Q.5 (A) Attempt any TWO of the following: [6]

(i) Evaluate: 1

3 2sin cosdx

x x

Ans. Let 13 2sin cos

I dxx x

2

2Put tan Then, ,2 1x t dx dt

t

2

2 2

2 1sin and cos1 1

t tx xt t

[1 Mark]

2

2

2 2

2 /(1 )2 13 2

1 1

dt tIt tt t

2

2 2

2

/(1 )23(1 ) 4 (1 )

1

dt tt t t

t

[1 Mark]

2 222 4 4 ( 1) 1

dt dtt t t

1tan ( 1)t c

1tan tan 1

2x c

[1 Mark]

Topic: Integration; Sub-topic:Method of substitution__L-2 __XII-HSC Board Test_Mathematics

(ii) If 3 3cos , sin ,x a t y a t

show that 13dy y

dx x

Ans. We have, / , / 0/

dy dy dt dx dtdx dx dt

...(1) [1 Mark]

Now, 133 3sin (sin ) sin yy a t a t t

a

3 2(sin ) 3(sin ) (sin )dy d da t a t tdt dt dt

23 sin cosa t t ...(2)

Also, 133 3cos (cos ) cos xx a t a t t

a


1616

23cos (cos )dx da t tdt dt

23 cos ( sin )a t t

23 cos sina t t ...(3) [1 Mark]From (1), (2) and (3),

12 3

2

3 sin cos sin3 cos sin cos

dy a t t t ydx a t t t x

[1 Mark]

Topic:Differentiation; Sub-topic:Parametric function__L-2 __XII-HSC Board Test_Mathematics

(iii) Examine the continuity of the function:log100 log(0.01 )( )

3xf x

x

100

3 for 0;at 0x x

Ans. f is continuous at 0x if 0lim ( ) (0).x

f x f

[1 Mark]

100R.H.S. (0) ........(given).........3

f ...(1) [1 Mark]

0 0 0

log100 log (0.01 ) log(1 100 )L.H.S. lim ( ) lim lim3 3x x x

x xf xx x

0

1 [log(1 100 )] 100lim 1003 100 3x

xx

From (1) and (2), L.H.S. = R.H.S.

0lim ( ) (0).x

f x f

f is continuous at 0.x [1 Mark]Topic: Continuity; Sub-topic:At a point__L-1 __XII-HSC Board Test_Mathematics

Q.5 (B) Attempt any TWO of the following: [8]

(i) Find the maximum and minimum value of the function:3 2( ) 2 21 36 20f x x x x

Ans. 3 2( ) 2 21 36 20f x x x x 2( ) 2(3 ) 21(2 ) 36(1) 0f x x x

2 26 42 36 6( 7 6)x x x x

6( 1)( 6)x x [1 Mark]f has a maxima/minima if ( ) 0f x

i.e. if 6( 1)( 6) 0x x

i.e. if 1 0 or 6 0x x


1717

i.e. if 0 or 6x x [1 Mark]Now, ( ) 6(2 ) 42(1) 12 42f x x x

(1) 12(1) 42 30f

(1) 0f

Hence, f has a maximum at 1,x by the second derivative test.Also, (6) 12(6) 42 30f

(6) 0f

Hence, f has a minimum at 6,x by the second derivative test.Now, maximum value of f at 1,

3 2(1) 2(1 ) 21(1 ) 36(1) 20f

2 21 36 20 3 [1 Mark]and minimum value of f at x = 6,

3 2(6) 2(6 ) 21(6 ) 36(1) 20f

432 756 216 20 128 [1 Mark]

Topic: AOD; Sub-topic:Maxima or Minima__L-1__XII-HSC Board Test_Mathematics

(ii) Prove that: 2

1 1 log2

a xdx ca x a a x

Ans. 2 2

1 1dx dxa x a x a x

[1 Mark]

12

a x a xdx

a a x a x

1 1 12

dxa a x a x

[1 Mark]

1 1 12

dx dxa a x a x

log1 1log log log2 1 2

a xa x c a x a x c

a a

[1 Mark]

1 log

2a x c

a a x

[1 Mark]

Topic: Integration; Sub-topic:Theorem__L-1 __XII-HSC Board Test_Mathematics


1818

(iii) Show that:

0

( ) 2 ( ) ,a a

a

f x dx f x dx

if ( )f x is an even function

0, if ( )f x is an odd functionAns. We shall use the following results :

b a

a b

f x dx f x dx ....(1)

b b

a a

f x dx f t dt ....(2)

If c is between a and b, then

b c b

a a c

f x dx f x dx f x dx ....(3)

Since 0 lies between -a and a, by (3), we have,

0

1 20

.... Saya a

a a

f x dx f x dx f x I I

[1 Mark]

In 1,I put .x t Then dx dt When ,x a t a t a

When 0, 0 0x t t

0 0 0

a a a

f x dx f t dt f t dt

0

.... 1a

f t dt By

0

.... 2a

f x dx By

0 0

a a a

a

f x dx f x dx f x dx

[1 Mark]

(i) If f is an even function, then f x f x in this case,

0 0 0

2a a a a

a

f x dx f x dx f x dx f x dx

[1 Mark]

(ii) If f is an odd function, then f x f x in this case,

0 0 0 0

0.a a a a a

a

f x dx f x dx f x dx f x dx f x dx

[1 Mark]

Topic: Definite integral; Sub-topic:Theorem__L-1 __XII-HSC Board Test_Mathematics


1919

Q.6 (A) Attempt any TWO of the following : [8]

(i) If 2

2

9( ) , for 33

5, for 32 3 , for 3

xf x xx

xx x x

is continuous at 3,x find and .

