CBSE 2015 All India 1

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CBSE 2015 ALL INDIA EXAMINATION DELHI

SET 1/2/3 (Series SSO Code No. 65/1/A) With Complete Explanations

Note that all the sets have same questions. Only their sequence of appearance is different.

Max. Marks : 100 Time Allowed : 3 Hours SECTION A

Q01. cos sin

If Asin cos

, then for any natural number n, find the value of nDet (A ) .

Sol. cos sin

Given Asin cos

ncos n sin n

Asin n cos n

n n 2 2cos n sin n

A i.e., Det(A ) cos n sin n 1sin n cos n

.

Q02. Find the sum of the order and degree of the given differential equation : 3 2

2

dy d yy x

dx dx

.

Sol. Here order and degree of the given differential equation are 2 and 1 respectively. Therefore, the required sum is 2 + 1 = 3.

Q03. Find the solution of the differential equation : 2 2x 1 y dx y 1 x dy 0 .

Sol. Given 2 2x 1 y dx y 1 x dy 0 2 2

1 2xdx 1 2ydy0

2 21 x 1 y

2 21 1

2 1 x 2 1 y C2 2

i.e., 2 21 x 1 y C .

Q04. In a triangle OAC, if B is the mid-point of the side AC and OA a

, OB b

, then what is OC

?

Sol. As B is the mid-point of side AC so, OA OC

OB2

i.e., a OC

b2

. Therefore, OC 2b a

.

Q05. Find a vector of magnitude 171 which is perpendicular to both of the vectors a i 2 j 3k

and

b 3i j 2k

.

Sol. We have

i j k a b 1 2 3 i 11j 7k

3 1 2

Required vector a b i 11j 7k 171 171 (i 11j 7k)

| a b | 1 121 49

.

Q06. Find the angle between the lines 2x 3y z and 6x y 4z .

Sol. Given lines are 2x 3y z i.e, x y z

3 2 6

and

x y z

2 12 3

And their respective d.r.s are 3,2, 6 and 2, 12, 3

13 2 2 ( 12) ( 6)( 3)

required angle between lines is cos9 4 36 4 144 9

1 16 24 18

cos cos (0)29 4 36 4 144 9

.

SECTION B Q07. There are 2 families A and B. There are 4 men, 6 women and 2 children in family A, and 2 men, 2 women and 4 children in family B. The recommended daily amount of calories is 2400 for men, 1900 for women, 1800 for children and 45 grams of protein for men, 55 grams for women and 33 grams for children. Represent the above information using matrices. Using matrix multiplication, calculate the total requirement of calories and proteins for each of the 2 families. What awareness can you create among people about the balanced diet from this question?

CBSE 2015 Annual Exam Paper (All India) Compiled By OP Gupta (+91-9650350480)


Sol. The given information can be expressed as :

Men Women Children Calories Proteins

2400 45Family A 4 6 2

1900 55Family B 2 2 4

1800 33

Family A 9600 11400 3600 180 330 66 24600 576

Family B 4800 3800 7200 90 110 132 15800 332

So 24600 calories and 579 grams of proteins are needed for Family A and 15800 calories and 332 grams of proteins are needed for Family B. Balanced diet is of utmost importance for proper functionality and growth of our body. So people must be aware to take food which provides all the necessary nutrients.

Q08. Evaluate : 11

tan 2 tan5 4

.

Sol. Let 11

y tan 2 tan5 4

1

2

5tan tan1 4125

1 5tan tan12 4

1

1

5tan tan tan

12 4y

51 tan tan tan

12 4

51

7125 171

12

.

Q09. Using properties of determinants, prove that

3

3

3

a 2 a

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

c 2 c

.

Sol. LHS : Let

3

3

3

a 2 a

b 2 b

c 2 c

By 1 1 2 2 2 3R R R , R R R

3 3

3 3

3

a b 0 a b

b c 0 b c

c 2 c

Taking (a b) & (b c) common from 1 2R and R respectively

2 2

2 2

3

a ab b 0 1

(a b)(b c) b bc c 0 1

c 2 c

By 1 1 2R R R

2 2

3

(a c)(a c) b(a c) 0 0

(a b)(b c) b bc c 0 1

c 2 c

Taking 1(c a) common from R

2 2

3

a b c 0 0

(a b)(b c)(c a) b bc c 0 1

c 2 c

Expanding along 1R

(a b)(b c)(c a) ( a b c)[0 2] 0 0 2(a b)(b c)(c a)(a b c) RHS .



