Cbse 2014 Delhi Maths

8/11/2019 Cbse 2014 Delhi Maths

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CBSE 2014 ANNUAL EXAMINATION DELHI

SET – 1 With Complete Explanations

Max. Marks : 100 Time Allowed : 3 HoursSECTION – A

Q01. Let * be a binary operation, on the set of all non-zero real numbers, given by a*bab

5 for all

a,b R {0} . Find the value of x, given that 2 * (x * 5) = 10.

Sol. We have a*b ab5

for all a,b R {0} .

Now 2 * (x * 5) = 10x.5

2 * 105

2* x 10

2.x10

5 x 25.

Q02. If sin 1 11sin cos x 1

5

, then find the value of x.

Sol. Given that sin1 11

sin cos x 1

5

1 1 1 11

sin sin sin cos x sin (1)

5

1 11 πsin cos x

5 2

1 1 11 π

sin cos x sin x5 2

1 11sin sin sinsin x

5

1

x5

.

Q03. If3 4 1 y 7 0

25 x 0 1 10 5

, then find (x – y).

Sol. We have3 4 1 y 7 0

25 x 0 1 10 5

6 8 1 y 7 0

10 2x 0 1 10 5

7 8 y 7 010 2x 1 10 5

+

By equality of matrices, we get :

8 y 0, 2x 1 5 + = 8 y, x 2 = x y 10 .

Q04. Solve the following matrix equation for x :1 0

[x 1] O2 0

.

Sol. We have1 0

[x 1] O2 0

[x(1) 1( 2) x(0) 1.0] 0 0 + [x 2 0] 0 0

By equality of matrices, we get : x 2 0 x 2 .

Q05. If2x 5 6 2

8 x 7 3

, write the value of x.

Sol. Given2x 5 6 2

8 x 7 3

(2x)(x) (5)(8) 6 3 7 ( 2)

22x 40 18 14

2x 36 x 6 .

Q06. Write the antiderivative of :1

3 xx

.

Sol. For the antiderivative of1

3 xx

, we take

13 x dx

x

3/2x3 2 x C

3/2

So, the required antiderivative is 2 x [x 1] C .

Q07. Evaluate :

3

2

0

dx

9 x .



CBSE 2014 Annual Exam Paper (Delhi) Compiled By OP Gupta (+91-9718240480)


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Sol. Let

33

1

2

00

dx 1 xI tan

9 x 3 3

3

1 1 1 1

0

1 3 0 1I tan tan [tan (1) tan (0)]

3 3 3 3

11 π 1 πI tan tan

3 4 3 4

π

I12

.

Q08. Find the projection of the vector ˆ ˆ î 3j 7k on the vector ˆ ˆ ˆ2i 3j 6k .

Sol. We know that the Projection of a vector a

on the vector b

is given by â .b

.

the projection of the vector ˆ ˆ î 3j 7k on the vector ˆ ˆ ˆ2i 3j 6k is

2 2 2

ˆ ˆ ˆ2i 3j 6k ˆ ˆ ˆ(i 3j 7k).2 ( 3) 6

ˆ ˆ ˆ2i 3j 6k 2 9 42ˆ ˆ ˆ(i 3j 7k). 57 7

.

Q09. If a and b

are two unit vectors such that a + b

is also a unit vector, then find the angle between

a and b

.

Sol. We have |a + b

|2 = |a

|2 + | b

|2 + 2 a .b

[ a + b, a and b are unit vectors

2 2

1 1 +1 2|a||b|cosθ, where θ is the angle between a and b

= +

1 2(1)(1)cosθ = 1

cosθ2

2π

θ3

.

Q10. Write the vector equation of the plane, passing through the point (a, b, c) and parallel to the planeˆ ˆ ˆr.(i j k) 2

+ + = .

Sol. Since required plane is parallel to the plane ˆ ˆ ˆr.(i j k) 2

+ + = so both will have their normal vector

as ˆ ˆ î j k + + [by comparing to r.m a.m

= , where m

is the normal vector].

Point on the required plane is (a, b, c). So, its position vector is ˆ ˆ â ai bj ck

= .

Required plane is ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆr.(i j k) (ai bj ck).(i j k)

+ + = + +

i.e., ˆ ˆ ˆr.(i j k) a b c

+ + = .

