KBPE Class 12th Maths Question Paper With

Solution 2015

QUESTION PAPER CODE 7018

Question 1[a]: Choose the correct statement related to matrices

(i) A3 = A, B3 ≠ 0 (ii) A3 ≠ A, B3 = B

(iii) A3 = A, B3 = B (iv) A3 ≠ A, B3 ≠ B

Answer: (iii)

[b] If M = then verify the equation M2 - 10M + 11I2 = 0.

Solution:

[c] Inverse of the matrix .

Solution:

There is the inverse matrix on the right.

Question 2[a]: Prove that = (x + y + z) (x - y) (y - z) (z - x).

OR

[b] Prove that = 4!.

Solution:

[a]

= (y - x) (z - x) [(z2 + zx + x2) - (y2 + xy + x2)]

= (y - x) (z - x) [z2 - y2 + x (z - y)]

= (y - x) (z - x) [(z + y) (z - y) + x (z - y)]

= (y - x) (z - x) [(z - y) (x + y + z)]

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

OR

[b]

Question 3: Solve the system of linear equations.

x + 2y + z = 8

2x + y - z = 1

x - y + z = 2

Solution:

Question 4[a]: What is the minimum number of ordered pairs to form a non-

zero reflexive relation on a set of n elements?

Solution:

[a] n

[b] On the set R of real numbers, S is a relation defined as S

= {(x, y) x ∈ R, y ∈ R, x + y = xy}. Find a ∈ R such that 'a' is

never the first element of an ordered pair in S. Also find b

∈ R such that 'b' is never the second element of an ordered

pair in S.

Solution:

R = {(x , y) / x ∈ R, y ∈ R, x + y = xy}

x + y = xy is not defined when x = 1, y = 1.

Hence, a = 1, b =1.

[c] Consider the function f (x) = 3x + 4 / x - 2; x ≠ 2. Find a

function g(x) on a suitable domain such that (g o f) (x) = x =

(f o g) (x).

Solution:

f (x) = 3x + 4 / x - 2

(g o f) (x) = x = (f o g) (x)

So, g (x) is the inverse of f (x).

y = 3x + 4 / x - 2

y (x - 2) = 3x + 4

xy - 2y = 3x + 4

xy - 3x = 4 + 2y

x (y - 3) = 4 + 2y

x = (4 + 2y) / (y - 3)

f -1 (y) = (4 + 2y) / (y - 3)

f -1 (x) = (4 + 2x) / (x - 3)

Question 5[a]: What is the value of sin-1 (sin 160°)?

(i) 160° (ii) 70° (iii) -20° (iv) 20°

Answer: (iv)

[b]: Prove that 2 tan-1 (1 / 2) + tan-1 (1 / 7) = tan-1 (31 / 17).

Solution:

2 tan-1 (1 / 2) = tan-1 {[2 * (1 / 2)] / [1 - (1 / 2)]2}

= tan-1 (1 / (1 - (1 / 4))

= tan-1 (4 / 3)

2 tan-1 (1 / 2) + tan-1 (1 / 7)

= tan-1 (4 / 3) + tan-1 (1 / 7)

= tan-1 {[(4 / 3) + (1 / 7)] / [1 - (4 / 3) * (1 / 7)]}

= tan-1 [28 + 3 / 21 - 4]

= tan-1 (31 / 17)

Question 6[a]: Find a and b if the function is a

continuous function on [-2, 2].

Solution:

[a] f (x) is continuous at x = 0

lim x→0- f (x) = lim x→0+ f (x)

lim x→0 sin x / x = limx→0 (a * 2x)

1 = a * 20

a = 1

f (x) is continuous at x = 1

lim x→1- f (x) = lim x→1+ f (x)

lim x→1- (a * 2x) = lim x→1+ b + x

a * (21) = b + 1

2a = b + 1

2 = b + 1

b = 1

[b] How many of the functions f (x) = |x|, g (x) = |x|2 and h(x) = |x|3 are not

differentiable at x = 0?

(i) 0 (ii) 1 (iii) 2 (iv) 3

Answer: (iv)

Question 7[a]: Find dy / dx:

[i] x3 + 2x2y + 3xy2 + 4y3 = 5

[ii] x = 2cos3 θ, y = 2sin3 θ

[iii] y = sin-1 (2x √1 - x2); -1 ≤ x ≤ 1

Solution:

[i] x3 + 2x2y + 3xy2 + 4y3 = 5

3x2 + 2[x2 * (dy / dx) + y * 2x] + 3[2xy (dy / dx) + y2] + 12y2 (dy / dx) = 0

3x2 + 2x2 (dy / dx) + 4xy + 6xy (dy / dx) + 3y2 + 12y2 (dy / dx) = 0

dy / dx = - [3x2 + 4xy + 3y2] / [2x2 + 6xy + 12y2]