Q10. Using elementary row operations (transformations), find the inverse of

0 1 2

1 2 3

3 1 0

.

OR If

0 6 7

A 6 0 8

7 8 0

,

0 1 1

B 1 0 2

1 2 0

,

2

C 2

3

, then calculate AC, BC and (A + B) C.

Also verify that (A + B) C = AC + BC.

Sol. Let

0 1 2

A 1 2 3

3 1 0

By using elementary row transformations, A IA

0 1 2 1 0 0

1 2 3 0 1 0 A

3 1 0 0 0 1

By 1 2R R

1 2 3 0 1 0

0 1 2 1 0 0 A

3 1 0 0 0 1

By 3 3 1R R 3R

1 2 3 0 1 0

0 1 2 1 0 0 A

0 5 9 0 3 1

By 3 3 2R R 5R

1 2 3 0 1 0

0 1 2 1 0 0 A

0 0 1 5 3 1

By 1 1 2R R 2R

1 0 1 2 1 0

0 1 2 1 0 0 A

0 0 1 5 3 1

By 2 2 3R R 2R

1 0 1 2 1 0

0 1 0 9 6 2 A

0 0 1 5 3 1

By 1 1 3R R R

1 0 0 3 2 1

0 1 0 9 6 2 A

0 0 1 5 3 1

1I A A

13 2 1

A 9 6 2

5 3 1

OR We have

0 6 7 2 0 12 21 9

AC 6 0 8 2 12 0 24 12

7 8 0 3 14 16 0 30

,

And,

0 1 1 2 0 2 3 1

BC 1 0 2 2 2 0 6 8

1 2 0 3 2 4 0 2

.



Also

0 6 7 0 1 1 0 7 8

A B 6 0 8 1 0 2 5 0 10

7 8 0 1 2 0 8 6 0

0 7 8 2 0 14 24 10

(A B)C 5 0 10 2 10 0 30 20

8 6 0 3 16 12 0 28

(i)

And,

9 1 10

AC BC 12 8 20 ...(ii)

30 2 28

By (i) & (ii), it is clear that (A + B) C = AC + BC.

Q11. Discuss the continuity and differentiability of the function f (x) x x 1 in the interval ( 1,2) .

Sol. Method 1 : Since the continuity and differentiability of modulus function is doubtful at the corner points. So well shall check continuity and differentiability at the critical points x = 0, 1 ( 1, 2) .

x (x 1) 2x 1, if x 0

f (x) x x 1 x (x 1) 1, if 0 x 1

x x 1 2x 1, if x 1

Continuity at x = 0 : We have f (0) = 1. LHL (at x = 0) :

x 0lim ( 2x 1) 2 0 1 1

RHL (at x = 0) : x 0lim 1 1

Since x 0lim f (x) f (0)

so f (x) is continuous at x = 0.

Continuity at x = 1 : We have f (1) = 2 (1) 1 = 1. LHL (at x = 1) :

x 1lim1 1

RHL (at x = 1) : x 1lim 2x 1 2 1 1 1

Since x 1limf (x) f (1)


Differentiability at x = 0 :

LHD (at x = 0) : x 0 x 0 x 0

| x | | x 1| 1 2x 1 1 2xlim lim lim 2

x 0 x x

RHD (at x = 0) : x 0 x 0 x 0

1 1 0lim lim lim 0 0 LHD (at x 0)

x 0 x

f (x) is not differentiable at x 0 .


LHD (at x = 1) : x 1 x 1 x 1

1 1 0lim lim lim 0 0

x 1 x 1

RHD (at x = 1) : x 1 x 1

x x 1 1 2(x 1)lim lim 2 LHD (at x 1)

x 1 x 1


Method 2 : Since the continuity and differentiability of modulus function is doubtful at the corner points. So well shall check continuity and differentiability at the critical points x = 0, 1 ( 1, 2) .

Continuity at x = 0 : We have f (0) = |0| + |0 1| = 1.

LHL (at x = 0) : x 0lim | x | | x 1| | 0 | | 0 1| 1

RHL (at x = 0) : x 0lim | x | | x 1| | 0 | | 0 1| 1

Since x 0lim f (x) f (0)


Continuity at x = 1 : We have f (1) = |1| + |1 1| = 1.