SECTION – BQ11. Let A = {1, 2, 3, …, 9} and R be the relation in A A defined by (a, b) R (c, d) if a + d = b + c for

(a, b), (c, d) in A A. Prove that R is an equivalence relation. Also obtain the equivalence class

[(2, 5)].

Sol. Reflexivity : Let (a, b) be an arbitrary element of A A. Then,

(a, b)A A a, bA.

So, a + b = b + a (a, b) R (a, b)

Thus, (a, b) R (a, b) (a, b)A A. Hence R is reflexive.

Symmetry : Let (a, b), (c, d)A A be such that (a, b) R (c, d). Then,

So, a + d = b + c c + b = d + a (c, d) R (a, b).Thus, (a, b) R (c, d) (c, d) R (a, b) (a, b), (c, d)A A. Hence R is symmetric.

Transitivity : Let (a, b), (c, d), (e, f)A A be such that (a, b) R (c, d) and (c, d) R (e, f). Then,

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

R (e, f). That is, (a, b) R (c, d) and (c, d) R (e, f) (a, b) R (e, f) (a, b), (c, d), (e, f)A A.

Hence R is transitive.

Since R is reflexive, symmetric and transitive so, R is an equivalence relation as well.

Equivalence class for [(2, 5)] = {(1,4), (2,5), (3,6), (4,7), (5,8), (6,9)}

Q12. Prove that :1 1 sin x 1 sin x x π

cot ; x 0,2 41 sin x 1 sin x

.

OR Prove that :1 1 11 5 2 π

2 tan sec 2 tan5 7 8 4

.





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Sol. LHS :1 1 sin x 1 sin x

cot1 sin x 1 sin x

1 1 sin x 1 sin x 1 sin x 1 sin xcot

1 sin x 1 sin x 1 sin x 1 sin x

21

2 2

[ 1 sin x 1 sin x ]cot

[ 1 sin x ] [ 1 sin x ]

1 1 sin x 1 sin x 2 1 sin x 1 sin xcot

[1 sin x] [1 sin x]

21 2 2 1 sin x

cot2sinx

1 2 2cos xcot

2sinx

1 1 cos xcot

sinx

2

1

x2cos

2cotx x

2sin cos2 2

1 x xcot cot RHS

2 2

. Hence Proved.

OR Let1 5 2 5 2

sec θ sec θ7 7

2

2 15 2 1 1tan θ sec θ 1 tan θ 1 θ tan

7 7 7

.

LHS : Let Y =1 1 11 5 2

2 tan sec 2 tan5 7 8

1 1 11 12 tan tan 2 tan

5 7 8

1 1 1

2 2

12 2

15 8tan tan tan71

1 15 8

[Using1 1

2

2x2 tan x tan

1 x

1 1 15 1 16tan tan tan

12 7 63

1 1

1 165 7 63tan tan

1 1612 17 63

1 15 175tan tan

12 425

1 15 7tan tan12 17

1

5 7

12 17tan5 7

112 17

1 169tan169

1 πtan (1) RHS

4

. Hence Proved.

Q13. Using properties of determinants, prove that:3

2y y z x 2y

2z 2z z x y (x y z)

x y z 2x 2x

.

Sol. LHS : Let

2y y z x 2y

2z 2z z x y

x y z 2x 2x

[Applying 1 1 2 2 2 3C C C ,C C C




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x y z y z x 2y

0 x y z z x y

x y z 0 2x

[Taking x + y + z common from C1 and C2 both

2

1 1 2y

(x y z) 0 1 z x y

1 0 2x

[Applying1 1 2

R R R

21 0 y z x

(x y z) 0 1 z x y

1 0 2x

+ [Expanding along R 1

21 z x y 0 z x y 0 1

(x y z) 1 0 (y z x)0 2x 1 2x 1 0

+

2(x y z) (2x 0) 0 (y z x)(0 1) +

3(x y z) RHS . Hence Proved.

Q14. Differentiate2

1 1 xtan

x

with respect to1 2cos [2x 1 x ] ; when x 0 .

Sol. Let2

1 1 xy tan

x

. Put x cosθ 1θ cos x...(i)

21 1 11 cos θ sinθ

y tan tan y tan tanθ θcosθ cosθ

1y cos x [By (i)

On differentiating w.r.t. x both sides, we get :2

dy 1

dx 1 x

.

