2016 12 Maths Sample Paper 01 Solution

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MATHEMATICS SAMPLE PAPER

SOLUTIONS

SECTIONA

1. Projection of a

on a.bbb

=

2 2

7 2 1 6 ( 4) 3 14 6 12 87492 6 3

+ + + = = =

+ +

2. Since a,b

and c

vectors are coplanar.

a b c 0 =

1 3 12 1 1 00 3

=

1( 3 ) 3(6 0) 1(2 0) 0 + + + + =

3 18 2 0 + + = 3 21 0 = 7 = 3. Let l, m, n, be direction cosine of given line.

ocos90 0; = =l o 1m cos602

= = and n = cos and n = cos

2 2 2m n 1+ + = l 2

210 cos 12

+ + =

23

cos4

= 3cos2

= ( is acute angle)

6pi =

4. 232 3 1 1

a2 2 2

= = =

5. Given family of curve is A Br

= +

Differentiating with respect to r, we get

2d Adr r

=

Again differentiating with respect to r, we get

2

2 3d v 2Adr r

= 2

2 2d v 2 A

.

dr r r =

2

2d v 2 dv

.

dr r dr

=

2

2d v dv

r 2dr dr

=

2

2d v dv

r 2 0dr dr

= + =



6. Give differential equation is

2 xe y dx 1

dyx x

=

2 xe y dx. 1dyx

=

2 x

dx xdy e y

=

2 xdx e y

dx x

=

2 xdy e y

dx x x

= 2 xdy 1 e

.ydx x x

+ =

dx1

x 2 x21IF e dx e ex

= = =

SECTION-B

7.

2 0 1A 2 1 3

1 1 0=

2A A A=

2 0 1 2 0 12 1 3 2 1 31 1 0 1 1 0

=

4 0 1 0 0 1 2 0 0 5 1 24 2 3 0 1 3 2 3 0 9 2 5

2 2 0 0 1 0 1 3 0 0 1 2

+ + + + +

= + + + + + =

+ + +

Now, 2A 5A 4I +

5 1 2 2 0 1 1 0 09 2 5 5 2 1 3 4 0 1 00 1 2 1 1 0 0 0 1

= +

1 1 31 3 105 4 2

=

Now given 2A 5A 4I X 0 + + =

1 1 31 3 10 X 05 4 2

+ =

1 1 3X 1 3 10

5 4 2

=

1 1 3X 1 3 10

5 4 2

=

OR



Given

1 2 3A 0 1 4

2 2 1

=

1 0 2A ' 2 1 2

3 4 1

=

A' 1( 1 8) 0 2( 8 3) 9 10 1 0= + = + = Hence, 1(A') will exist.

111 2

A 1 8 9;4 1

= = = 122 2

A ( 2 6) 83 1

= = =

132 1

A 8 3 5;3 4

= = + = 210 2

A (0 8) 84 1

= = + =

221 2

A 1 6 7;3 1

= = + = 231 0

A (4 0) 43 4

= = =

310 2

A 0 2 2;1 2

= = =

321 2

A (2 4) 22 2

= = =

331 0

A 10 12 1

= = =

Adj(A)

79 8 5 9 8 28 7 4 8 7 22 2 1 5 4 1

= =

1

9 8 2 9 8 21(A ') 8 7 2 8 7 21

5 4 1 5 4 1

=

8. Here f(x) = 2

a 1 0ax a 1ax ax a

Taking a common from 1C ,we get

f(x) = 2

a 1 0ax a 1ax ax a

applying 2 2 1,C C C + we get

f(x) =a2 2

1 0 0x a x 1x ax x a

+

+

Expanding along 1R , we get



f(x) =a[1( 2 2a ax ax x ) 0 0}+ + + + f(x) =a( 2 2a 2ax x+ + )

f(x) =a(a+x)2

Now, f(2x)-f(x)=a(a+2x)2-a(a+x)2

=a{(a+2x)2-(a+x)2}=a(a+2x+a+x)(a+2x-a-x)

=ax(2a+3x)