LHL (at x = 1) : x 1lim | x | | x 1| |1| |1 1| 1

RHL (at x = 1) : x 1lim | x | | x 1| |1| |1 1| 1

Since x 1limf (x) f (1)



LHD (at x = 0) : x 0 x 0 x 0

| x | | x 1| 1 x x 1 1 2xlim lim lim 2

x 0 x x

RHD (at x = 0) : x 0 x 0 x 0

| x | | x 1| 1 x x 1 1 0lim lim lim 0 LHD (at x 0)

x 0 x x



LHD (at x = 1) : x 1 x 1 x 1

| x | | x 1| 1 x x 1 1 0lim lim lim 0

x 1 x 1 x 1

RHD (at x = 1) : x 1 x 1 x 1

| x | | x 1| 1 x x 1 1 2(x 1)lim lim lim 2 LHD (at x 0)

x 1 x 1 x 1


Q12. If x a(cos 2t 2t sin 2t) and y a(sin 2t 2t cos2t) , then find 2

2

d y

dx.

Sol. Given x a(cos 2t 2t sin 2t) and y a(sin 2t 2t cos2t)

dx

a( 2sin 2t 4t cos2t 2sin 2t) 4atcos 2tdt

& dy

a(2cos2t 4t sin 2t 2cos2t) 4at sin 2tdt

dy dy dt 4at sin 2t

So, tan 2tdx dt dx 4at cos 2t

2

2

2

d dy d y dt2sec 2t

dx dx dx dx

2 3

2

2

d y 1 sec 2t2sec 2t

dx 4at cos2t 2at .

Q13. If y/x(ax b)e x , then show that 22

3

2

d y dyx x y

dx dx

.

Sol. We have y/xx (a bx) e y/xlog x log[(a bx) e ]

y/xlog x log(a bx) log e y

log x log(a bx) log ex

x

or, y x loga bx

y x logx log(a + bx) (A)

dy 1 b

x [log x log(a bx)].1dx x a + bx

[By (A), y

logx log(a + bx)x ]

dy a y

So ...(i)dx a bx x

ax

xy ya bx

2

(a bx).a ax(b)xy + y y

(a bx)

22

2

a yxy y

(a bx) x

[By (i), y a

yx a bx

]

That is,

2

2

xy yxy

x

3 2x y (xy y)

223

2

d y dyx x y

dx dx

.

Q14. Evaluate : 3

2

xdx

(x 1)(x 1) .

OR Evaluate : sin x x cos x

dxx(x sin x)

.

Sol. Let 3

2

xI dx

(x 1)(x 1)

2

2

x x 1I 1 dx

(x 1)(x 1)



2

2 2

x 1 xI 1 dx dx

(x 1)(x 1) (x 1)(x 1)

2

1 xI x dx dx

x 1 (x 1)(x 1)

1I x log x 1 I (A) Now 1 2x

I dx(x 1)(x 1)

Consider 2 2 2

x A 2Bx C

(x 1)(x 1) x 1 x 1 x 1

2x A(x 1) 2Bx(x 1) C(x 1)

On equating the coefficients of like terms on both the sides, we get : 1 1 1

A ,B ,C2 4 2

.

2 2

1 dx 1 2xdx 1 dxI x log x 1

2 x 1 4 x 1 2 x 1

2 11 1 1

I x log x 1 log x 1 log x 1 tan x C2 4 2

Therefore, 2 11 1 1

I x log x 1 log x 1 tan x C2 4 2

.

OR Let sin x x cos x

I dxx(x sin x)

x sin x x x cos xI dx

x(x sin x)

x sin x x x cos x

I dxx(x sin x) x(x sin x)

1 1 cos xI dx dx

x x sin x

Put ndx sin x t (1 cos x)dx dt in 2 integral dt

I log | x |t

I log | x | log | t | C I log | x | log | x sin x | C .

Q15. Evaluate : /2 2

2

0

cos xdx

1 3sin x

.

Sol. Let /2 2

2

0

cos xdxI

1 3sin x

/2 2

4 2 2

0

sec xdxI

sec x 3sec x tan x

On dividing Nr & Dr by 4cos x

/2 2

2 2 2

0

sec xdxI

sec x(1 tan x 3tan x)

/2 2

2 2

0

sec xdxI

(1 tan x)(1 4 tan x)

Put 2tan x t sec xdx dt. When x 0 t 0 & when x t2

2 2

0

dtSo, I

(1 t )(1 4 t )

2 2

0

1 1 4I dt

3 1 t 1 4 t

1

1

0

1 tan (2t)I tan t 4

3 2

1I 2 0 0

3 2 2 6

.