Also let

1 2

z cos [2x 1 x ]

. Put x sinθ

1

θ sin x...(ii)

1 2 1z cos [2sinθ 1 sin θ] cos [2sin θcosθ] 1 1π

z cos [sin 2θ] sin [sin 2θ]2

πz 2θ

2

1πz 2sin x

2

[By (ii)

On differentiating w.r.t. x both sides, we get :2 2

dz 1 20 2

dx 1 x 1 x

.

Now,2

2

dy dy dx 1 1 x

dz dx dz 21 x

dy 1

dz 2 .

Q15. Ifxy x , then prove that

22

2

d y 1 dy y0

dx y dx x

.

Sol. Given y = xx … (i)

xlog(x ) x logxy y e e

On differentiating w.r.t. x both the sides, we get :x logxdy d

( )dx dx

e x logxdy d

(x logx)dx dx

e

xdy d dx (x) (log x) (log x) (x)

dx dx dx

xdy 1x (x) log x

dx x

xdy

x 1 log xdx

…(ii)

Again diff. w.r.t. x both sides, 2

x x

2

d y d dx 1 log x 1 log x (x )

dx dx dx

2

x x

2

d y 1x 1 log x (x ) 1 log x

dx x [By using (ii)





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2

2

d y y 1 dy dy

dx x y dx dx

[By using (i) & (ii)

So,

22

2

d y 1 dy y0

dx y dx x

. Hence Proved.

Q16. Find the intervals in which the function4 3 2(x) 3x 4x 12x 5 f is

(a) strictly increasing (b) strictly decreasing.

OR Find the equations of tangent and normal to the curve 3x a sin θ and y a cos θ at

πθ

4 .

Sol. Given 4 3 2(x) 3x 4x 12x 5 f 3 2(x) 12x 12x 24x f

For2(x) 0 12x(x x 2) 0 f 12x(x 1)(x 2) 0 x 1,0,2 .

Interval Sign of (x) f (x) f is strictly

( , 1 ] Negative Decreasing

[ 1,0 ] Positive Increasing

[0, 2] Negative Decreasing

[ 2, ) Positive Increasing

OR Given3

x a sin θ and3y a cos θ

On differentiating w.r.t. θ , we get :2dx

3a sin θcosθdθ

and2dy

3acos θ sinθdθ

2

2

dy dy dθ 3a cos θ sinθcotθ

dx dθ dx 3a sin θ cosθ

θ = π/4

dy πcot 1

dx 4

at

= mT & m N 1

Now equation of tangent :1 T 1

y y m (x x )

y – acos3 π

4

= (–1)[x – a sin3 π

4

] 2x 2y a

And equation of normal :1 N 1

y y m (x x )

y – acos3 π

4 = (1)[x – a sin

3 π

4] y x .

Q17. Evaluate :6 6

2 2

sin x cos xdx

sin x.cos x

. OR

2(x 3) x 3x 18 dx .

Sol. Let6 6

2 2

sin x cos xI dx

sin x.cos x

2 3 2 3

2 2

(sin x) (cos x)I dx

sin x. cos x

2 2 4 4 2 2

2 2

(sin x cos x)(sin x cos x sin xcos x)I dx

sin x. cos x

+ +

2 2 2 2 2

2 2

[sin x cos x] 3sin xcos xI dx

sin x. cos x

+

2 2

1I [ 3]dx

sin x cos x

2 2

2 2

sin x cos xI 3 dx

sin xcos x

2 2I sec x cosec x 3 dx I tan x cot x 3x C

OR Let 2I (x 3) x 3x 18 dx

Let x – 32d

A (x 3x 18) Bdx

x 3 A(2x 3) B

On equating the coefficients of like terms, we get : A = 1/2, B = –9/2.

21 9I (2x 3) x 3x 18 dx2 2

2 2

1 9I (2x 3) x 3x 18 dx x 3x 18 dx

2 2

Put 2x 3x 18 t in 1st integral (2x 3)dx dt + =





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2 21 9i.e., I t dt (x 3/2) (9/2) dx

2 2 +

3/2 22 2 2 21 t 9 (x 3/2) (9/2)

I (x 3/2) (9/2) log | x 3/2 (x 3/2) (9/2) | C2 3/2 2 2 2

++ +

2 3/2 2 21 9 (2x 3) 81 3I (x 3x 18) x 3x 18 log | x x 3x 18 | C

3 2 4 8 2

++ + +

Q18. Find the particular solution of the differential equation

x 2 y

e 1 y dx dy 0x

, given that y = 1when x = 0.