9. Here

2 2

1 dxsin x sin 2x

1 1I dx I dxsin x 2sin x cos x sin x(1 2cos x)

sin x sin xI dx I dxsin x(1 2cos x) (1 cos x)(1 2cos x)

=+

= =+ +

= =+ +

Let cosx = z - sinx dx = dz

2

dz dzI I(1 z )(1 2z) (1 z)(1 z)(1 2z)

= = + + +

Here , integrand is proper rational function. Therefore by the form of partial function ,

we can write 1 A B C

(1 z)(1 z)(1 2z) 1 z 1 z 1 2z= + ++ + + + (i)

1 A(1 z)(1 2z) B(1 z)(1 2z) C(1 z)(1 z)(1 z)(1 z)(1 2z) (1 z)(1 z)(1 2z)

+ + + + + + =

+ + +

1= A(1-z)( 1+2z) +B(1+z)(1+2z)+C(1+z)( 1 - z ) (ii)

Putting the value of z=-1 in (ii) we get

11 2A 0 0 A2

= + + =

Again, putting the value of z=1 in (ii),we get

1=0+B.2.(1+2)+0 1=6B B=16

Similarly, putting the value of z=12

in(ii) ,we get

1 31 0 0 C2 2

= + +

1=

34

C c=43

Putting the value of A,B,C in (i) we get

1 1 1 4

(1 z)(1 z)(1 2z) 2(1 z) 6(1 z) 3(1 2Z)= + ++ + + +

1 1 4I dz2(1 z) 6(1 z) 3(1 2Z)

= + + + +



1 1 4I dz2(1 z) 6(1 z) 3(1 2Z)

=

+ +

1 1 4I log 1 z log 1 z log 1 2z C2 6 3 2

= + + + +

putting the value of z , we get

1 1 2log 1 cos x log 1 cos x log 1 2cos x C2 6 3

= + + + +

OR

Let

2 2

2 2

2)

2 2 2

2

2 2

2 1 1 2

1 2 2

x 3x 1 x 1 2 3xdx dx1 x 1 x

(1 x dx xdxdx 2 31 x 1 x 1 x

dx xdx1 x dx 2 31 x 1 x

1 1x 1 x sin x 2sin x 3 1 x C

2 23 1

sin x x 1 x 3 1 x C2 2

+ + = =

= +

= +

= + + +

= + +

10. Here 2(cos ax sinbx) dxpi

pi

=

2 2bx

2 2

cos ax sin 2cosaxsinbx)dx

cos axdx sin axdx 0

pi

pi

pi pi

pi pi

= +

= +

[First two integranda are even function while third is odd function]

2 2

0 0

0 0

0 0 0 0

I 2 2cos axdx 2sin bxdx

(1 cos2ax)dx (1 cos2bx)dx

dx cos2axdx dx cos2bxdx

pi pi

pi pi

pi pi pi pi

= +

= + +

= + +

[ ]00 0

sin2ax sin2bxI 2 x2a 2b

sin2a sin2b22a 2b

pi pipi

= +

pi pi = pi+

11. Let E,F and A three events such that

E = selection of Bag A and F=selection of bag B

A= getting one red and one black ball of two



Here ,p(E)=P(getting 1 or 2 in a throw of die)=2 16 3

=

1 2p(F) 13 3

= =

Also, P(A/E)=P (getting one red and one black if bag A is selected)=6 4

1 110

2

C C 2445C

=

and P(A/F)=P(getting one red and one black if bag Black if bag B is selected)=3

1 110

2

C 7C 2145C

=

Now, by theorem of total probability,

p(A)=P(E).P(A/E)+P(F).P(A/F)

1 24 2 21 8 14 22p(A)3 45 3 45 45 45

+ = + = =

OR

Let number of head be random variable X in four tosses of a coin .X may have values

0,1,2,3 or 4 obviously repeated tosses of a coin are Bernoulli trials and thus X has

binomial distribution with n=4 and p= probability of getting head in one toss=12

q=probability of getting tail (not head) in one toss= 1 - 12

=12

since, we know that P(X=r)= n r n rrC q , r 0,1, 2..........n

=

therefore,

P(X=0)= 0 4 0 4

40

1 1 1 1C 1 12 2 2 16

= =

( )

