Class 12 Mathematics: CBSE Sample Question Paper- Term I (2021-22)- 2

# Class 12 Mathematics: CBSE Sample Question Paper- Term I (2021-22)- 2 | Mathematics (Maths) Class 12 - JEE PDF Download

 Table of contents Class-XII Time: 90 Minutes Max. Marks: 40 Section - A Section - B Section - C

## Max. Marks: 40

General Instructions :

1. This question paper contains three sections – A, B and C. Each part is compulsory.
2. Section - A has 20 MCQs, attempt any 16 out of 20.
3. Section - B has 20 MCQs, attempt any 16 out of 20.
4. Section - C has 10 MCQs, attempt any 8 out of 10.
5. There is no negative marking.
6. All questions carry equal marks

## Section - A

Q.1: What is the principal value branch of cosec–1 x ?
(a) (-1, 1)
(b) [-1,1]
(c)
(d)

The cosec function is periodic so to calculate its inverse function we need to make the function bijective. For that we have to consider an interval in which all values of the function exist and do not repeat. For cosec function this interval is considered as  Thus when we take the inverse of the function the domain becomes range and the range becomes domain. Hence the principal value branch is the range of cosec–1 x, that is

Q.2: Find the value of k if derivative of the function exists
(a) k = 4/5
(b) k = 4/7
(c) k = 4/9
(d) k =4/11

Differentiate and substitute the given value.

Q.3: The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:
(a) 9
(b) 28
(c) 512
(d) 64

The given matrix of the order 3 × 3 has 9 elements and each of these elements can be either 0 or 1. Now, each of the 9 elements can be filled in two possible ways. Therefore, by the multiplication principle, the required number of possible matrices is 29 = 512.

Q.4: Calculate the determinant of the given matrix
(a) 1/2
(b) -1/2
(c) 3/2
(d) None of the above

Q.5: Which of the following is satisfied by the function, f (x)= 2 sin x−x+1 in the interval

(a) Function has local minimum at x = π/3
(b) Function has local maximum at x = π/3
(c) Function has no critical points in the given interval
(d) Function is increasing in the given interval

f'(x) > 0 for x < π/3 and f"(x) < 0 for x < π/3

Q.6: if  then the value of x is:

(a) –6
(b) –36
(c) 6
(d) 36

Q.7: If the set A contains 3 elements and the set B contains 6 elements, then the number of bijective mappings from A to B is:
(a) 520
(b) 10
(c) 0

(d) None of these

We know that, if A and B are two non-empty finite sets containing m and n elements, respectively, then the number of one-one and onto mapping (bijective mappings) from A to B is
n! if m = n 0,
if m ≠ n
Given that, m = 5 and n = 6 ⇒ m ≠ n
Number of one-one and onto mapping = 0

Q.8: If
(a) Null Matrix
(b) I
(c) A
(d) –A

Q.9: The normal to curve y = 7x2 – x4 at x=2 passes through
(a) (30,4)
(b) (20,4)
(c) (10,5)
(d) (5,20)

Tangent is perpendicular to the normal.

Q.10: What is the domain of the sin–1 x ?
(A) [–∞,∞]
(B) (∞,-∞)
(C) (–1, 1)
(D) [–1, 1]

The sine function is periodic so to calculate its inverse function we need to make the function bijective. For that we have to consider an interval in which all values of the function exist and do not repeat. Now for the inverse of a function the domain becomes range and the range becomes domain. Thus the range of sine function, that is, [-1,1] becomes the domain of inverse function.

Q.11: Let f : R → R be defined as f (x) = 5x. Choose the correct answer.
(a) f is one-one onto
(b) f is many-one onto
(c) f is one-one but not onto
(d) f is neither one-one nor onto

f : R R is defined as f(x) = 5x. Let x, y ∈ R such that
f(x) = f(y) ⇒ 5x = 5y
⇒ x = y
Therefore, f is one-one. Also, for any real number (y) in co-domain R, there exists y/5 in R such that

Therefore, f is onto. Hence, function f is one-one and onto.

Q.12: Find dy/dx where y = cosec x log x.

(a)

(b)

(c)

(d)

y = cosecx logx

Q.13: The value of x for the given determinant  is

(a) ± 3

(b) 3, 1/2

(c) 0

(d) -3, 1/2

In the given determinant, to find the value of x,

Q.14: What is second derivative of the function,
(a)
(b)
(c)
(d)

Q.15: Skew symmetric matrix is also called:
(a) symmetric
(b) identical matrix
(c) square matrix
(d) anti symmetric

In mathematics, particularly in linear algebra, a skew symmetric (or anti symmetric or antisymmetric) matrix is a square matrix whose transpose equals its negative; that is, it satisfies the condition A = – At.

Q.16: What are the local maximum and minimum for the function 3x4 −54x2 −108x+4x3 in the interval [–5 5]
(a) x =−1, 3,−3
(b) x =−1, 3,−2
(c) x =−1, 4,−4
(d) x = 2, 3,−3

Max and min points are the points at which derivative has value 0.