Ans. f is continuous at 3x 3 3lim ( ) lim (3)x x

f x f

...(1) [1 Mark]

Now, 2

0 33

9 ( 3)( 3)lim ( ) lim lim3 ( 3)x xx

x x xf xx x

0lim[( 3) ]x

x

(3 3) 6

and 2 2

33lim ( ) lim 2 3 2(3) 3(3) 18 9 27

xxf x x x

Also, (3) 5 ......(given)f

From (1), we get, 6 27 5 [1 Mark]6 5 and 27 5

5 6 1 and 5 27 22

1 and 22 [1 Mark]Topic: Continuity; Sub-topic:At a point__L-2__XII-HSC Board Test_Mathematics

(ii) Find dydx if

12

5 1tan3 6

xyx x

Ans. Let 1

2

5 1tan3 6

xyx x

1

2

5 1tan1 2 6

xx x

[1 Mark]

1 5 1tan

1 (3 2)(2 1)x

x x

1 (3 2) (2 1)tan

1 (3 2) (2 1)x x

x x

1 1tan (3 2) tan (2 1)y x x [1 Mark]Differentiate w.r.t. x

2 2

3 21 (3 2) 1 (2 1)

dydx x x


2020

2 2

3 21 9 12 4 1 4 4 1x x x x

2 2

3 19 12 5 2 2 1x x x x

[1 Mark]

Topic:Differentiation; Sub-topic:Inverse function__L-3__XII-HSC Board Test_Mathematics

(iii) A fair coin is tossed 9 times. Find the probability that it shows head exactly 5 times.Ans. Let X = no. of heads shows.

1 192 2

n p q [1 Mark]

( ) .( ) 0,1,.......,n n n xxP X x C p q X n [1 Mark]

5 49

51 1( 5)2 2

P X C

9

9 8 7 6 14 3 2 1 2

126512

0.2460 [1 Mark]Topic:Binomial distribution; Sub-topic:Binomial distribution_L-1__XII-HSC Board Test_Mathematics

Q.6 (B) Attempt any TWO of the following : [8](i) Verify Rolle’s theorem for the following function:

2( ) 4 10f x x x on [0, 4]

Ans. Since ( )f x is a polynomial,(i) It is continuous on [0, 4] [1 Mark](ii) It is differentiable on (0, 4) [1 Mark](iii) (0) 10, (4) 16 16 10 10f f

(0) (4) 10f f [1 Mark]Thus all the conditions on Rolle’s theorem are satisfied.The derivative of ( )f x should vanish for at least one point c in (0, 4). To obtain the value of c, weproceed as follows

2( ) 4 10f x x x

( ) 2 4 2( 2)f x x x

( ) 0 ( 2) 0f x x

2x

2 in (0, 4)c

We know that 2 (0, 4) [1 Mark]Thus Rolle’s theorem is verified.

Topic:AOD; Sub-topic:Roll’s theorem L-1__XII-HSC Board Test_Mathematics


2121

(ii) Find the particular solution of the differential equation:

(1 log ) log 0dxy x x xdy

when 2y e and .x eAns. Given equation is

(1 log ) log 0dxy x x xdy

(1 log ) logdxy x x xdy

(1 log ) logy x dx x x dy Separating the variables,1 1 log

logxdy dx

y x x

[1 Mark]

Integrating, we have,1 1 log

logxdy dx

y x x

log log log logy x x c [1 Mark]

log log logy cx x [1 Mark]

logy cx x is the general solution.

Given : 2,x e y e 2 . .loge c e e 2 . .e c e

c e

logy ex x [1 Mark]Topic:Differential equation; Sub-topic:Variable separable method_L-1_XII-HSC Board Test_Mathematics

(iii) Find the variance and standard deviation of the random variable X whose probability distribution isgiven below :

0 1 2 31 3 3 1( )8 8 8 8

x

P X x


2222

Ans. 2

10 0 083 3 318 8 83 6 1228 8 81 3 938 8 8

12 24Total 38 8

i i i i i ix p p x p x

[1 Mark]

12 3( )8 2i iE X p x [1 Mark]

2 2

1

( )n

i ii

Var X p x

233

2

934

34

2 3( )4

Var X [1 Mark]

Standard deivation of 3 34 2xX [1 Mark]

Topic: Probability distribution; Sub-topic:Expected value_L-1__XII-HSC Board Test_Mathematics

xii hsc - board - 2018...rao iit academy/ xii hsc - board exam 2018 / mathematics / qp + solutions 3...

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