Sol. We havex 2 y

e 1 y dx dy 0x

x

2

yxe dx dy 0

1 y

Applying By Parts in 1st integral. Also in 2

nd integral, put 1 – y

2 = t

1ydy dt

2

i.e.,x xd 1 1

x e dx x e dx dx dt 0dx 2 t

x x 1xe e 2 t C

2

x x 2xe e 1 y C

As it is given that y = 1 when x = 0 so,0 0 20e e 1 1 C C= 1

Hence the required solution is :x x 2xe e 1 1 y .

Q19. Solve the differential equation :2

2

dy 2(x 1) 2xy

dx x 1

.

Sol. We have2

2

dy 2(x 1) 2xy

dx x 1

2 2 2

dy 2x 2y

dx (x 1) (x 1)

It is a linear differential equation of the formdy

P(x)y Q(x)dx

So, here 2 2 22x 2P(x) , Q(x)(x 1) (x 1)

= .

Now I.F.22

2xdx

log( x 1) 2( x 1) (x 1)

e e

Solution is given by,2 2

2 2

2y(x 1) (x 1) dx

(x 1)

2

2

2y(x 1) dx

(x 1)

2 1 x 1y(x 1) 2 log C

2 x 1

2 x 1y(x 1) log C

x 1

, is the required solution.

Q20. Prove that, for any three vectors a, b, c

, [a +b, b+ c, c+ a] 2[a, b, c]

= .

OR Vectors a, b and c

are such that a + b c 0

= and |a| 3, |b| 5 and |c| 7

= = = . Find the angle

between a and b

.

Sol. LHS : [a +b, b+c, c+a] (a+b).[(b+c) (c+a)]

=

(a + b).[b c b a c c+c a]

= + (a +b).[b c b a 0 c a]

= + +

a.(b c) a.(b a) a.(c a) b.(b c) b.(b a) b.(c a)

= +

[a b c] [a b a] [a c a] [b b c] [b b a] [b c a]

= +

[a b c] 0 0 0 0 [a b c]

= + 2[a b c] RHS

= . Hence Proved.

OR We have a + b c 0

= a + b c

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

=





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2 2 2| a| +| b| +a.b+ b.a | c |

= 2 2 23 +5 +2a.b 7

= 9+25+2a.b 49

=

15| a||b|cosθ , where θ is the angle between a & b

2

=

153 5 cosθ

2 =

1cosθ

2 =

πθ

3 = .

Q21. Show that the linesx 1 y 3 z 5

3 5 7

and

x 2 y 4 z 6

1 3 5

intersect. Also find their point

of intersection.Sol. Let L1:

x 1 y 3 z 5

3 5 7

and L2:

x 2 y 4 z 6μ

1 3 5

1Coordinates of random point on the line L is P(3 1,5 3,7 5) and on the line L2 is

Q(μ 2,3μ 4,5μ 6) .

If the lines intersect then P and Q must coincide.

i.e., 3 1 μ 2,5 3 3μ 4,7 5 5μ 6 =

3 μ 3...(i),5 3μ 7...(ii),7 5μ 11...(iii) =

Solving (i) and (ii), we get :1 3

,μ2 2

.

Substituting the values1 3

,μ2 2

in LHS of (iii) :1 3

7 5 11 RHS2 2

.

Hence the lines L1 and L2 intersect each other.

And their point of intersection is1 1 3

, ,2 2 2

.

Q22. Assume that each born child is equally likely to be a boy or a girl. If a family has two children,

what is the conditional probability that both are girls? Given that (i) the youngest is a girl.

(ii) at least one is a girl.

Sol. Let B and G represent the boy and the girl child respectively. If a family has two children, the

sample space will be S = {(B, B), (B, G), (G, B), (G, G)}.

Let A be the event that both children are girls. That means A = {(G, G)}

(i) Let E be the event that the youngest child is a girl. That gives E = {(B, G), (G, G)}

That implies, A E = {(G, G)}.

P(A E)P(A|E)

P(E)

1/4 1P(A|E)

2/4 2 .

(ii) Let F be the event that at least one child is a girl. That gives F = {(B, G), (G, B), (G, G)}

That implies, A F = {(G, G)}.

P(A F)P(A|F)

P(F)

1/4 1P(A|F)

3/4 3 .

