# HBSE Class 12 Maths Question Paper 2020 Answer Key

HBSE Class 12 Maths Question Paper 2020 Answer Key

HBSE Class 12 Maths Previous Year Question Paper with Answer. HBSE Board Solved Question Paper Class 12 Maths 2020. HBSE 12th Question Paper Download 2020. HBSE Class 12 Maths Paper Solution 2020. Haryana Board Class 12th Maths Question Paper 2020 Pdf Download with Answer.

Subjective Questions (Coming Soon…)

SET-A

Q1. Let f : R → R is defined as f(x) = x³ then f is :

(A) One-one, into

(B) One-one, onto

(C) Many-one, onto

(D) Many-one, into

Ans. (B) One-one, onto

Q2. The principal value of tan-¹x is :

(A) [0, π/2]

(B) [0, π]

(C) [-π/2, π/2]

(D) None of these

Ans. (D) None of these (-π/2, π/2)

Q3. If X + Y = [5    2]  and X – Y = [3     6] then matrix X:

[0    9]                      [-2   1]

(A) [4      4]

[-1    5]

(B) [8      8]

[-2  10]

(C) [1    -2]

[1      4]

(D) None of these

Ans. (A) [4      4]

[-1    5]

Q4. If det. |2     4| = |2x   4| then the value of x is :

|5     1|    |6     x|

(A) 6

(B) ±6

(C) -6

(D) None of these

Ans. (D) None of these (x = ± √3 )

2(1) – 5(4) = 2x(x) – 6(4)

2 – 20 = 2x² – 24

-18 = 2x² -24

2x² -24 = -18

2x² = -18 + 24 = 6

x² = 6/2 = 3

x = ± √3

Q5. Differentiate sec(tan√x) with respect to x.

Sol. sec(tan√x).tan(tan√x).sec²√x.1/2√x

Q6. f(x) = x³ – 3x + 4  has a maxima at x is equal to :

(A) -1

(B) 1

(C) 0

(D) None of these

Ans. (A) -1

f'(x) = 3x² – 3 = 0,    x = ±1

f”(x) = 6x at x = -1,  f”(x)<0

Maxima at x = -1

Q7. f(x) = log(sinx) is strictly decreasing in interval :

(A) (0, π/2)

(B) (π/2, π)

(C) (0, π)

(D) None of these

Ans. (B)  (π/2, π)

f'(x) = cosx/sinx = cotx < 0  in (π/2, π).

Q8. Find the value of ∫tan-¹x/1+x² dx .

Ans. (tan-¹x)²/2 + c

Q9. Evaluate ∫sin³x.cos² dx. (Note- use Limit -π/2 to π/2)

Sol.

Q10. Find the degree and order of the differential equation x³(d²y/dx²)³ + (dy/dx)² + x.dy/dx + y = 0.

Ans. Degree = 3, Order = 2

Q11. Solve the differential equation :

(1+x²)dy/dx = 1+y²

Sol. dy/1+y² = dx/1+x²

tan-¹y = tan-¹x + c

Q12. A bag contains 4 white and 6 black balls. Two balls are drawn at random with replacement. Find the probability both the balls are black.

Sol. P(BB) = 6/10 × 6/10 = 36/100 = 9/25

Q13. A and B are independent event such that P(A) = 0.3 and P(B) = 0.4, find the P(A/B ).

Sol. P(A/B) = P(A) = 0.3 (A & B independent)

Q14. A random variable X has the following probability distribution :

Find k.

Sol. ∑P(x) = 1

0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1

10k² + 9k = 1

10k² + 9k – 1 = 0

(10k – 1)(k + 1) = 0

k = 1/10 or k ≠ -1

Q15. Find a unit vector in the direction of the sum of the vectors a = 2i + 2j – 5k and b = j – k.

Sol. a + b = 2i + 2j – 5k + j – k = 2i + 3j – 6k

|a + b| = √(2² + 3² + 6²) = √(4 + 9 + 36) = √49 = 7

unit vector = 2/7i + 3/7j – 6/7k

Q16. Write the equation of line passing through the point with position vector i + 2j + 3k and in the direction 3i + 2j – 2k in vector form.

Sol. a = i + 2j + 3k, b = 3i + 2j – 2k

r = i + 2j + 3k + λ(3i + 2j – 2k)

SET-B

Q1. Let f : R → R+ is defined as f(x) = x⁴ then f is :

(A) One-one, onto

(B) One-one, into

(C) Many-one, onto

(D) Many-one, into

Ans. (C) Many-one, onto

Q2. The principal value of cos-¹x is :

(A) [0, π]

(B) [-π/2, π/2]

(C) (-π/2, π/2)

(D) None of these

Ans. (A) [0, π]

Q3. If 2X + Y = [1     0] & 2X – Y = [3    4] then matrix X:

[-3   2]                   [-1  2]

(A) [4      4]

[-4    4]

(B) [1      1]

[-1     1]

(C) [-1    -2]

[-1      0]

(D) None of these

Ans. (B) [1      1]

[-1     1]

Q4. If det. |3x    5|  =  |2     0| then the value of x is :

|1      x|       |-1   2

(A) +2/3

(B) 2

(C) ±√3

(D) 0

Ans. (C) ±√3

3x(x) – 5(1) = 2(2) – 0(-1)

3x² – 5 = 4 – 0

3x² = 4 + 5

3x² = 9

x² = 9/3 = 3

x = ±√3

Q5. Differentiate log(sec√x) with respect to x.

Sol. d/dx log(sec√x) = (1/sec√x).sec√x.tan√x.(1/2√x) = (tan√x)/(2√x)

Q6. f(x) = sinx + cosx  has a local maxima at x is equal to :

(A) 0

(B) π/6

(C) π/4

(D) π/2

Ans. (C) π/4

f(x) = sinx + cosx

f'(x) = cosx – sinx = 0

sinx = cosx

sinx/cosx = tanx = 1

x = π/4, 5π/4

f”(x) = – sinx – cosx

f”(x) is (-ve) at x = π/4 (maxima)

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