Q.17: Matrix
(a) Skew-symmetric matrix
(b) Symmetric matrix
(c) Scalar matrix
(d) None of these

The given matrix is not a skew symmetric matrix as A’ ≠ -A. By Definition; we know, A matrix is a skew- symmetric matrix if A’ = -A.

Q.18: Find dy/dx where
(a)
(b)
(c)
(d)

Q.19: For the constraints of a LPP problem given by
x1 + 2x2 ≤ 2000, x1 + x2 ≤ 1500, x2 ≤ 600 and x1, x2 ≥ 0, the points does not lie in the positive bounded region.
(a) (1000,0)
(b) (0, 500)
(c) (2, 0)
(d) (2000,0)

From the graph, it is clear that the point (2000, 0) is outside.

Q.20: If y = f (x)= sin(log x), then dy/dx =
(a)
(b)
(c)
(d)

y = f(x) = sin(log x)

## Section - B

Q.21: The function f : R → R defined as f(x) = x3 is:
(a) One-one but not onto
(b) Not one-one but onto
(c) Neither one-one nor onto
(d) One-one and onto

Let f(x1) = f(x2) ∀ x1, x2 ∈ R

⇒ f is one-one Let f(x) = x3 = y ∀ y ∈ R
⇒ x = y1/3 every image y ∈ R has a unique pre image in R
⇒ f is onto
∴ f is one-one and onto.

Q.22: If x = a sec θ, y = b tan θ, then
(a)

(b)
(c)
(d)

x = a secθ

Q.23: In the given graph, the feasible region for a LPP is shaded. The objective function Z = 2x – 3y, will be minimum at:

(a) (4, 10)
(b) (6, 8)

(c) (0, 8)
(d) (6, 5)

Z is minimum –24 at (0, 8)

Q.24: The derivative of  w.r.t  is:
(a) 2
(b) π/2 - 2
(c) π/2
(d) -2

and

⇒ sin v = x   ...(i)

Using (i), we get

⇒ sin–1(2sin v cos v)

⇒ u = 2v
Differentiating with respect to v, we get:
du/dv = 2

Q.25: if , then:

(a) A–1 = B
(b) A–1 = 6B

(c) B–1 = B
(d) B–1 = 1/6A

AB = 6I
B–1 = 1/6A

Q.26: The real function f(x) = 2x3 – 3x2 – 36x + 7 is:
(a) Strictly increasing in (− ∞,−2) and strictly decreasing in ( −2, ∞)

(b) Strictly decreasing in ( −2, 3)

(c) Strictly decreasing in (− ∞, 3) and strictly increasing in (3, ∞)

(d) Strictly decreasing in (− ∞, −2) ∪ (3, ∞)

f'(x) = 6(x2 – x – 6)

= 6(x – 3)(x + 2)

As f'(x) < 0∀ x ∈ (–2, 3)

⇒ f(x) is strictly decreasing in (–2, 3)

Q.27: Simplest form of
(a) π/2 - x/2
(b) 3π/2 - x/2
(c) -x/2
(d) π - x/2

Q.28: Given that A is a non-singular matrix of order 3 such that A2 = 2A, then value of |2A| is:
(a) 6
(b) 8
(c) 64
(d) 16

A2 = 2A

⇒ |A2| = |2A|

⇒ |A|2 = 23|A|
as |kA| = kn|A| for a matrix of order n.

⇒ either |A| = 0 or |A| = 8

But A is non-singular matrix

∴ |2A| = 82 = 64

Q.29: The value of b for which the function f(x) = x + cosx + b is strictly decreasing over R is:
(a) b < 1
(b) No value of b exists
(c) b ≤ 1
(d) b ≥ 1

f'(x) = 1 – sin x

⇒ f'(x) > 0 ∀ x ∈ R

⇒ no value of b exists

Q.30: Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b > 6}, then:
(a) (2, 4) ∈ R
(b) (3, 8) ∈ R
(c) (6, 8) ∈ R
(d) (8, 7) ∈ R

a = b – 2 and b > 6

⇒ (6, 8) ∈ R

Q.31: The point(s), at which the function f given by  is continuous, is/are:

(a) −6, −12, −18
(b) −6, −4, −9

(c) −6, 4, 9
(d) −6, 12, 18

⇒ f(x) = –1∀ x ∈ R

⇒ f(x) is continuous ∀ x ∈ R as it is a constant function

Q.32: If  then the values of k, a and b respectively are:
(a) −6, −12, −18
(b) −6, −4, −9
(c) −6, 4, 9
(d) −6, 12, 18

⇒ k = –6, a = –4
and b = –9

Q.33: A linear programming problem is as follows: Minimize Z = 30x + 50y subject to the constraints,
3x + 5y ≥15
2x + 3y ≤ 18
x ≥ 0, y ≥ 0
In the feasible region, the minimum value of Z occurs at
(a) a unique point
(b) no point
(c) infinitely many points
(d) two points only

Minimum value of Z occurs at two points

Q.34: The area of a trapezium is defined by function f and given by  then the area when it is maximised is:
(a) 75 cm2
(b) 7√3 cm2
(c) 75√3 cm2
(d) 5cm2