SECTION – CQ23. Two schools P and Q want to award their selected students on the values of Discipline, Politeness

and Punctuality. The school P wants to award `x each, `y each and `z each for the three

respective values to its 3, 2 and 1 students with a total award money of `1000. School Q wants to

spend `1500 to award its 4, 1 and 3 students on the respective values (by giving the same award

money for the three values as before). If the total amount of awards for one prize on each value is

`600, using matrices, find the award money for each value.

Apart from the above three values, suggest one more value for awards.

Sol. Let the award money spent on the values of Discipline, Politeness and Punctuality be `x each, `y

each and `z each respectively.

3x + 2y + z = 1000, 4x + y + 3z = 1500, x + y + z = 600.

The given situation can be expressed as :

3 2 1 x 1000

4 1 3 y 1500

1 1 1 z 600





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3 2 1 x 1000

where A 4 1 3 , X y , B 1500

1 1 1 z 600

AX B 1

X A B …(i)

Now, |A| = 3(1 – 3) – 2(4 – 3) + 1 (4 – 1) = –5 0, so A –1

exists.

Consider Aij as the cofactors of the element aij of matrix A.

11 12 13

21 22 23

31 32 33

A 2, A 1, A 3,

A 1, A 2, A 1,

A 5, A 5, A 5

So,

2 1 5

.A 1 2 5

1 5

adj 1

2 1 51 1

A .A 1 2 5| A | 5

3 1 5

adj

By (i),

2 1 5 10001

X 1 2 5 15005

3 1 5 600

x 2000 1500 3000 1001

y 1000 3000 3000 2005

z 3000 1500 3000 300

By equality of matrices, we get : x 100, y 200, z 300 .

Hence the award money for the values of Discipline, Politeness and Punctuality is `100, `200 and`300 respectively.

Also, the value of Obedience can also be included for the awards.

Q24. Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is

1 1cos

3

.

Sol. Given slant height AB, l =2 2r h

2 2 2h r l …(i)

Volume of the cone, V21

πr h3

2 21V π[ h ]h

3 l [By using (i)

2 31V π[ h h ]

3 l

On diff. w.r.t. h both sides,2 2dV 1

π[ 3h ]dh 3

l

Again diff. w.r.t. h both sides,2

2

d V2πh

dh .

For local points of maxima and minima,2 2dV 1

0 π[ 3h ] 0 hdh 3 3

l

l

2

2

h3

d V2π 0

dh 3

l at

l

So, V is maximum at h3

l

.

Now inh 1

AOB, cosθ3

=l

, where θ is the semi-verticle angle of cone.

1 1θ cos

3

. Hence Proved.

Q25. Evaluate :

π/3

π/6

dx

1 cot x .

A

θ

h l

O r B





Page | 9

Sol. Let

π/3

π/6

dxI

1 cot x

π/3

π/6

sin dxI

sin cosx

x

x…(i)

Using

b b

a a

(x)dx (a b x)dx f f + , we have :

π/3

π/6

sin(π/6 π/3 x) dxI

sin(π/6 π/3 x) cos(π/6 π/3 x)

π/3

π/6

cosx dxI

cosx sin x

…(ii)

Adding (i) & (ii), we get :

π/3 π/3

π/6 π/6

sin dx cosx dx2Isin cosx cosx sin x

x

x

π/3

π/ 3

π/ 6

π/6

1 1 1 π πI 1dx [x]

2 2 2 3 6

=

πI

12 .

Q26. Find the area of the region in the first quadrant enclosed by the x-axis, the line y x and the

circle2 2x y 32 .

Sol. We have y = x …(i) and 2 2x y 32 …(ii)

4 2

2 4 2 2 1

0

4

1 x 32 x[x ] (4 2) x sin

2 2 2 4 2

1 11 1[16 0] 0 16sin (1) 8 16sin

2 2

Required Area 4π Sq.units .

Q27. Find the distance between the point (7, 2, 4) and the plane determined by the points A(2, 5,–3),

B(–2,– 3, 5) and C(5, 3,–3).

OR Find the distance of the point (–1,–5,–10) from the point of intersection of the line

ˆ ˆ ˆ ˆ ˆ ˆr 2i j 2k (3i 4 j 2k)

= + + + and the plane ˆ ˆ ˆr.(i j k) 5

+ = .