[b] x = 2cos3 θ, y = 2sin3 θ

dx / dθ = - 6cos2θ sinθ

dy / dθ = 6 sin2θ cosθ

dy / dx = [dy / dθ] / [dx / dθ]

= [6 sin2θ cosθ] / [- 6cos2θ sinθ]

= - tan θ

[c] y = sin-1 (2x √1 - x2)

Put x = sinθ; θ = sin-1 x

y = sin-1 (2 sinθ √1 - sin2θ)

= sin-1 [2 sinθ cos θ]

= sin-1 [sin 2θ]

= 2θ

= 2 sin-1x

dy / dx = 2 * (1 / √1 - x2)

= 2 /√1 - x2

Question 8[a]: Which of the following functions is always increasing?

(i) x + sin 2x (ii) x - sin 2x (iii) 2x + sin 3x (iv) 2x - sin x

Solution: (iv)

(b) The radius of a cylinder increases at a rate of l cm / s and its height

decreases at a rate of 1 cm/s. Find the rate of change of its volume when the

radius is 5 cm and the height is 15 cm. If the volume should not change even

when the radius and height arc changed, what is the relation bet~een the

radius and height?

Solution:

dr / dt = 1 cm / second

dh / dt = - 1 cm / second

V = 𝛑r2h

dV / dt = 𝛑 [r2 (dh / dt) + 2rh (dr / dt)]

r = 5, h = 15

= 𝛑 [25 * (-1) + 15 * 2 * 5 * 1]

= 125𝛑

V = constant, dV / dt = 0

[r2 (dh / dt) + 2rh (dr / dt)]

r2 (-1) + 2rh + 1 = 0

2rh - r2 - 0,

r (2h - r) = 0

r = 2h

(c) Write the equation of tangent at (1, 1) on the curve 2x2 + 3y2 = 5.

Solution:

2x2 + 3y2 = 5

2 * 2x + 3 * 2y (dy / dx) = 0

4x + 6y (dy / dx) = 0

dy / dx = - 4x / 6y = - 2x / 3y

(dy / dx) at (1, 1) = Slope of tangent = - 2 / 3

The equation of the tangent is y - 1 = (- 2 / 3) (x - 1)

2x + 3y - 5 = 0

Question 9[a]: Integrate the following:

[i] (x - 1) / (x +1)

[ii] sinx / sin (x - a)

[iii] 1 / √3 - 2x - x2

Solution:

[i] (x - 1) / (x +1)

= (x + 1) - 2 / (x + 1)

= 1 - (2 / (x + 1))

On integration,

x - 2 log |x + 1| + c

[ii] ∫[sinx / sin (x - a)] dx

Put x - a = t

x = t + a

dx = dt

∫[sin (t + a) / sin t] dt

= ∫[sin t cos a + cos t sin a] / [sin t] dt

= ∫[cos a + sin a cot t] dt

= cos a t + sin a * log |sin (x - a)| + c

= cos a (x - a) + sin a log |sin (x - a)| + c

[c] ∫1 / √3 - 2x - x2

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

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

= - (x2 + 2x + 1 - 4)

= 22 - (x + 1)2

∫1 / √22 - (x + 1)2 = sin-1 ([x + 1] / 2) + c

Question 10[a]: What is the value of ∫01 x (1 - x)9 dx?

(i) 1 / 10 (ii) 1 / 11 (iii) 1 / 90 (iv) 1 / 110

Answer: (iv)

[b] Find ∫01 (2x + 3) dx as the limit of a sum.

Solution:

∫01 (2x + 3) dx

Let f (x) = 2x + 3

a = 0, b = 1, nh = b - a = 1 - 0 = 1

h = (b - a) / n = 1 - 0 / n = 1 / n

= (b - a) lim n→∞ [1 / n] [f (a) + f (a + h) + f (a + 2h) …… f (a + [n - 1]h)

= (1) lim n→∞ [1 / n] [f (0) + f (h) + f (2h) + ….. f ([n - 1]h]

= lim n→∞ [1 / n] [3 + f (1 / n) + f (2 * (1 / n)) + f [(n - 1) * (1 / n)]

= lim n→∞ [1 / n] [3 + [(2 / n) + 3] + [(4 / n) + 3] + …… [2 (n - 1) / n + 3]]

= lim n→∞ n (1 / n) {(3 + 3 + 3 + ....n times) + [(2 / n) + (4 / n) + ....+ 2(n - 1) / n]}

= lim n→∞ (1 / n) [3n + (2 / n) (1 + 2 + ... + n - 1)]

= lim n→∞ (1 / n) [3n + (2 / n) (n - 1)(n - 1 + 1) / 2]

= lim n→∞ (1 / n) [3n + (2 / n) (n - 1) (n / 2)]

= lim n→∞ (1 / n) [3n + n - 1]

= lim n→∞ (1 / n) (4n - 1)

= lim n→∞ (n / n) [4 - 1 / n]

= lim n→∞ [4 - 1] / n

= 4 - 0

= 4

Question 11[a]: Consider the functions: f (x) = |x| - l and g (x) = 1 - |x|

(a) Sketch their graphs and shade the closed region between them.

