Madhya Pradesh Board Question Papers for Class 12th Maths 2019 With Solutions In PDF

MPBSE 12th Maths Question Paper 2019 with Answers – Free Download

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MPBSE Class 12th Maths Question Paper With Solutions 2019

QUESTION PAPER CODE E – 236

SECTION – A

Question 1: Choose and write the correct options. 1* 5 = 5M

[i] Let A = {1, 2, 3}, then number of equivalence relations containing

(1, 2) is:

(A) l (B) 2 (C) 3 (D) 4

Solution: (B)

Total possible pairs = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}

Reflexive means (a, a) must be in relation.

Hence (1, 1), (2, 2), (3, 3) must be in a relation.

Symmetric means if (a, b) is in relation, then (b, a) must be in relation.

Therefore, if (1, 2) is in relation, then (2, 1) must also be in relation.

Transitive means if (a, b) is in relation, and (b, c) is in relation, then (a, c) is in relation.

If (1, 2) is in relation, and (2, 1) is in relation, then (1, 1) must be in relation.

Relation R₁ = {(1, 2), (2, 1), (1, 1), (2, 2), (3, 3)}

Total possible pairs = {(1, 1) , (1, 2), (1, 3), (2, 1) , (2, 2), (2, 3), (3, 1) , (3, 2), (3, 3)}

The smallest relation is R₁ = {(1, 2), (2, 1), (1, 1), (2, 2), (3, 3)}

Add (3, 2) also, as it is symmetric but, as (1, 2) and (2, 3) are there, (1, 3) is to be added as it is transitive.

Since we are adding (1, 3), (3, 1) also is to be added, as it is symmetric.

Relation R₂ = {(1, 2), (2, 1), (1, 1), (2, 2), (3, 3), (2, 3), (3, 2), (1, 3), (3, 1)}

So, only 2 possible relations are there which are equivalence.

[ii] If sin^-1 x = y, then

(A) 0 ≤ y ≤ 𝛑 (B) – 𝛑 / 2 ≤ y ≤ 𝛑 / 2

(C) 0 < y < 𝛑 (D) – 𝛑 / 2 < y < 𝛑 / 2

Solution: (B)

sin^-1 x = y

y = sin^-1 x

The range of principal value of sin is [- 𝛑 / 2, 𝛑 / 2].

So, – 𝛑 / 2 ≤ y ≤ 𝛑 / 2 is the correct answer.

[iii] If and A + A’ = I, then the value of ɑ is

(A) 𝛑 / 6 (B) 𝛑 / 3

(C) 𝛑 (D) 3𝛑 / 2

Solution: (B)

A + A’ = I

Since the matrices are equal, the corresponding elements can be equated.

2 cos ɑ = 1

cos ɑ = 1 / 2

cos ɑ = cos 60^o

ɑ = 𝛑 / 3

[iv] Let A be a nonsingular square matrix of order 3 x 3, then adj |A| is equal to

(A) |A|² (B) |A| (C) |A|³ (D) 3|A|

Solution: A

A (adj A) = |A| I

Taking determinants both sides

|A (adj A)| = ||A| I|

|A (adj A)| = |A| |adj A| (|𝐴𝐵|=|𝐴||𝐵|)

||A|I| = |A|³|I| = |A|³

|A (ad jA)| = ||A|I|

Putting values

|A| |adj (A)| = |A|³

|adj (A)| = |A|³ / |A|

|adj (A)| = |A|²

B is the correct answer.

[v] Function f (x) = |x| at x = 0 is

(A) Continuous but not differentiable

(B) Discontinuous and not differentiable

(C) Discontinuous and differentiable

(D) Continuous and differentiable

Solution: (A)

f (x) = |x|

f is continuous at x = 0 if the left-hand limit is equal to the right-hand limit.

lim _x→0- f (x) = lim _x→0+ f (x) = f (0)

LHL = lim _x→0- f (x)

= lim _x→0- |x|

= lim _x→0 (-x)

= – (0)

= 0

RHL = lim _x→0+ f (x)

= lim _x→0+ |x|

= lim _x→0 (x)

= 0

LHL = RHL and f (0) = 0.

Hence lim _x→0- f (x) = lim _x→0+ f (x) = f (0).

f is continuous at x = 0.