( )

( )

( )

1 4 1 1 44

1

2 4 2 2 24

2

2 4 3 3 14

3

44

1 1 1 1 4C 42 2 2 2 161

1P X 1 4

3P X 2 8

1P X 3 4

P X 4

1 1 1 6C 62 2 2 2 161 1 1 1 4C 42 2 2 2 161

C2

= =

= =

= =

= = =

= = =

= = =

= =

4 4 4 4 01 1 1

112

6

2 2 1

= =

Now required probability distribution of X is

x 0 1 2 3 4

4P(x) 116

14

38

14

1

16

Required mean = i ix p =

=

1 1 3 1 10 1 2 3 416 4 8 4 16

1 3 3 1 8 24 4 4 4 4

+ + + +

= + + + = =



variance = 2

2 2 2x i i i i i ix p x p X p

= =

= 2 2 2 2 21 1 3 1 10 1 2 3 4 2

16 4 8 4 16

+ + + +

1 3 9 1 44 4 4

= + + +

1 3 9 34 4 41 6 9 12 4 1

4 4

= + +

+ + = = =

12. Here

Now

( )( ) r i r j xy {(xi yj zk) i}.{(xi yj zk) j} xy ( yk zj).(xk zi) xy (0i zj yk) ( zi 0 j xk) xy

0 0 xy xy 0

+ = + + + + +

= + + = + + + +

= + + =

13. Let P ( , , ) be the point of intersection of the given line (i) and plane (ii)

x 2 y 1 z 23 4 12 +

= = ..(i)

and x y + z = 5 .(ii)

since ,point P ( , , ) lies on line (i)( therefore it satisfy(i)

2 1 23 4 12

3 2; 4 1; 12 2

+ = = =

= + = = +

Also point P ( , , ) lie on plane (ii) 5 + = (iii) putting the value of , , in (iii) we get

3 2 4 1 12 2 511 5 5 0

2; 1; 2

+ + + + = + = = = = =

hence the coordinate of the point of intersect ion p is (-2,-1,2)

therefore ,required distance= d = 2 2 2(2 1) ( 1 5) (2 10)+ + + + + 9 16 144 169 13units+ + = = 14. Here sin[ 1cot (x 1) + ]= 1cos(tan x) let 1cot (x 1) + = cot x 1 = + 2 2 2cos ec 1 cot 1 (x 1) x 2x 2 = + = + + = + +



12 2

1 1sin sin

x 2x 2 x 2x 2

= =

+ + + +

1 12

1cot (x 1) sin

x 2x 2

+ =

+ +

1

2 2

1

2 2

1 1

2

tan x tan x

sec 1 tan 1 x

1 1cos cos

1 x 1 x1

tan cos1 x

again

= =

= + = +

= =

+ +

=

+

now equation (i) becomes

1 12 2

1 1sin sin cos cos

x 2x 2 1 x

= + + +

2 2

2 2

2 2

1 1x 2x 2 1 x

x 2x 2 1 xx 2x 2 1 x 2x 2 1

1x

2

= + + = ++ + +

+ + = + + =

=

Or

Here

21 2 1 2

21 2 1 2

2 21 2 1 2 1

2 21 2 1

21 2 1

5(tan x) (cot x)8

5(tan x) ( tan x)8

5(tan x) (tan x) tan x4 8

52(tan x) tan x 04 832(tan x) tan x 88

pi+ =

pi + pi =

pi pi + + pi =

pi pi pi + =

pi pi =

let 1tan x y =

22 2 2

2 2

1 1

32y y 0 16y 8 y 3 08

16y 12 y 4 y 3 0 4y(4y 3 ) (4y 3 ) 03(4y 3 )(4y ) 0 y or y

4 43

tan x [ does not belongs to domain of tan x i.e, , ]4 4 2 4

x tan 14

pi pi = pi pi =

pi + pi pi = pi + pi pi =

pi pi pi + pi = = =

pi pi pi pi =

pi = =

15.