Sol. Plane determined by three non-collinear points is :

1 1 1

2 1 2 1 2 1

3 1 3 1 3 1

x x y y z z

x y y z z 0 .

x x y y z z

So, plane determined by the points A(2, 5,–3), B(–2,–3, 5) and C(5, 3,–3) is given as :

x 2 y 5 z 3

2 2 3 5 5 3 0

5 2 3 5 3 3

x 2 y 5 z 3

4 8 8 0

3 2 0

Taking –4 common from R 2,

x 2 y 5 z 3

4 1 2 2 0

3 2 0

Yy x

2 2x y 32

A

O B C X

Curve (ii) represents a circle of radius 4 2 and

centre at (0, 0)Solving (i) and (ii), we get : x = 4, –4; y = 4, –4

So, Required area = ar (OACBO)

=

4 4 2

(i) (ii)

0 4

y dx y dx +

=

4 4 2

2 2

0 4

xdx (4 2) x dx +





Page | 10

Applying 2 2 3R R R + ,

x 2 y 5 z 3

4 0 2 0

3 2 0

Expanding along R 1, ( x 2 ) [ 0 4 ] ( y 5 ) [ 0 6 ] ( z 3 ) [ 8 0 ] 0

i.e., 2 ( x 2 ) 3 ( y 5 ) 4 ( z 3 ) 0 2 x 3 y 4 z 7 0 = …(i)

Now distance of (7, 2, 4) from plane (i) is,2 2 2

| 2 ( 7 ) 3 ( 2 ) 4 (4 ) 7 |

2 3 4

p Units

i.e., 2 9 p Units [Using 1 1 1

2 2 2

| A x B y C z D |

A B C

p Units.

OR We have ˆ ˆ ˆ ˆ ˆ ˆr 2i j 2k (3i 4 j 2k)

= + + + …(i) and the plane ˆ ˆ ˆr.(i j k) 5

+ = …(ii)

Solving (i) and (ii) simultaneously, we get :

ˆ ˆ ˆ ˆ ˆ î(3 2) j(4 1) k(2 +2) .(i j k) 5 + + = (3 2) (4 1) (2 +2) 5 0 + =

Substituting value of 0 in (i), we get : ˆ ˆ ˆr 2i j 2k

= + = Position vector of point of intersection.

Therefore the point of intersection of the line and plane is P(2, –1, 2) say. Let Q(–1,–5,–10).2 2 2PQ ( 1 2) ( 5 1) ( 10 2) Units = 13 Units.

Q28. A dealer in rural area wishes to purchase a number of sewing machines. He has only `5760 to

invest and has space for at most 20 items for storage. An electronic sewing machine cost him `360

and a manually operated sewing machine `240. He can sell an electronic machine at a profit of

`22 and a manually operated sewing machine at a profut of `18. Assuming that he can sell all the

items that he can buy, how should he invest his money in order to maximize his profit? Make it as

a LPP and solve it graphically.

Sol. Let the number of electronic operated machine be x and the number of manually operated sewing

machines be y.

Z is maximum at x = 8 and y = 12.

The dealer should invest in 8 electronic and 12 manually operated machines.

Q29. A card from a pack of 52 playing cards is lost. From the remaining cards of the pack three cards are

drawn at random (without replacement) and are found to be all spades. Find the probability of the

lost card being a spade.

OR From a lot of 15 bulbs which include 5 defectives, a sample of 4 bulbs is drawn one by one

with replacement. Find the probability distribution of number of defective bulbs. Hence find the

mean of the distribution.

Sol. Let E1 : the lost card is a spade,1

E : the lost card is a non-spade card.

Let E : the three cards drawn are all spades.

So, P(E1)13 1

52 4 , P( 1E )

39 3

52 4 ,

Also P(E|E1)

12

351

3

C 12 11 10C 51 50 49

, P(E| 1E )

13

351

3

C 13 12 11C 51 50 49

By Bayes’ Theorem, P(E1|E) 1 1

1 1 1 1

P(E|E )P(E )

P(E|E )P(E ) P(E|E )P(E )

To maximize: Z = `(22x + 18y)

Subject to constraints:

x y 20 …(i)

360x 240y 5760 i.e., 3x 2y 48 …(ii)

and x 0, y 0 .

Corner points of the feasible region are A(0, 20),

P(8, 12), B(16, 0) and O(0, 0).

Now, ZA = `360, ZB = `352, ZP = `392, ZO = `0.





