Class12 CBSE TERM1 MATHS SPECIMEN 2022
BOARD -
CLASS -
SUBJECT -
CBSE
12th
MATHS
Paper Pattern for MCQ Term-I
TIME -
MARKS -
1 Hour 30 Minutes
40
Visit CBSE OFFICIAL PAGE for Regulations and Syllabus of Class 12th CBSE
Solved Specimen Paper Semester-I 2021
SECTION- A
Q.1 sin[π⁄ 3]- sin-1 (-1/2)
- 1/2
- 1/3
- -1
- 1
Solution
Q.2 The value of k (k < 0) for which the function π defined as
is continuous at x = 0 is:- ±1
- -1
- ±1/2
- 1/2
Solution
Q.3 If A = [aij] is a square matrix of order 2 such that , then A2 is:
Solution
Q.4 Value of π, for which A =
k | 8 |
4 | 2k |
- 4
- -4
- ±4
- 0
Solution
Q.5 Find the intervals in which the function f given by f (x) = x2 β 4x + 6 is strictly increasing:
- (-∞,2) U (2, ∞)
- (2, ∞)
- (-∞,2)
- (-∞,2]U (2, ∞)s
Solution
Q.6 Given that A is a square matrix of order 3 and | A | = - 4, then | adj A | is equal to:
- -4
- 4
- -16
- 16
Solution
Q.7 A relation R in set A = {1,2,3} is defined as R = {(1, 1), (1, 2), (2, 2), (3, 3)}. Which of the following ordered pair in R shall be removed to make it an equivalence relation in A?
- (1, 1)
- (1, 2)
- (2, 2)
- (3, 3)
Solution
Q.8 If
2a+b | a-2b |
5c-d | 4c+3d |
4 | -3 |
11 | 24 |
- 8
- 10
- 4
- -8
Solution
Q.9 The point at which the normal to the curve y = π₯ + 1/x, x > 0 is perpendicular to the line 3x β 4y β 7 = 0 is:
- (2, 5/2)
- (Β±2, 5/2)
- (- 1/2, 5/2
- (1/2, 5/2)
Solution
Q.10 sin (tan-1x), where |x| < 1, is equal to:
- x ⁄ √ 1-x2
- 1 ⁄ √ 1-x2
- 1 ⁄ √ 1+x2
- x ⁄ √ 1+x2
Solution
Q.11 Let the relation R in the set A = {x β Z : 0 β€ x β€ 12}, given by R = {(a, b) : |a βb| is a multiple of 4}. Then [1], the equivalence class containing 1, is:
- {1, 5, 9}
- {0, 1, 2, 5}
- ∅
- A
Solution
Q.12 If ex + ey = ex+y , then ππ¦/ππ₯ is:
- e y - x
- ex+y
- - e y - x
- 2e x - y
Solution
Q.13 Given that matrices A and B are of order 3Γn and mΓ5 respectively, then the order of matrix C = 5A +3B is:
- 3Γ5
- βΉ5Γ3
- 3Γ3
- 5Γ5
Solution
Q.14 If y = 5 cos x β 3 sin x, then π2π¦ ⁄ ππ₯2is equal to:
- -y
- y
- 25y
- 9y
Solution
Q.15 For matrix A =
2 | 5 |
-11 | 7 |
-
2 -5 11 -7 -
7 5 11 2 -
7 11 -5 2 -
7 -5 11 2
Solution
Q.16 The points on the curve π₯2⁄9 + y2⁄16 = 1 at which the tangents are parallel to y axis are:
- (0,Β±4 )
- (Β±4,0)
- (Β±3,0)
- (0, Β±3)
Solution
Q.17 Given that A = [πππ] is a square matrix of order 3Γ3 and |A| = β7, then the value of β3i=1 ππ2π΄π2, where π΄ππ denotes the cofactor of element πππ is:
- 7
- -7
- 0
- 49
Solution
Q.18 If y = log(cos ππ₯), then ππ¦/ππ₯is:
- cosππ₯β1
- e-xcos e x
- ex sin ex
- -ex tan ex
Solution
Q.19 Based on the given shaded region as the feasible region in the graph, at which point(s) is the objective function Z = 3x + 9y maximum?