2 21

2 2

2 2 2 21

2 2 2 2

1 x 1 xy tan1 x 1 x

1 x 1 x 1 x 1 xtan

1 x 1 x 1 x 1 x

+ + = +

+ + + + = + + +

=

4 41 1

2 2 2

41

2

2 2 1 x 2 2 1 xtan tan

1 x 1 x 2x

1 1 xtan

x

+ + = + +

+ =

let x2 = sin 1 2sin (x ) = putting the value of x2,we get

21

2

1 1

1 1

1 2

1 1 sintan

sin

2cos1 cos 2tan tansin 2sin cos

2 2

tan cot tan tan2 2 2

sin x2 2 2

+ =

+

= =

pi = =

pi pi= =

20 x 1

sin 0 sin sin2

0 02 2 2

02 4

2 2 2 2 4

2 2 2 4

, ,

2 2 4 2 2 2

pi <

pi pi pi > >

pi pi pi pi pi

differentiating both sides with respect to x, we get

4

dy 2xdx 2 1 x

=

4

dy xdx 1 x

=

16. Given x= a cos +bsin

dx

a sin b cosd

Also, y a sin b cos

= +

=

dya cos bsin

ddy

dy a cos bsinddxdx a sin b cosd

= +

+ = = +

dy xdx y

= 2

2 2

dyy x.d y dxdx y

=

2 2

2 22 2

d y dy d y dyy y x y x y 0dx dxdx dx

= + + =

17. Let A be the area and a be the side of an equilateral triangle.



23A a

4=

Differentiating with respect to t we get

dA 3 da2adt 4 dt

=

dA 3 2a 2dt 4

= [Given dadt

= 2cm/sec.]

dA 3adt

= a 20cm

da 20 3sq cm / sdt

=

=

18. Let 2I (x 3) 3 4x x dx= +

Let x + 3 = 2dA (3 4x x ) B

dx +

x + 3 = A (-4 2x) +B x + 3 =- 4A =-2Ax + B x + 3= (-4A + B) =- 2Ax [By comparing coefficients]

-2A = 1 A = 12

=

Again, -4A + B = 3

14 B 32

+ = 2 +B = 3 B = 1

Here, 1

x 3 ( 2 4) 12

+ = +

21I ( 2x 4) 1 3 4x x dx2

= +

2 21I ( 2x 4) 3 4x x dx 3 4x x dx2

= +

1 21I I I2

= + (i), where..

Now 21I ( 2x 4) 3 4x x dx Let 3-4x x2 = z (-2x 4) dx = dz

23

1 12I zdz (z) C3

= + 3

2 22 1

2I (3 4x x ) C3

+=

Again 22I 3 4x x dx

22I (x 4x 3)dx= + { }22I x 2) 7 dx= + 2 22I ( 7) (x 2) dx= +

2 12 21 7 x 2I (x 2) 3 4x x sin C2 2 7

+= + + +

Putting the value of 1I and 2I in (i), we get



3

2 2 1122

2

C1 2 7 x 2I (3 4x x ) (x 2) 3 4x x sin C2 3 C 2 7

+= + + +

3

2 2 122

1 1 7 x 2I (3 4x x ) (x 2) 3 4x x sin C3 2 2 7

+= + + + +

2 2 11 1 7 x 23 4x x (3 4x x ) (x 2) sin C3 2 2 7

+ = + + + +

2 1x 7 x 23 4x x (2x 11) sin C,6 2 7

+= + + + where 12

CC C2

=

19. The number of handmade fans, mats and plates sold by three school A, B and can be

represented by 3 3 matrix as

A 40 50 20X B 25 40 30

C 35 50 40

=

And their selling price can be represented by 3 1 matrix as

25 Handmade fansY 100 Mats

50 Plates

=

Now, the total funds collected by each school is given by the matrix multiplication as

A 40 50 20 25XY B 25 40 30 100

C 35 50 40 50

=

40 25 50 100 20 50 7000XY 25 25 40 100 30 50 XY 6125

35 25 50 100 40 50 7875

+ +

= + + = + +

Hence, total funds collected by school A = Rs.7000

Total funds collected by school B = Rs.6125

Total funds collected by school C = Rs.7875

Total funds collected for the purpose = Rs.(7000+6125+7875)

= Rs. 21000

Value: Students are motivated for social service.