Page | 11

12 11 10 1

51 50 49 412 11 10 1 13 12 11 3

51 50 49 4 51 50 49 4

10 10

10 39 49

.

OR Let X denotes the number of defective bulbs. So, X = 0, 1, 2, 3, 4.

Total bulbs = 15, No. of defective bulbs = 5. So, no. of non-defective bulbs = 15 – 5 = 10.

X 0 1 2 3 4

P(X)

410

15

16

81

3

4

1

5 10C

15 15

32

81

2 2

4

2

5 10C

15 15

24

81

3

4

3

5 10C

15 15

8

81

4

4

4

5C

15

1

81

Mean = X P(X)16 32 24 8 1

0 1 2 3 481 81 81 81 81

4

3 .

NOTE: Only those Questions from Set 2 and 3 are given here which are not in common with Set 1.

SET – 2 With Complete Explanations

Q09. Evaluate :1cos (sin x)dx

.

Sol. We have,1 1 π

cos (sin x)dx cos cos x dx2

2π π 1x dx x x C

2 2 2

= , where C is integral constant.

Q10. If vectors a and b

are such that,2

|a| 3, |b| and a b3

= = is a unit vector, then write the angle

between a and b

.

Sol. Since â b |a||b|sinθ

n ,

where θ is the angle between vectors a and b

; and ˆ is the unit vector in the direction of a b

n .

ˆ|a b| |a||b||sin θ|| |

n 2

1 3 sin θ(1)3

1

sinθ2

= π

θ6

= .

Q19. Using properties of determinants, prove that :

3

a b 2c a b

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

.

Sol. LHS : Let

a b 2c a b

c b c 2a b

c a c a 2b

2(a b c) a b

2(a b c) b c 2a b

2(a b c) a c a 2b

[1 1 2 3

Applying C C C C

1 a b2(a b c) 1 b c 2a b

1 a c a 2b

[Taking 2(a + b + c) common from C1





Page | 12

1 a b

2(a b c) 0 b c a 0

0 0 c a b

[2 2 1 3 3 1

Applying R R R , R R R

b c a 0 a b a b2(a b c) 1 0 0

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

[Expanding along C1

22(a b c) 1.(a b c) 0 0

32(a b c) RHS . Hence Proved.

Q20. Differentiate1

2

xtan

1 x

with respect to

1 2sin [2x 1 x ] .

Sol. Let1

2

xy tan

1 x

. Put x sinθ 1θ sin x...(i)

1 1 1

2

sinθ sinθy tan tan y tan tanθ θ

cosθ1 sin θ

1y sin x [By (i)


dy 1

dx 1 x

.

Also let1 2z sin [2x 1 x ] . Put x sinθ 1θ sin x...(ii)

1 2 1z sin [2sinθ 1 sin θ] sin [2sin θ cosθ] 1z sin [sin2θ] 2θ

1z 2sin x

[By (ii)


dz 12

dx 1 x

.

Now,2

2

dy dy dx 1 1 x

dz dx dz 21 x

dy 1

dz 2 .

Q21. Solve the differential equation :2 2dy

cosec x log y x y 0

dx

.

Sol. We have2 2dy

cosec x log y x y 0dx

2

2

log y xdy dx 0

y cosec x

2 2log y y dy x sin x dx 0 [Integrate using By Parts in both integrals

2 2 2 2d dlog y y dy [log y] y dy dy x sin x dx [x ] sin x dx dx 0

dy dx

2

2

log y 1dy x ( cos x) 2xcos xdx 0

y y

2log y 1 dx cos x 2x cos x dx [2x] cos x dx dx 0y y dx

2log y 1x cos x 2x sin x 2 cosx 0

y y

2[1 log y] yx cos x 2 y(x sin x cosx) , which is the required solution.

Q22. Show that the lines5 x y 7 z 3

4 4 5

and

x 8 2y 8 z 5

7 2 3

are coplanar.

Sol. Let L1 :5 x y 7 z 3

4 4 5

x 5 y 7 z ( 3)

4 4 5

and L2 :

x 8 y 4 z 5

7 1 3

.

for L1,1 1 1

(x ,y ,z ) (5,7, 3)=1 1 1

; d.r.'s a ,b ,c 4,4, 5= and for L2,2 2 2

(x ,y ,z ) (8,4,5)= ;

2 2 2d.r.'s a ,b ,c 7,1,3= .