Question 2: Fill in the blanks. 1 * 5 = 5M

Solution:

y = x + e^x

dy / dx = 1 + e^x

d²y / dx² = e^x

Solution:

∫₁² x² dx

= (x³ / 3)₁²

= 1 / 3 (2³ – 1³)

= 7 / 3

Solution:

2x + y – z = 5

Since it cuts the x-axis, y and z = 0.

2x = 5

x = 5 / 2

Question 3: Write true/false in the following statements. 1 * 5 = 5M

(i) If E₁ and E₂ are exclusive events, then P (E₁ ⋂ E₂) is 0. [True]

(ii) If P (A) = 1 / 2, P (B) = 0, then P (A / B) is not defined. [True]

(iii) The objective function of an LPP is always linear. [True]

(iv) The feasible region of a linear programming problem is always a linear

Polygon. [True]

(v) The value of ∫₀^𝛑 cos³ x dx is 0. [True]

Question 4: Match the correct pairs. 1 * 5 = 5M

Solution:

(i) – (c)

(ii) – (a)

(iii) – (d)

(iv) – (f)

(v) – (g)

Question 5: Write the answers in one word/sentence each. 1 * 5 = 5M

(i) The maximum value of x^1/x is

Solution:

y = x^1/x

log y = (1 / x) log x

(1 / y) (dy / dx) = (1 – logx) / x²

dy / dx = 0

1 – logx = 0

log x = 1

x = e

The maximum value is e^1/e.

(ii) Rate of change in the area of a circle having radius r when r = 5 cm.

Solution:

Area of circle = A = 𝛑r²

dA / dr = d (𝛑r²) / dr

= 2𝛑r

For r = 5,

dA / dr = 2𝛑 * 5 = 10𝛑

(iii) The slope of the normal to the curve y = 2x² + 3 sin x at x = 0.

Solution:

y = 2x² + 3 sin x

dy / dx = 4x + 3 cos x

Slope of tangent * Slope of normal = -1

(4x + 3 cos x) * Slope of normal = -1

Slope of normal = -1 / (4x + 3 cos x)

At x = 0,

Slope of normal = -1 / (4 * 0 + 3 cos 0)

= -1 / 3

(iv) The minimum value of 3 sin θ + 4 cos θ is

Solution:

Let f (θ) = 3 sin θ + 4 cos θ

f ‘ (θ) = 3 cos θ – 4 sin θ —- (1)

f ‘’ (θ) = – 3 sin θ – 4 cos θ —- (2)

Now for maximum of minimum value of f (θ), f ′(θ) = 0.

3 cos θ – 4 sin θ = 0

sin θ = ± 3 / 5; cos θ = ± 4 / 5

From (2), sin θ = – 3 / 5; cos θ = – 4 / 5

f ‘’ (θ) > 0

The minimum value of f (θ) = – (3² + 4²) / 5 = -5.

(v) Derive the equation of tangent line at (1, 1) on curve y = x³.

Solution:

dy / dx = 3x²

At (1, 1), dy / dx = 3

y – 1 = 3 (x – 1)

y – 1 = 3x – 3

y = 3x – 2

SECTION – B

Question 6: [i] If then find the value of A . B.

OR

[ii] If then find the value of A – B.

Solution:

OR

Question 7: [i] Check the continuity of the function f given by f (x) = 2x + 3 at x = 1.

OR

[ii] Prove that the modulus function f (x) = |x| is not differentiable at x = 0.

Solution:

LHS = lim _x→1 f (x)

= lim _x→1 (2x + 3)

= 2 * 1 + 3

= 5

RHS = f (1)

= 2 * 1 + 3

= 5

LHS = RHS

The function is continuous.

OR

– x x < 0

f (x) is not differentiable at x = 0.

f ‘ (x) = x / |x|

Question 8: [i] Find the following integrals ∫(x^2/3 + 1) dx.

OR

[ii] Find the following integrals ∫[(1 – sinx) / cos² x] dx.

Solution:

= ∫(x^2/3 + x⁰) dx

= ∫x^2/3 dx + x⁰ dx

= ∫x^{(2/3) + 1} / [(2 / 3) + 1] + ∫x^{0 + 1} / [0 + 1] + c

= x^(5/3) / [(5 / 3)] + x + c

= (3x^5/3 / 5) + x + c

OR

Question 9: [i] Find the unit vector in the direction of vector a = 2i + 3j + k.