- Point B
- Point C
- Point D
- every point on the line segment CD
Solution
Q.20 The least value of the function π(π₯) = 2πππ π₯ + π₯ in the closed interval [0,π/2] is:
- 2
- π⁄6 + β3
- π/2
- The least value does not exist
Solution
Q.21 The function π: RβΆR defined as π(π₯) = π₯3 is:
- One-on but not onto
- Not one-one but onto
- Neither one-one nor onto
- One-one and onto
Solution
Q.22 If x = a sec π, y = b tan π, then π2π¦ ⁄ ππ₯2 at π =π/6 is:
- β3β3π⁄ π2
- β2β3π⁄ π
- β3β3π⁄ π
- -b⁄ 3β3π2
Solution
Q.23 In the given graph, the feasible region for a LPP is shaded. The objective function Z = 2x β 3y, will be minimum at:
- (4, 10)
- (6, 8)
- (0, 8)
- (6, 5)
Solution
Q.24 The derivative of sin-1 (2π₯√ 1 β π₯2) w.r.t sin-1x, β1/β2 < π₯ < 1/β2, is:
- 2
- (π/2)-2
- π/2
- β2
Solution
SECTION- B
Q.25 If A =
1 | -1 | 0 |
2 | 3 | 4 |
0 | 1 | 2 |
2 | 2 | -4 |
-4 | 2 | -4 |
2 | -1 | 5 |
- A-1 = B
- A-1 = 6B
- B-1 = B
- A-1 = 1/6 A
Solution
Q.26 The real function f(x) = 2x3 β 3x2 β 36x + 7 is:
- Strictly increasing in (ββ, β2) and strictly decreasing in ( β2, β)
- Strictly decreasing in ( β2, 3)
- Strictly decreasing in (ββ, 3) and strictly increasing in (3, β)
- Strictly decreasing in (ββ, β2) βͺ (3, β)
Solution
Q.27 Simplest form of tan-1 , π < π₯ < 3π⁄2 is:
- βΉ20 Per share
- βΉ18 Per share
- βΉ22 Per share
- βΉ8 Per share
Solution
Q.28 Given that A is a non-singular matrix of order 3 such that A2 = 2A, then value of |2A| is:
- 4
- 8
- 64
- 16
Solution
Q.29 The value of π for which the function π(π₯) = π₯ + πππ π₯ + π is strictly decreasing over R is:
- π < 1
- No value of b exists
- π β€ 1
- π β₯ 1
Solution
Q30. Let R be the relation in the set N given by R = {(a, b) : a = b β 2, b > 6}, then
- (2,4) β R
- (3,8) β R
- (6,8) β R
- (8,7) β R
Solution
Q.31 The point(s), at which the function f given by is continuous, is/are:
- π₯ π R
- π₯ = 0
- π₯ π R β{0}
- π₯ = β1 and 1
Solution
Q.32 If A =
0 | 2 |
3 | -4 |
0 | 3a |
2b | 24 |
- β6, β12, β18
- β6, β4, β9
- 6, 4, 9
- β6, 12, 18
Solution
Q.33 A linear programming problem is as follows: πππππππ§π π = 30π₯ + 50π¦ subject to the constraints, 3π₯ + 5π¦ β₯ 15 2π₯ + 3π¦ β€ 18 π₯ β₯ 0, π¦ β₯ 0 In the feasible region, the minimum value of Z occurs at
- a unique point
- no point
- infinitely many points
- two points only
Solution
Q.34 The area of a trapezium is defined by function π and given by π(π₯) = (10 + π₯)√ 100 β π₯2 , then the area when it is maximised is:
- βΉ75ππ2
- 7β3ππ2
- 75β3ππ2
- 5ππ2
Solution
Q.35 If A is square matrix such that A2 = A, then (I + A)Β³ β 7 A is equal to:
- A
- I + A
- I β A
- I
Solution
Q.36 If tan-1 x = y, then:
- β1 < y < 1
- βπ/2 β€ y β€ π/2
- βπ/2 < y < π/2
- y π{βπ/2,π/2}
Solution
Q.37 Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let π = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Based on the given information, π is best defined as:
- Surjective function
- Injective function
- Bijective function
- function
Solution
Q.38 For A =
3 | 1 |
-1 | 2 |
- 14
2 -1 1 3 - 5
4 -2 2 6 - 2
2 -1 1 β3 - 2
-3 -1 1 β2
Solution
Q.39 The point(s) on the curve y = x3 β 11x + 5 at which the tangent is y = x β 11 is/are:
- (-2,19)
- (2, - 9)
- (Β±2, 19)
- (-2, 19) and (2, -9)
Solution
Q.40 Given that A =
πΌ | π½ |
πΎ | βπΌ |
- 1 + πΌ2 + π½πΎ = 0
- 1 - πΌ2 - π½πΎ = 0
- 3 - πΌ2 - π½πΎ = 0
- 3 + πΌ2 + π½πΎ = 0
Solution
Q.41 For an objective function π = ππ₯ + ππ¦, where π, π > 0; the corner points of the feasible region determined by a set of constraints (linear inequalities) are (0, 20), (10, 10), (30, 30) and (0, 40). The condition on a and b such that the maximum Z occurs at both the points (30, 30) and (0, 40) is:
- π β 3π = 0
- π = 3b
- π + 2π = 0
- 2π β π = 0
Solution
Q.42 For which value of m is the line y = mx + 1 a tangent to the curve y2 = 4x?
- 1/2
- 1
- 2
- 3
Solution
Q.43 The maximum value of [π₯(π₯ β 1) + 1]1/3, 0 β€ π₯ β€ 1 is:
- 0
- 1/2
- 1
Solution
Q.44 In a linear programming problem, the constraints on the decision variables x and y are π₯ β 3π¦ β₯ 0, π¦ β₯ 0, 0 β€ π₯ β€ 3. The feasible region
- is not in the first quadrant
- is bounded in the first quadrant
- is unbounded in the first quadrant
- does not exist
Solution
Q.45 Let A =
1 | sinΞ± | 1 |
-sinΞ± | 1 | sinΞ± |
1 | sinΞ± | 1 |
- |A|=0
- |A| π (2, β)
- |A| π (2,4)
- |A| π [2,4]
Solution
CASE STUDY The fuel cost per hour for running a train is proportional to the square of the speed it generates in km per hour. If the fuel costs βΉ 48 per hour at speed 16 km per hour and the fixed charges to run the train amount to βΉ 1200 per hour. Assume the speed of the train as π£ km/h.Based on the given information, answer the following questions.
Q.46 Given that the fuel cost per hour is π times the square of the speed the train generates in km/h, the value of π is:
- 16/3
- 1/3
- 3
- 3/16
Solution
Q.47 If the train has travelled a distance of 500km, then the total cost of running the train is given by function:
Solution
Q.48 The most economical speed to run the train is:
- 18km/h
- 5km/h
- 80km/h
- 40km/h
Solution
Q.49 The fuel cost for the train to travel 500km at the most economical speed is:
- βΉ 3750
- βΉ 750
- βΉ 7500
- βΉ 75000
Solution
Q.50The total cost of the train to travel 500km at the most economical speed is:
- βΉ 3750
- βΉ 75000
- βΉ 7500
- βΉ 15000