SECTION-C

20 Reflexivity: By commutative law under addition and multiplication

B + a = a +b a,b N Ab=ba a,b N Ab(b+a)=ba(a+b) a,b N (a,b)R(a,b) Hence, R is reflexive

Symmetry: Let(a,b)R(c,d)

(a,b)R(c,d) ad(b + c) = bc (a + d) bc(a + d) = ab(b + c)



cb(d + a) da(c + b) [By commutative law under addition and multiplication]

(c + d) R(a, b) Hence, R is symmetric.

Transitivity: Let(a, b) R (c, d) and (c, d) R (e, f)

Now, (a, b) R (c, d) and (c, d) R (e, f)

ad( b +c) = bc(a + d) and cf(d + e) de( c + f)

b c a dbc ad+ +

= and d e c fde cf+ +

=

1 1 1 1c b d a

+ = + and 1 1 1 1e d f c

+ = +

Adding both, we get

1 1 1 1 1 1 1 1c v e d d a f c

+ + + = + + +

1 1 1 1b e a f

+ = + e b f abe af+ +

=

af(b+e) = be(a +f) (a, b) R (e, f) [c,d 0] Hence, r is transitive.

In this way, r is reflexive symmetric and transitive

Therefore, r is an equivalence relation.

21. Given circle is 2 2x y 4+ =

dy2x 2y 0dx

+ = [By differentiating]

dy xdx y

=

Now, slope of tangent at (1, 3 )

dy 1(1, 3) .dx 3

= =

Slope of normal at (1, 3) 3= Therefore, equation of tangent is

y 3 1x 1 3

=

x 3y 4+ = Again, equation of normal is



y 3 3x 1

=

y 3x 0 =

To draw the graph of the triangle formed by the lines x-axis, (i) and (ii), we find the

intersecting of these three lines which give vertices of required triangle. Let O, A, B be

the intersecting of these lines.

Obviously, the coordinate of O, A, B are (0, 0), ( )1, 3 and (4, 0) respectively. Required area = area of triangle OAB

= area of region OAC + area of region CAB

1 40 1y dx y dx= + [Where in 1st integrand y 3x= and in 2nd 4 xy

3

= ]

1 42 21 10 0

0 1

4 x x 1 (4 x)3xdx dx 32 23 3

= + =

3 1 90

2 23

=

3 9 12 2 3

2 2 3 2 3= + = = sq units.

Or

3 3 3

2 3x 2 2 3x 2

1 1 1

(e x 1)dx e e dx (x 1)dx + + = + +

2 1 2e .I I= + . (i), 3 3

3x 22 2

1 1

I e dx;I (x 1)dx= = +

We have, b

h 0a

f (x)dx lim h{(a b) f (a 2h) ... f (a nh)}

= + + + + + +

For 1I 3xf (x) e ,= a = 1, b = 3

b a 2h h nh 2n n

= = =

Now, 3

3x

h 01

e dx lim h{f (1 h) f (1 2h) ...f (1 nh)}

= + + + + +

{ }3(1 h) 3(1 2h) 3(1 nh)h 0lim h e e ... e + + +

= + + +

{ }3 3h 3 6(1 2h) 3nhh 0lim h e .e e .e ... e +

= + + +

{ }3h 3h 2 3h nh 0lim h e (e ) ... (e )

= + + +

3h 3h n

33hh 0

e ( 1 (e )e .lim h

1 e

=

3h 3nh

33hh 0

e (1 e )e .lim h

1 e

=

[Applying formula for sum of GP]



3h 6

3 3 6 03h3hh 0

3h 0

e ((1 e ) 1e .lim e (1 e ).e .