OR

[ii] Find the projection of the vector a = 2i + 3j + 2k on the vector b = i + 2j + k.

Solution:

|a| = √2² + 3² + 1²

= √4 + 9 + 1

= √14

Unit vector = a / |a|

= 2i + 3j + 1k / √14

= (2 / √14) i + (3 / √14) j + (1 / √14) k

OR

b = i + 2j + k

(a . b) = (2 * 1) + (3 * 2) + (2 * 1)

= 2 + 6 + 2

= 10

|b| = √1² + 2² + 1

= √1 + 4 + 1

= √6

Projection of vector a on b = (a . b) / |b|

= (1 / √6) (10)

= 10 / √6

= (10 / √6) (√6 / √6)

= (10 / 6) (√6)

= (5 / 3) (√6)

Question 10: [i] Find the equation of the plane whose intercepts on the coordinate axes are -4, 2, 3.

OR

[ii] Find the angle between the line (x + 1) / 2 = (y / 3) = (z – 3) / 6 and the plane 3x + y + z = 7.

Solution:

(x / – 4) + (y / 2) + (z / 3)

OR

The plane is 3x + y + z = 7.

sin ɸ = (2 * 3 + 3 * 1 + 6 * 1) / (√4 + 9 + 36) (9 + 1 + 1)

= (6 + 3 + 6) / 7 * √11

= 15 / 7√11

ɸ = sin^-1 (15 / 7√11)

Question 11: [i] The radius of an air bubble is increasing at the rate 1 / 2 cm per second. At what rate is the volume of the bubble increasing when the radius is 1 cm.

OR

[ii] Find the slope of the normal to the curve x = 1 – a sin θ, y = bcos² θ at

θ = 𝛑 / 2.

Solution:

It is given that dr / dt = (1 / 2) cm / s — (1)

The bubble is in the shape of a sphere.

So, the volume of the bubble = volume of the sphere = (4 / 3) 𝛑r³

dv / dt at r = 1 cm

dv / dt = d ((4 / 3) 𝛑r³) / dt

= [(4 / 3) 𝛑] 3r² (dr / dt)

= 4𝛑r² (1 / 2)

= 2𝛑r²

When radius is 1 cm, dv / dt = 2𝛑 (1)²

= 2𝛑 cm³ / sec

OR

dx / dθ = 0 − a cos θ

dy / dθ = − 2b cos θ sin θ

dy / dx = (dy / dθ) / (dx / dθ)

= (− 2b cos θ sin θ) / (− a cos θ)

= (2b / a) sin θ

The slope of the tangent at θ = 𝛑 / 2 is given by,

(dy / dx)_{θ = 𝛑 / 2} = [(2b / a) sin θ]_{θ = 𝛑 / 2}

= (2b / a) sin (𝛑 / 2)

= 2b / a

The slope of the normal at θ = 𝛑 / 2 is given by,

1 / (slope of the tangent at θ = 𝛑 / 4) = – 1 / (2b / a)

= – a / 2b

Question 12: [i] For what value of x is y = x (5 – x) maximum or minimum?

OR

[ii] Use differentials to find the value of √49.5.

Solution:

y = 5x – x²

dy / dx = 5 – 2x

dy / dx = 0

5 – 2x = 0

x = 5 / 2

d²y / dx² = – 2 (negative)

x = 5 / 2 is maximum and no minimum.

OR

y = x^1/2

dy / dx = 1 / 2√x

= 1 / 2 * (49)

= 1 / (2 * 7)

= 1 / 14

δy = (dy / dx) (δx)

= (1 / 14) (0.5)

= 1 / 28

√49.5 = y + δy

= 7 + (1 / 28)

= 7.036

Question 13: [i] If a + b + c = 0, then prove that a x b = b x c = c x a.

OR

[ii] Find the area of parallelogram whose adjacent sides are given by the vectors a = 3i + j + 4k and b = i – j + k.