(b) Find the area of their shaded region.

Solution:

[a]

[b] Area = ∫-11 (1 - |x|dx) + |∫-1

1 (|x| - 1)dx

= ∫-10 (1 - |x|dx) + ∫0

1 (1 - |x|dx)

= ∫-10 (1 + x) dx + ∫0

1 (1 - x) dx + |∫-10 (- x - 1) dx + ∫0

1 (x - 1) dx

= 1 / 2 + 1 / 2 + |(- 1 / 2) + (- 1 / 2)|

= 1 + 1

= 2 square units

Question 12[a]: Consider the family of all circles having their centre at the

point (1, 2). Write the equation of the family. Write the corresponding

differential equation.

[b] Write the integrating factor of the differential equation cos x (dy / dx) + y

= sin x; 0 ≤ x ≤ 𝛑 / 2.

Solution:

[a] (x - 1)2 + (y - 2)2 = r2

Differentiating,

2 (x - 1) + 2 (y - 2) (dy / dx) = 0

(x - 1) + (y - 2) (dy / dx) = 0

[b] cos x (dy / dx) + y = sin x

dy / dx + sec xy = tan x

IF = e∫P dx

= e∫secx dx

= elog |secx + tanx|

= sec x + tan x

Question 13[a]: If a, b, c, d respectively are the position vectors representing

the vertices A, B, C, D of a parallelogram, then write d in terms of a, b and c.

(b) Find the projection vector of b = i + 2j + k along the vector a = 2 i + j + 2k.

Also, write b as the sum of a vector along with a and a vector perpendicular to

a.

(c) Find the area of a parallelogram for which the vectors 2i + j and 3i + j + 4k

are adjacent sides.

OR

(a) Write the magnitude of a vector a in terms of the dot product.

(b) If a, b and a + b are unit vectors, then prove that the angle between a and

b is 2𝛑 / 3.

(c) If 2i + j - 3k and mi + 3 j - k are perpendicular to each other, then find m.

Also, find the area of the rectangle having these two vectors as sides.

Solution:

[a] Since ABCD is a parallelogram,

AB = DC

b - a = c - d

b + d = a + c

d = a + c - b

[b] Projection of b on a = (a . b) / |a|

= (2i + j + 2k) (i + 2j + k) / √22 + 12 + 22

= 6 / 3

= 2

[c] Area = a x b = 4i - 8j - k

|a x b| = √42 + (-8)2 + (-1)2

= √81

= 9

OR

[a] Dot product = |a| |a| cos 0

[b] Given a, b and a + b are unit vectors.

|a| = 1

|b| = 1

|a + b| = 1

|a + b|2 = 1

|a|2 + |b|2 + 2 |a| . |b| = 1

1 + 1 + 2 |a| . |b| = 1

|a| . |b| = - 1 / 2

cos θ = [|a| . |b|] / [|a| |b|]

= - 1 / 2 * 1 * 1

= - 1 / 2

- cos θ = 1 / 2

cos (𝛑 - θ) = 1 / 2

(𝛑 - θ) = cos-1 (1 / 2)

(𝛑 - θ) = 𝛑 / 3

θ = 𝛑 - (𝛑 / 3)

θ = 2𝛑 / 3

[c] If two vectors are perpendicular to each other then their dot product is 0.

cos 90o = 0

(2i + j - 3k) . (mi + 3j - k) = 0

2m + 3 + 3 = 0

m = -3

a = 2i + j - 3k

b = - 3i + 3j - k

|a| = √22 + 12 + (-3)2 = √14

|b| = √(-3)2 + 32 + (-1)2 = √19

Area of the rectangle = |a| . |b| . sin θ = √14 . √19 = √266

Question 14[a]: Write the Cartesian equation of the straight line through the

point (1, 2, 3) and along the vector 3i + j + 2k.

(b) Write a general point on this straight line.

(c) Find the point of intersection of this straight line with the plane 2x + 3y - z

+ 2 = 0.

(d) Find the distance from (1, 2, 3) to the plane 2x + 3y - z + 2 = 0.