Page | 13

If L1 and L2 are coplanar then

2 1 2 1 2 1

1 1 1

2 2 2

x x y y z z

a b c 0

a b c

.

LHS :

2 1 2 1 2 1

1 1 1

2 2 2

x x y y z z 8 5 4 7 5 ( 3)

a b c 4 4 5

a b c 7 1 3

3 3 8

4 4 5

7 1 3

[Expanding along R 1

3[4 3 ( 5)] ( 3)[4 3 7( 5)] 8[4 1 7 4]

51 141 192 0 RHS .

Hence the lines L1 and L2 are coplanar lines.

Q28. Evaluate :

π

0

xtanxdx

secx.cosecx .

Sol. Letπ

0

xtanxI dx

secx.cosecx

π2

0

I x sin xdx …(i) [Usinga a

0 0

(x)dx (a x)dx f f

π

2

0

I (π x)sin (π x)dx

π π

2 2

0 0

I πsin x dx x sin x dx

π

2

0

I π sin xdx I [By using (i)

π/2

2

0

2I π 2 sin x dx

2 2 2

2a a

0 0

Let (x) sin x (π x) sin (π x) sin x (x)

using (x)dx 2 (x)dx, if (2a x) (x)

f f = f

f f f f

π/2

2

0

I π sin xdx ...(ii)

π/2 π/2

2 2

0 0

I π sin (π/2 x)dx π cos x dx ...(iii)

Adding (ii) and (iii), we get :

π/2

2 2

0

πI [sin x cos x]dx

2

π/2

0

πI 1dx

2

π/2

0

π π πI x 0

2 2 2

2π

I2

.

Q29. Prove that the semi-vertical angle of the right circular cone of given volume and least curved

surface area is 1cot 2 .

Sol. Given volume, V =21

π r h3

23V π r h …(i)

Also2 2 2

r h l

2

2 2

2

3Vr

π r

l …(ii) [By (i)

Curved surface area of the cone, C π r l

2

2

2

3VC π r r

π r

[By using (ii)

2 6 2 22 2 2 2 4

2 4 2

π r +9V 9VC π r π r + (r) say

π r r

f

A

θ

h l

O r B







Page | 15

2 2 2

2 2 2

2 2 2

x 1 x xxyz

y y 1 yxyz

z z z 1

[1 1 2 3

Applying R R R R

2 2 2 2 2 2 2 2 2

2 2 2

2 2 2

1 x y z 1 x y z 1 x y z

y y 1 y

z z z 1

+ + + + + +

[2 2 2

1Taking 1 x y z common from R + +

2 2 2 2 2 2

2 2 2

1 1 1

(1 x y z ) y y 1 y

z z z 1

+ + [1 1 2 2 2 3

Applying C C C ,C C C

2 2 2 2

2

0 0 1

(1 x y z ) 1 1 y

0 1 z 1

+ + [Expanding along R 1

2 2 2 2 2 21 1

(1 x y z ) 0 0 1 (1 x y z )[1(1 0)]0 1

+ + + +

2 2 2(1 x y z ) RHS + + . Hence Proved.

Q20. Differentiate2

1 1 x 1tan

x

with respect to1

2

2xsin , when x 0

1 x

.

Sol. Let2

1 1 x 1y tan

x

. Put x tanθ 1θ tan x...(i)

21 1 11 tan θ 1 secθ 1 1 cosθ

y tan tan tantanθ tanθ sinθ

2

1 1

θ2sinθ θ2y tan tan tan

θ θ 2 22sin cos2 2

11

y tan x2

[By (i)


dy 1 1

dx 2 1 x

.

Also let 1

2

2xz sin

1 x

. Put x tanθ 1θ tan x...(ii)

1 1

2

2 tanθz sin sin sin 2θ 2θ

1 tan θ

1z 2 tan x

[By (ii)


dz 12

dx 1 x

.

Now,2

2

dy dy dx dy 1 1 1 x

dz dx dz dx 2 1 x 2

dy 1

dz 4 .

Q21. Find the particular solution of the differential equationdy x(2 log x + 1)

dx sin y y cos y

, given that

πy(1)

2 .

Sol. We havedy x(2 log x + 1)

dx sin y y cos y

[sin y y cos y]dy x(2 log x + 1)dx

sin ydy y cos ydy 2x log x dx xdx + [Integrate using By Parts





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Cbse 2014 Delhi Maths