Solution:

a x (a + b + c) = a x 0

a x a + a x b + a x c = a x 0

0 + a x b + a x c = a x 0

a x b + a x c = 0

a x b = – a x c

a x b = c x a

Similarly

a + b + c = 0

b x (a + b + c) = b x 0

b x a + b x b + b x c = b x 0

b x a + b x b + b x c = b x 0

b x a + b x c = 0

b x c = – b x a

b x c = a x b

a x b = b x c = c x a

OR

b = i – j + k

Area of the parallelogram = |a x b|

a x b =

= i (1 × 1 – (−1) × 4) − j (3 × 1 – 1 × 4) + k (3 × −1 − 1 × 1)

= i (1 − (– 4)) − j (3 − 4) + k (−3 −1)

= i (1 + 4) − j (−1) + k (− 4)

= 5i + j − 4k

Magnitude of a x b = √(5² + 1² + (−4)²)

|a x b| = √(25 + 1 + 16)

= √42

Area of parallelogram = |a x b| = √42

Question 14: [i] Find the minimum distance between the line l₁ and l₂ given by

r = i + 2j – 4k + ƛ (2i + 3j + 6k)

r = 3i + 3j – 5k + µ (2i + 3j + 6k)

OR

[ii] Find the distance of the plane 2x – 3y + 4z – 6 = 0 from the origin.

Solution:

r = 3i + 3j – 5k + µ (2i + 3j + 6k)

The above two lines passes through the points having position vectors,

a₁ = i + 2j – 4k

a₂ = 3i + 3j – 5k and are parallel to the vector b = 2i + 3j + 6k

a₂ – a₁ = 2i + j – k

(a₂ – a₁) b = (2i + j – k) (2i + 3j + 6k)

=

= 9i – 14j + 4k

|(a₂ – a₁) b| = √9² + (- 14)² + 4²

= √81 + 16 + 16

= √293

|b| = √2² + 3² + 6²

= √4 + 9 + 36

= 7

Shortest distance = |(a₂ – a₁) b| / |b|

= √293 / 7

OR

2x – 3y + 4z = 6 — (1)

The direction ratios are given by a = 2, b = – 3 and c = 4.

√a² + b² + c² = √2² + (- 3)² + 4²

= √4 + 9 + 16

= √29

The direction cosines are l = 2 / √29, m = – 3 / √29, n = 4 / √29.

The equation of the plane is lx + my + nz = d.

(2 / √29) x + (- 3 / √29) y + (4 / √29) z = d

2x – 3y + 4z = d √29

On comparing with equation (1),

6 = d √29

d = 6 / √29

Question 15: [i] Show that relation “is equal to” in sets is an equivalence relation.

OR

[ii] If f (x) = x² and g (x) = x + 3. x ∈ R , then find the value of (g o f) x , (f o g) x , (f o g)².

Solution:

a. Reflexivity: x = x,

b. Symmetry: if x = y then y = x,

c. Transitivity: if x = y and y = z then x = z.

All of these are true.

OR

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

= f (x) + 3

= x² + 3

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

= (x + 3)²

= x² + 9 + 6x

(f o g)² = (2 + 3)²

= 5²

= 25

Question 16: [i] Show that sin^-1 (3 / 5) – sin^-1 (8 / 17) = cos^-1 (84 / 85).

OR

[ii] Prove that cos^-1 x = 2 cos^-1 √(1 + x) / 2.

Solution:

sinx = 3 / 5 ; siny = 8 / 17

cosx = √(1 – sin²x)

= √(1 – (3 / 5)²)

= √(1 – 9 / 25)

= 4/5

cosy = √(1 – sin²y)

= √(1 – (8 / 17)²)

= √(1 – 64 / 289)

= 15 / 17

cos (x – y) = cos x cos y + sin x sin y

= 4 / 5 x 15 / 17 + 3 / 5 x 8 / 17 = 60 / 85 + 24 / 85 = 84 / 85

⇒ x – y = cos^-1 (84 / 85)

⇒ sin^-1 (3 / 5) – sin^-1 (8 / 17) = cos^-1 (84 / 85)

OR

Put x = cos θ

cos^-1 cos θ = 2 cos^-1 √(1 + cos θ) / 2

θ = 2 cos^-1 √[1 + 2 cos² θ / 2] / 2

= 2 cos^-1 cos (θ / 2)

= θ

θ = θ

LHS = RHS

Question 17: [i] Prove that .