e 11 e lim 33h

= =

3 6e (1 e )

3

=

For 2I F(x) = x2 + 1, a = 1, b = 3 b ah

n

= 2hn

= b

nh=2

Now, { }3

2

h 01

(x 1)dx lim h f (1 h) f (1 2h) .... f (1 nh)

+ = + + + + + +

( ){ } ( ){ } ( ){ }2 2 2h 0limh 1 h 1 1 1 2h 1 ... 1 nh 1 = + + + + + + + + + ( ) ( ) ( )2 2 2 2 2

h 0limh n 1 h 2h 1 4h 4h 1 9h 6h ... (1 n h 2nh)

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

( ) ( )2 2 2 2 2h 0limh n n h 1 2 3 ...n 2h 1 2 3... n

= + + + + + + + + +

2

h 0

h n(n 1)(2 1) 2h n(n 1)lim h 2n6 2

+ + += + +

h 0

h nh(nh h)(2nh h)lim 2nh nh(nh h)6

+ + = + + + +

h 0

2(2 h)(2 2 h) 2(2 0)(4 0)lim 4 nh(nh h) 4 2(2 0)6 6

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

16 8 324 4 86 3 3

= + + = + =

Putting the value of 1I and 2I in (i), we ge,

2 3 6 1 6 1 7e .e (1 e ) 32 e (1 e ) 32 32 (e e )I

3 3 3 3 3

+ = + = + =

22. The given differential equation can be written as

1

2 2

dx x tan ydy 1 y 1 y

+ =+ +

Now (i) is linear differential equation of the form 1 1dx P x Qdy

+ = ,

where , 1

1 12 2

1 tan yp and Q1 y 1 y

= =

+ +

Therefore, I.F = 12

1 dytan y1 ye e

+=

Thus , the solution of the given differential equation is

1 1

1tan y tan y

2

tan yxe e dy C

1 y

= +

+

Let 1

1tan y

2

tan yI e dy1 y

=

+

substituting 1tan y t = so that 21 dy dt, we get

1 y

= +



t t t t t tI t e dt t e 1.e dt t e e e (t 1)= = =

1tan y 1I e (tan y 1) = substituting the valiue of I in the equation (ii) , we get

1 1tan y tan y 1

x.e e (tan y 1) C = + or 11 tan yx (tan y 1) Ce = + which is the gernal solution of the given differential equation

OR

Given differential equation is

2 2dy xydx x y

=

+ ..(i)

Let y= x dy d

xdx dx

= +

Now (i) becomes

2 2

2 2 2 2 2

d x d xx x

dx dxx x x (1 v )

+ = + =+ +

2 2d d

x xdx dx1 x 1

+ = =

+ +

3 3

2 2

d dx x

dx dx1 1

= =+ +

2

3dx 1 dx

+ =

Integrating both sides, we get

2

3 31 dx d d dxd

x xv

+ = + =

21 log log x C

2 + = +

putting the value of yx

= ,we get

2

2

x ylog log x Cx2y

+ + = 2

2

x log y log x log x C2y

+ + =

2

2

x log y C2y

+ =

put y=1 and x=0 in(ii) 0+log 1 =C C = 0

Therefore required particular solution is 2

2

x log y 02y

+ =

23. Let the given lines

x 1 y 1 z 1

2 3 4 +

= = (i)

and x 3 y k z

1 3 1

= = (ii)intersect at P ( , , ) P lie in (i)

1 1 1

2 3 2 +

= = = (say)



2 1, 3 1, 4 1 = + = = +

again, P lie on (ii)also

3 k

1 2

= =

2 1 3 3 1 k 4 1

1 2 1 + +

= =

2 2 3 1 k 4 1

1 2 1 +

= =

I II III

from I and II

3 1 k2 2 4 4 3 1 k

2

= =

k=3 -1-4 +4 k=- +3

k =32

+3 =92

Now, we know that equation of plane containing lines.