Solution:

[a] Equation of a line passing through a point (x1, y1, z1) and parallel to a line with

direction cosines a, b, c is (x - x1) / a = (y - y1) / b = (z - z1) / c

Since line passes through (1, 2, 3),

x1 = 1, y1 = 2, z1 = 3

Also, the line is parallel to 3i + j + 2k.

a = 3, b = 1, c = 2

Equation of the line in cartesian form is given by

(x - 1) / 3 = (y - 2) / 1 = (z - 3) / 2

[b] Let (x - 1) / 3 = (y - 2) / 1 = (z - 3) / 2 = k

x = 3k + 1

y = k + 2

z = 2k + 3 is a general point on this straight line [here k is a constant].

[c] The equation of the plane is 2x + 3y − z + 2 = 0

Substituting the general point in the equation of the plane gives

⇒ 2 (3k + 1) + 3 (k + 2) − (2k + 3) + 2 = 0

⇒ k = −1

∴ The point of intersection is (−2,1,1).

[d] The direction cosines of the line and plane are (3, 1, 2) and (2, 3,

−1).

∴ The angle between the line and the plane is

(3, 1, 2) . (2, 3, −1) = |(3, 1, 2)| × |(2, 3, −1)| cosα

cos α = 1 / 2

Distance between the points (1, 2, 3) and (−2, 1, 1) is √14

units.

sin α = (perpendicular distance from the plane) / √14 The perpendicular distance from the plane is √21 / 2 units.

Question 15[a]: Consider the linear inequalities 2x + 3y ≤ 6,

2x + y ≤ 4, x ≥ 0, y ≥ 0.

(a) Mark the feasible region.

(b) Maximise the function z = 4x + 5y subject to the given constraints.

Solution:

[a]

[b] Z = 4x + 5y

At (2, 0), Z = 4 * 2 + 5 * 0 = 8

At (0, 2), Z = 4 * 0 + 5 * 2 = 10

At (3 / 2, 1), Z = 4 * (3 / 2) + 5 * 1 = 6 + 5 = 11

Z is maximum at (3 / 2, 1).

Question 16: In a factory, there are two machines A and B producing toys.

They respectively produce 60 and 80 units in one hour. A can run a maximum

of 10 hours and B a maximum of 7 hours a day. The cost of their running per

hour respectively amounts to 2,000 and 2.500 rupees. The total duration of

working these machines cannot exceed 12 hours a day. If the total cost cannot

exceed Rs. 25,000 per day and the total daily production is at least 800 units,

then formulate the problem mathematically.

Solution:

Let machine A work for x hours a day and machine B for y hours a day.

Total production per day = 60x + 80y,

x ≤ 10, y ≤ 7

Cost of working = 2000x + 2500y,

x + y ≤ 12,

2000x + 2500 y ≤ 25000,

60x + 80y ≥ 800

Minimise Z = x + y, subject to the above said linear constraints.

Question 17[a]: For two independent events A and B, which of the following

pair of events need not be independent?

(i) A' . B' (ii) A . B' (iii) A' . B (iv) A - B, B - A

Answer: (iv)

(b) If P (A) = 0.6, P (B) = 0.7 and P (A ⋃ B) = 0.9, then find P (A /

B) and P (B / A).

Solution:

P (A ⋃ B) = P (A) + P (B) - P (A ⋂ B)

0.9 = 0.6 + 0.7 - P (A ⋂ B)

P (A ⋂ B) = 0.4

P (A / B) = P (A ⋂ B) / P (B) = 0.4 / 0.7 = 4 / 7

P (B / A) = P (A ⋂ B) / P (A) = 0.4 / 0.6 = 4 / 6 = 2 / 3

Question 18[a]:

X 1 2 3 4 5

P (X) 1 / 2 1 / 4 1 / 8 1 / 16 p

The probability distribution of a random variable X taking values I, 2, 3, 4, 5

is given

(a) Find the value of P

(b) Find the mean of X

(c) Find the variance of X

Solution:

[a] [1 / 2] + [1 / 4] + [1 / 8] + [1 / 16] + p = 1

[8 + 4 + 2 + 1] / 16 + p = 1

(15 / 16) + p = 1

p = 1 / 16

[b] 𝛍 = ∑xip(xi)

= 1 * (1 / 2) + 2 * (1 / 4) + 3 * (1 / 8) + 4 * (1 / 16) + 5 * (1 / 16)

= 1 / 2 + 1 / 2 + 3 / 8 + 9 / 16

= 31 / 16

= 1.9375

[c] E (X2) = ∑x2ip(xi)

= 1 * (1 / 2) + 4 * (1 / 4) + 9 * (1 / 8) + 16 * (1 / 16) + 25 * (1 / 16)

= 1 / 2 + 1 + 9 / 8 + 1 + 25 / 16

= 83 / 16

= 5.1875

Var (X) = E (X2) - [E (X)]2

= 5.1875 - (1.9375)2

= 1.4335