Q16. Evaluate : /4

0

sin x cos xdx

3 sin 2x

.

Sol. Let /4

0

sin x cos xI dx

3 sin 2x

Put sin x cos x t (sin x cos x)dx dt

And, 2 2(sin x cos x) t 21 t sin 2x . Also when x 0 t 1 and when x t 04

0

2

1

dtI

4 t

0

1

1 2 tI log

2(2) 2 t

1 2 0 2 1

I log log4 2 0 2 1

1

I 0 log 34

log3

I4

.



Q17. Let a i 4j 2k

, b 3i 2 j 7k

and c 2i j 4k

. Find a vector d

which is perpendicular to

both a

and b

and c.d 27

.

Sol. Since d

is perpendicular to both a

and b

d is parallel to a b

.

i j k So d (a b) 1 4 2 32 i j 14 k

3 2 7

Also c.d (2i j 4k).(32 i j 14 k) 27

3 .

Therefore, r 96i 3j 42k

.

Q18. Find the shortest distance between the following lines :

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

and r 2i 4 j 5k (4i 6j 8k)

.

OR Find the equation of the plane passing through the line of intersection of the planes

2x y z 3 and 5x 3y 4z 9 0 and is parallel to the line x 1 y 3 5 z

2 4 5

.

Sol. Let 1 L : r i 2 j 3k (2i 3j 4k)

and 2

L : r 2i 4 j 5k (4i 6j 8k)

Here 1 1 a i 2j 3k, b 2i 3j 4k;

2 2 a 2i 4j 5k, b 4i 6 j 8k

Note that 2 1b 2b

that implies 1 2b b so, the lines are parallel.

Therefore, S.D. between two parallel lines, 2 1| (a a ) b |

d| b |

| (i 2 j 2k) (2i 3j 4k) |

d | 2i 3j 4k |

| 2i k | 5 5units

294 9 16 29

.

OR Plane passing through the intersection of the planes 2x y z 3 and 5x 3y 4z 9 0

is : : 2x y z 3 (5x 3y 4z 9) 0 : x(2 5 ) y(1 3 ) z( 1 4 ) 3 9 0 .

Since plane is parallel to the line x 1 y 3 5 z

2 4 5

with d.r.s 2, 4, 5 so, the normal to the

plane will be perpendicular to the line. Therefore, 2(2 5 ) 4(1 3 ) 5( 1 4 ) 0 10 12 20 4 4 5 0

18 3 1/6

Therefore the required plane is 5 1 2 3

: x 2 y 1 z 1 3 06 2 3 2

That is 7x 9y 10z 27 0 .

Q19. A man takes a step forward with probability 0.4 and backward with probability 0.6. Find the probability that at the end of 5 steps, he is one step away from the starting point. OR Suppose a girl throws a die. If she gets a 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? Sol. Given probability that the man takes a step forward = 0.4 = p so, q = 1 p = 0.6. Let E1 be the event that out of 5 steps the man takes exactly 3 steps are forward and 2 steps backward. Let E2 be the event that out of 5 steps the man takes exactly 2 steps are forward and 3 steps backward. Also let E be the event that at the end of 5 steps, the man is one step away from the starting point.

1 2P(E) P(E ) P(E ) P(3) P(2) + = +5 3 2 5 2 3

3 2C (0.4) (0.6) C (0.4) (0.6) + [n r n r

rP(r) C (p) (q)

3 2 2 3 2 2 210(0.4) (0.6) 10(0.4) (0.6) 10 (0.4) (0.6) [0.4 0.6] 10 (0.24) 0.576 + .

OR Let A : getting 1 or 2, B : getting 3, 4, 5 or 6, E : getting exactly one tail.

So, P(A) = 2/6, P B 4/6, 1 3 1

31

1 1 3P E|A C (using binomial distribution),

2 2 8

1

P E|B2

.



By Bayes theorem,

P E|B P(B)

P B|E =P E|A P(A) P E|B P(B)

1 48 82 6P B|E =

3 2 1 4 3 8 118 6 2 6

.