1 1 1

1 1 1

x x y y z za b c

= =

and 2 2 2

2 2 2

x x y y z za b c

= = is

1 1 1

1 1 1

2 2 2

x x y y z za b c 0a b c

=

Therefore , required equation is

x 1 y 1 z 12 3 4 01 2 1

+

=

(x-1)(3-8)-(y+1)(2-4)+(z-1)(4-3)=0

-5(x-1)+2(y+1)+(z-1)=0

-5x+2y+z+6=0 5x-2y-z-6=0

24. we know that if A and B are two independent events then

P(A B) = P(A).P(B)

Also, since A and B are two independent events A ,B and A, B are also independent events .

P( A B) = P( A ).P(B) P(A B ) = P( A ).P( B ) Now, let P(A) = x and P(B) = y

P( A ) = 1 x and P( B ) = 1 y



Given P( A B) = 215

and P(A B ) = 16

P( A ).P(B) = 215

and P(A).P( B )= 16

(1-x).y =2

15 and x.(1 y )=

16

y xy =2

15 ..(i) and x xy =

16

(ii)

From(i) y. (1-x) =2

15

2y15(1 x)=

Putting the value of y in (ii) ,we get

x - x2

15(1 x) = 16

215x 15x 2x 1

15 15x 6

=

6(-15x2+13x)=15-15x -90x2+78x=15-15x

- 90x2 + 93x 15 = 0 30x2 31 x + 5=0

30x2 25x 6 x + 5=0 5x( 6 x 5 ) 1 (6 x 5 )=0

(6x-5)(5x-1)=0 x = 56

or x= 15

Now, x = 56

y = 2 12 4

5 15 515 16

= =

and x =15

2 1

1 61y

1

55

=

=

Hence P(A)= 56

and P(B)= 45

or P(A) =15

P(B) = 16

25. Given f(x) = sinx cosx f(x)=cosx+sinx

for critical points

f(x)=0 cosx+sinx=0

sinx = -cosx tanx=-1

tanx = tan34pi

x = 3

n4pi

pi + ,nZ

x = 34pi

,74pi

[other value does not belong to (0, 2pi )]

Now f(x) = - sinx + cosx

( ) 3x

4 f '' x pi

=

= - sin34pi

+cos 34pi

=1 1 2 2 02 2 2

= =



1 1 2 22 2 2

= =

Again ( ) 7x

4 f '' x pi

=

= - sin74pi

+ cos 74pi

=1 1 2 2 02

2 2

+ = = >

i.e., f(x) is minimum at x=74pi

Local minimum value of f(x) =f(74pi

)=sin74pi

- cos74pi

=1 1 2 22 2 2

= =

Therefore, local maximum and local minimum values are 2 and - 2 respectively. 26. Given constraints are

2x+4y 8 ..(i) 3x+y 4 (ii) x+y 4 ..(iii) x 0, y 0 ..(iv) from graph of 2x+4y 8

we draw the graph of 2x+ 4y = 8 as

X 0 4

y 2 0

2 0 4 0 8 + (0,0) origin satisfy the constraints.

Hence , fesible region lie origin side of line 2x + 4y = 8

For graph 3x+y 4 we draw the graph of line 3x+ y =6

X 0 2

y 6 0

3 0 0 6 +



origin (0,0) satisfy 3x+ y 6 hence , fesible region lie origin side of line 3x + y =6

for graph of x+ y 4 we draw the graph of line x+ y=4

X 0 4

y 4 0

0 0 4+ origin (0,0) satisfy x + y 4 hence feasible region lie origin side of line x+4= 4

also x 0, y 0 says feasible region is in ist quadrant. therefore, OABC is required feasible region.

Having corner point O(0,0) , (0,2) , B (8 6

,

5 5) , C(2,0)

here feasible region is bounded.

NOW the value of objective function Z = 2x + 5y is obtained as

Corner point Z=2x+5y

O(0,0) 0

(0,2) 2 0 5 2 =10

B (8 6

,

5 5)

8 62 55 5

=9.2

C(2,0)

2 2 5 0 =4

Hence, maximum value of Z is 10 at x =0, y =2

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2016 12 Maths Sample Paper 01 Solution