OR

[ii] Find the area of the triangle whose vertices are (3, 8), (- 4, 2) and (5, 1).

Solution:

= 0 + 2c (b (a + b) – cb) – 2b (cb – c(c + a))

= 2c (ab + b² – cb) – 2b (cb – c² – ca)

= 2abc + 2cb² – 2bc² – 2b²c + 2bc² + 2abc

= 2abc + 2abc + 2cb² – 2cb² – 2bc² + 2bc²

= 4abc + 0 + 0

= 4abc

OR

Δ = (1 / 2)

= (1 / 2) (3 (2 – 1) – 8 (- 4 – 5) + 1 (- 4 – 10))

= (1 / 2) (3 + 72 – 14)

= 61 / 2

Question 18: [i] Find the equation of the plane the coordinate point (1, – 1, 2) and is perpendicular to each of the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8.

OR

[ii] Show that the angle between any two diagonals of a cube is cos^-1 (1 / 3).

Solution:

a (x – 1) + b (y + 1) + c (z – 2) = 0 —- (1)

It is given that (1) is perpendicular to the plane 2x + 3y – 2z = 5.

2a + 3b – 2c = 0 —- (2)

It is given that (1) is perpendicular to the plane x + 2y – 3z = 8.

a + 2b – 3c = 0 —- (3)

On solving (1), (2), (3),

– 5 (x – 1) + 4 (y + 1) + 1 (z – 2) = 0

5x – 4y – z = 7

OR

There exist our diagonals for a cube.

The direction ratios of the diagonals are given by

(a – 0), (a – 0), (a – 0) = a, a, a and (0 – a), (a – 0), (a – 0) = a, a, a

Let θ be the angle between the diagonals.

cos θ = a (-a) + a (a) + a (a) / [√a² + a² + a² √- a² + a² + a²]

= a² / 3a²

= 1 / 3

cos θ = 1 / 3

θ = cos^-1 (1 / 3)

Question 19: [i] Draw the graph of the inequality 3x + 2y ≤ 6.

OR

[ii] Find the minimum value of P = 2x + 4y, subject to constraints:

4x + 3y ≤ 12

x + 2y ≥ 4

x, y ≥ 0

Solution:

OR

The value is minimum at the origin.

Question 20: [i] If P (A) = ½, P (B) = ¼ and P (A ⋂ B) = ¼, then find the following:

(i) P (A / B)

(ii) P (B / A)

OR

[ii] In four throws of two dice what is the probability of getting the same figure on both dice?

Solution:

P (B) = ¼

P (A ⋂ B) = ¼

(i) P (A / B) = P (A ⋂ B) / P (B)

= (1 / 4) / (1 / 4)

= 1

(ii) P (B / A) = P (A ⋂ B) / P (A)

= (1 / 4) / (1 / 2)

= 1 / 2

OR

p = 6 / 36 = 1 / 6

q =1 – (1 / 6) = 5 / 6

P (success) = 1 – P (no doublet)

= 1 – P (X = 0)

= 1 – ⁴C₀ (5 / 6)^4-0 (1 / 6)⁰

= 1 – (5 / 6)⁴

= 1 – (625 / 1296)

= 671 / 1296

Question 21: [i] A family has two children. What is the probability that both the children are boys given that at least one of them is a boy?

OR

[ii] Find the probability distribution of numbers of doublets in three throws of a pair of dice.

Solution:

A: at least one of the children is boy: {BB, BG, GB}

B: both are boys: {BB}

P (B / A) = P (A ⋂ B) / P (A)

= (1 / 4) / (3 / 4)

= 1 / 3

OR

(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

P (getting a doublet) = 6 / 36 = 1 / 6

P (not getting a doublet) = 1 – 1 / 6 = 5 / 6

When two dies are thrown thrice, 0 doublet or 1 doublet or 2 doublets or 3 doublets can be obtained.

X can take values 0, 1, 2, 3.