But x > 0

⇒ x = 5

Maximum area of trapezium is 75√3 cm2 when x = 5

Q.35: If A is square matrix such that A2 = A, then (I + A)³ – 7 A is equal to:

(a) A

(b) I + A

(c) I − A

(d) I

(I + A)3 – 7A

= I + A + 3A + 3A – 7A = I

Q.36: If tan–1 x = y, then:
(a) −1< y <1
(b)
(c)
(d)

Q.37: Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Based on the given information, f is best defined as:
(a) Surjective function
(b) Injective function
(c) Bijective function
(d) function

As every per-image x ∈A has a unique image y ∈ B
⇒ f is injective function

Q.38: For  then 14A–1 is given by:
(a)
(b)
(c)
(d)

Q.39: The point(s) on the curve y = x3 – 11x + 5 at which the tangent is y = x – 11 is/are:
(a) (–2, 19)
(b) (2, – 9)
(c) (±2, 19)
(d) (–2, 19) and (2, – 9)

y = x3 – 11x + 5

Slope of line y = x – 11 is 1

⇒ 3x2 – 11 = 1

⇒ x = ± 2

∴ point is (2, –9) as (–2, 19) does not satisfy given line

Q.40: Given that  and A2 = 3I, then:

(a) 1 + α2 + βg = 0
(b) 1 – α2 – βg = 0

(c) 3 – α2 – βg = 0
(d) 3 + α2 + βg = 0

A2 = 3I

## Section - C

Q.41: The feasible region for an LPP is shown in the given Figure. Let F = 3x – 4y be the objective function. Maximum value of F is

(a) 0
(b) 8
(c) 12
(d) –18

The feasible region as shown in the figure, has objective function F = 3x – 4y.

Hence, the maximum value of F is 12.

Q.42: If the curve ay + x2 = 7 and x3 = y, cut orthogonally at (1, 1), then the value of a is:
(a) 1
(b) 0
(c) –6
(d) 6

Given that, ay + x2 = 7 and x3 = y

On differentiating both equations with respect to x, we get

Since, the curve cuts orthogonally at (1, 1).

Q.43: If x is real, the minimum value of x2 – 8x + 17 is
(a) -1
(b) 0
(c) 1
(d) 2

Let,

f (x) = x2 - 8x + 17

On differentiating with respect to x, we get

Now, Again on differentiating with respect to x, we get
f"(x) = 2 > 0,∀x

So, x = 4 is the point of local minima. Minimum value of f(x) at x = 4
f(4) = 4.4 − 8.4 + 17 = 1

Q.44: Corner points of the feasible region determined by the system of linear constraints are (0, 3), (1, 1) and (3, 0). Let Z = px + qy, where p, q > 0. Condition on p and q so that the minimum of Z occurs at (3, 0) and (1, 1) is
(a) p = 2q
(b) p = q/2
(c) p = 3q
(d) p = q

So, condition of p and q, so that the minimum of Z occurs at (3, 0) and (1, 1) is

Q.45: if  Then A-1 exist if

(a) λ = 2
(b) λ ≠ 2

(c) λ ≠ –2
(d) None of these

Given that,

Expanding along R1,
|A| = 2(6-5) - λ(-5) -3(-2)
= 2 + 5λ + 6

We know that A-1 exists, if A is nonsingular matrix, i.e., |A|≠0

So, A−1 exists if and only if

Questions 46-50 are based on a Case-Study

Case- Study

Manjit wants to donate a rectangular plot of land for a school in his village. When he was asked to give dimensions of the plot, he told that if its length is decreased by 50 m and breadth is increased by 50 m, then its area will remain same, but if length is decreased by 10 m and breadth is decreased by 20 m, then its area will decrease by 5300 m2. Based on the given information, answer the following questions.

Q.46: The equations in terms of x and y are
(a) x – y = 50, 2x – y = 550
(b) x – y = 50, 2x + y = 550
(c) x + y = 50, 2x + y = 550
(d) x + y = 50, 2x + y = 550

Let length of the plot be x and breadth be y.
According to question,
xy = (x – 50)(y + 50)
xy = xy + 50x – 50y – 2500
x – y = 50

And
(xy – 5300) = (x – 10)(y – 20)
xy – 5300 = xy – 20x – 10y + 200

2x + y = 550 ...(ii)

Q.47: Which of the following matrix equation is represented by the given information
(a)
(b)
(c)
(d)

Q.48: The value of x (length of rectangular field) is
(a) 150 m
(b) 400 m
(c) 200 m
(d) 320 m

We have,

Q.49: The value of y (breadth of rectangular field) is

(a) 150 m
(b) 200 m
(c) 430 m
(d) 350 m

Q.50: How much is the area of rectangular field?
(a) 60000 sq.m.
(b) 30000 sq.m.
(c) 30000 m
(d) 3000 m

Area of rectangular field
= xy
= 200 × 150
= 30000 sq.m

The document Class 12 Mathematics: CBSE Sample Question Paper- Term I (2021-22)- 2 | Mathematics (Maths) Class 12 - JEE is a part of the JEE Course Mathematics (Maths) Class 12.
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## Mathematics (Maths) Class 12

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## Mathematics (Maths) Class 12

204 videos|288 docs|139 tests

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