SECTION C Q20. Determine whether the relation R defined on the set R of all real numbers as R {(a, b):a, b R

and a b 3 S , where S is the set of all irrational numbers}, is reflexive, symmetric and transitive. OR Let A R R and * be the binary operation on A defined by (a, b) * (c, d) = (a + c, b + d). Prove that * is commutative and associative. Find the identity element for * on A. Also write the inverse element of the element (3, 5) in A.

Sol. We have R {(a, b):a, b R and a b 3 S , where S is the set of all irrational numbers}

Reflexivity : Let a be any real number. a a 3 3 S (a, a) R

So, R is reflexive.

Symmetry : Let ( 3,2) R 3 2 3 2 3 2 S

But if (2, 3) R 2 3 3 2 S

That is, (a, b) R doesnt necessarily imply (b,a) R a,b R So, R isnt symmetric.

Transitivity : Let (2, 5), ( 5, 3) R 2 5 3 S and 5 3 3 5 S

But if (2, 3) R 2 3 3 2 S

That is, (a, b) and (b,c) R doesnt necessarily imply (a,c) R a, b,c R

So, R isnt transitive as well. OR Given A R R and * is a binary operation on A defined by (a, b)*(c,d) (a c, b d) .

Commutativity : Let (a, b), (c,d) A .

Then (a, b)*(c,d) (a c, b d) (c a, d b) (c,d)*(a, b) .

Hence, (a, b)*(c,d) (c,d)*(a, b) . * is commutative.

Associativity : Let (a, b), (c,d), (e, f ) A . Then we have,

[(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 (c e), b (d f ) (a, b)*(c e,d f ) (a, b)*[(c,d)*(e,f )]

Hence, [(a,b)*(c,d)]*(e, f ) (a, b)*[(c,d)*(e, f )] . * is associative.

Now, Let (x, y) be identity element for * on A. Then (a, b)*(x, y) (a, b) (a x, b y) (a, b)

a x a, b y b x 0, y 0 . So, the identity element for the binary operation * is (0, 0).

Let inverse of element (3, 5) be (e, f) so, (3, 5) * (e, f) = (0, 0) (3 e, 5 f ) (0,0)

3 e 0, 5 f 0 e 3, f 5 inverse of (3, 5) is ( 3,5) .

Q21. Tangent to the circle 2 2x y 4 at any point on it in the first quadrant makes intercepts OA and

OB on x and y axes respectively, O being the centre of circle. Find the minimum value of (OA + OB).

Sol. Given circle 2 2x y 4 (i) Let P (, ) 2 2 4...(ii)

dy

Also 2x 2y 0dx

dy x

dx y T

at P

dy m

dx

Equation of tangent : y (x )

2 2x y

By (ii), x y 4 x y

14/ 4/

4 4

x-intercept OA, y-intercept OB



Let 4 4

OA OB s

2

4 4s

4

2 2 3/2

ds 4 4

d (4 )

2 3/2 2 1/22

2 3 2 3

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

d s 8 2d (4 )

2 2

2 3 2 5/2

d s 8 8( 2)

d (4 )

For local maxima and/or minima, 2 2 3/2

ds 4 40

d (4 )

3 2 3/2(4 )

6 2 3 2 4 6(4 ) 64 48 12 3 2 2m 6m 24m 32 0, where m

2(m 2) 0 or (m 4m 16) 0 2m 2, m 4m 16 0 m R

st2 [ P lies in 1 quadrant 2

2 5/2

at 2

d s 8 8(2 2)6 2 0

d (4 2)2 2

So, s is minimum at 2 .

Therefore, the minimum value of s 4 4

4 22 4 2

.

Q22. If the area bounded by the parabola 2y 16ax and the line y 4mx is 2a

sq.units12

, then using

integration, find the value of m.

Sol. Let 2y 16ax (i), y 4mx (ii)

By (i) & (ii), 2(4mx) 16ax 2x(m x a) 0

2

ax 0 or x

m

Acc. to quest., area bounded by the curves

2

a

2m

i ii

0

a[y y ]dx

12

2 2

a a

2m m

0 0

a4 a x dx 4m xdx

12

2 2a a 2

3/2 2m m0 0

8 aa x 2m x

3 12

3/2 2 2

3 4

8 a a aa 0 2m 0

3 m m 12

2 2

3

8 a a2

3 m 12

3

2 1 1

3 m 12 m 2 .

Q23. Show that the differential equation dy

(x y) x 2ydx

is homogeneous and solve it also.