P (X = 0)

P (X = 0) = P (0 doublet on three throws)

= 5 / 6 × 5 / 6 × 5 / 6

= 125 / 216

P (X = 1)

P (X = 1) = P (one doublet on three throws)

= 1 / 6 × 5 / 6 × 5 / 6 + 5 / 6 × 1 / 6 × 5 / 6 + 5 / 6 × 5 / 6 × 1 / 6

= 3 × 5 / 6 × 5 / 6 × 1 / 6

= 75 / 216

P(X = 2)

P(X = 2) = P(two doublet on three throws)

= 1 / 6 × 1 / 6 × 5 / 6 + 1 / 6 × 5 / 6 × 1 / 6 + 5 / 6 × 1 / 6 × 1 / 6

= 3 × 1 / 6 × 1 / 6 × 5 / 6

= 15 / 216

P(X = 3)

P(X = 3) = P(three doublets on three throws)

= 1 / 6 × 1 / 6 × 1 / 6

= 1 / 216

The probability distribution is

Question 22: [i] If , then prove that A’ . A = I.

OR

[ii] If then verify that

(a) (A’)’ = A

(b) (A + B)’ = A’ + B’

Solution:

OR

(b)

Question 23: [i] Discuss the continuity of the following function

OR

[ii] Prove that the function

is not differentiable at x = 1.

Solution:

f (x) is continuous at x = a ⇔ lim _x→a f (x) = f (a)

To show that lim _x→0 (x) sin (1 / x) = f (0)

Let z = 1 / x, then as x → 0, z → ∞

lim _x→0 (x) sin (1 / x) = lim _{z → ∞} (1 / z) sin z

= lim _{z → ∞} sin z / z

= 0

OR

y = 1 – x

y = x² – 1

x² = y + 1

From the graph, the function is not differentiable at x = 1.

Question 24: [i] Evaluate ∫(xe^x) / (1 + x²) dx.

OR

[ii] Evaluate ∫₀¹ (tan^-1 x) / (1 + x²) dx.

Solution:

I = ∫(x + 1 – 1)e^x / (1 + x)² dx

I = ∫e^x (1 + x) dx – ∫e^x (1 + x)² dx

Applying integration by parts in first integral,

= e^x / (1 + x) – ∫- (1 / (1 + x)²) e^x dx – ∫e^x / (1 + x)² dx + C

= e^x / (1 + x) + ∫ (1 / (1 + x)²) e^x dx – ∫e^x / (1 + x)² dx + C

I = e^x / 1 + x) + C

OR

dx / 1 + x² = dt

When x = 0, t = tan⁻¹ 0 = 0

When x = 1, t = tan⁻¹ 1 = π / 4

I =∫₀^π/4 t dt

= [t² / 2]

= (1 / 2) [t²]

= (1 / 2) (π / 4)² − 0

= π² / 32

Question 25: [i] Find the area enclosed by circle x² + y² = a².

OR

[ii] Find the area of region bounded by the curves y₁ = sin x and y₂ = cosx

between x = 0 and x = π / 4.

Solution:

y² = a² – x²

y = √a² – x²

Area of circle = 4 × Area of first quadrant

= 4∫^a₀ y dx

= 4∫^a₀√(a² – x²)dx

= 4 [x / 2 (√(a² – x²)) + (a² / 2) sin^-1 x / a]^a₀

= 4 [0 + (a² / 2) sin^-1 a / a – ( 0 + (a² / 2) sin^-1 0 / a) ]

= 4 [(a² / 2) sin^-1 1 – 0 ]

= 2a² (π / 2)

= πa² square units

OR

Area = ∫₀^π/4 (cosx – sinx) dx

= [sinx + cosx]₀^π/4

= sin (π / 4) + cos (π / 4) – [sin 0 + cos 0]

= (1 / √2) + (1 / √2) – [0 + 1]

= (2 / √2) – 1

= √2 – 1 square units

Question 26: [i] Verify that the function y = a cosx + b sin x. where a, b ∈ R is a solution of the differential equation d²y / dx² + y = 0.

OR

[ii] Solve the differential equation dy / dx = x · log x.

Solution:

dy / dx = – a sinx + b cosx

d²y / dx² = – a cos x – b sinx

LHS = d²y / dx² +y

= – a cosx – b sinx + a cosx + b sinx

= 0

= RHS

OR

y = [x³ / 2] log x − x² + c

y = (1 / 2) x² + (1 / 2) x² log x + c

y = (x² / 2) logx − x² / 4 + c