OR Find the differential equation of the family of curves 2 2 2(x h) (y k) r , where h and k

are arbitrary constants.

Sol. We have dy

(x y) x 2ydx

dy x 2y

f (x, y) saydx x y

Put x x, y y , x 2 y x 2y

f ( x, y) f (x, y)x y x y

Hence the differential equation is homogenous.

Now put dy dv

y vx v xdx dx

dv x 2vx 1 2v

v xdx x vx 1 v

2dv 1 2v v v

xdx 1 v

2

1 v dxdv

v v 1 x

Let 2d

1 v A [v v 1] Bdv

1 v A[2v 1] B



On equating the coefficients of like terms, we get : 1 3

A , B2 2

2 2

3 1(2v 1) dxdv

2(v v 1) 2(v v 1) x

2

22

3 1 1dv log v v 1 log x

2 21 3v

2 2

2 2

1

2

3 2 2v 1 1 y xy xtan log log x C

2 2 x3 3

1 2 22y x 1

3 tan log y xy x C23 x

.

OR We have 2 2 2(x h) (y k) r (i) 2(x h)(1 0) 2(y k)(y 0) 0

(x h) (y k)y (ii) 21 (y k)y (y )

2[1 (y ) ]

(y k)y

Replacing value of (y k) in (ii),

2[1 (y ) ](x h) y

y

Substituting values of (x h) and (y k) in (i), we get :

2 22 22[1 (y ) ] [1 (y ) ]y r

y y

22 2 2 2[1 (y ) ]y [1 (y ) ] (ry )

2 2 2 2(y ) 1 [1 (y ) ] (ry ) 2 3

2

2

[1 (y ) ]r

y

Hence 2 3 2 21 2[1 y ] r y is the required differential equation of all circles of radius r.

Q24. Find the equation of a plane passing through the point P(6, 5, 9) and parallel to the plane determined by the points A(3, 1, 2), B(5, 2, 4) and C(1, 1, 6). Also find the distance of this plane from the point A.

Sol. Let AB 5i 2j 4k 3i j 2k 2i 3j 2k

, BC i j 6k 5i 2j 4k 6i 3j 2k

So, normal to the plane determined by the points A, B and C is m AB BC

i j k

m 2 3 2

6 3 2

12i 16j 12k d.r.s of normal to the required plane : 3, 4,3

[Note that the required plane and plane through A, B, C are parallel.]

Also the plane passes through P(6, 5, 9) i.e., a 6i 5j 9k

: r.m a.m

r.(3i 4j 3k) (6i 5j 9k).(3i 4 j 3k)

r.(3i 4j 3k) 18 20 27

r.(3i 4j 3k) 25

or 3x 4y 3z 25 (i)

Now distance of (i) from A(3, 1, 2) 2 2 2

3 3 4 ( 1) 3 2 25 6units

343 ( 4) 3

.

Q25. An urn contains 5 red and 2 black balls. Two balls are randomly drawn, without replacement. Let X represent the number of black balls drawn. What are the possible values of X? Is X a random variable? If yes, find the mean and variance of X. Sol. No. of red balls = 5 and no. of black balls = 2. As X : No. of black balls drawn possible values of X : 0, 1, 2. Clearly, X is a random variable.

X 0 1 2

P(X) 5

27

2

C 10

C 21

5 21 17

2

C C 10

C 21

22

72

C 1

C 21

10 10 1 12 4

Mean, X P(X) 0 1 221 21 21 21 7



And, variance 2

2 2 2 2 2 210 10 1 4 50X P(X) ( ) 0 1 221 21 21 7 147

.

Q26. Solve the following linear programming problem graphically. Minimise z 3x 5y

subject to the constraints x 2y 10, x y 6, 3x y 8, x 0, y 0 .

Sol. To minimize : z = 3x + 5y Subject to the constraints x 2y 10, x y 6, 3x y 8, x 0, y 0

Critical Points Value of z

A(10, 0) 30

B(2, 4) 26Min.Value

C(1, 5) 28

D(0, 8) 40

As the feasible region is unbounded; therefore 26 may or may not be the minimum value of z. In order to check, we plot the graph of inequality 3x 5y 26 .

It can be seen easily that the feasible region has no common point with 3x 5y 26 therefore, the minimum value of z is 26.

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3x+5y=26

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CBSE 2015 All India 1