Class 12 Mathematics: CBSE Sample Question Paper- Term I (2021-22)- 3

# Class 12 Mathematics: CBSE Sample Question Paper- Term I (2021-22)- 3 | Mathematics (Maths) Class 12 - JEE PDF Download

## 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 domain of the cos–1 x?
(a) [–∞, ∞]
(b) (–∞, ∞)
(c) (–1, 1)
(d) [–1, 1]

The cosine 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 cosine function, that is, [-1,1] becomes the domain of inverse function.

Q.2: Find dy/dx  where x = a cos2 θ and y = b sin2 θ.
(a) - b/α
(b) − b/α secθ
(c) − b/α tan θ
(d) − b/α cot θ

x = αcos2 θ, y = bsin2θ

Q.3: Evaluate the determinant
(a) 0
(b) 27
(c) –27
(d) 5

If any one of the row or column of a square matrix has all elements as 0 then the determinant of the matrix is 0.

Q.4: If matrix A = [12 3]. Write A. A' where A' is the transpose of matrix A.
(a) 7
(b) 21
(c) 14
(d) 28

A.A' = 1 + 4+ 9
A.A' = 14

Q.5: At what angle, the tangent to curve y = 3x2 − x3 at x = 2 makes with positive x-axis?
(a) 45°
(b) 0°
(c) 30°
(d) 60°

y = 3x2 – x3
Slope = dy/dx = 6x – 3x2

Slope = tanθ
0 = tanθ
tan 0° = tanθ
θ = 0°

Q.6: For the given matrix , the values of y is

(a) x = 3
(b) x = 14
(c) y = 7
(d) y = –2

Since corresponding entries of equal matrices are equal,
So x = 3 --------(1)
3x – y = 2 --------(2)
2x +z = 4 --------(3)
3y – w = 7 --------(4)
Put the value of x = 3 from equation 1 in equation 2
3x – y = 2
3(3) – y = 2
9 – y = 2
y = 9 – 2
y = 7

Q.7: Set A has 3 elements and set B has 4 elements. Then the number of injective mappings that can be defined from A to B is
(a) 144
(b) 12
(c) 24
(d) 64

We know that, if A and B are two non-empty finite sets containing m and n elements, respectively, then
Number of one-one functions = nPm (if n≥m)
Number of one-one functions = 0
(if n < m)
So, by using the formula,

Number of one-one/ injective functions

Q.8: Find the minimum and maximum values of the given determinant

(a) 2 and 4
(b) –1 and 1
(c) 0 and 1
(d) –1 and 2

The value of determinant comes out to be ∆ = 2 (1 + sin2 θ). As the minimum and maximum value of sin θ is known, it is manipulated according to the required equation.
-1 ≤ sin θ ≤ 1
0 ≤ sin2 θ ≤ 1

0 + 1 ≤ + sin2 θ ≤ 1 + 1
2 ≤ 2(1 + sin2 θ) ≤ 4

Q.9: Which of the function is continuous in R?
(a) log x
(b) sin x
(c) 3x2
(d) All of these

Every polynomial function is continuous on R and every rational function is continuous on its domain. For Example: The constant function f(x) = 1 and the identity function g(x) = x are continuous on R.

Q.10: If a relation R on the set {1, 2, 3} be defined by R = {(1, 1)}, then R is
(a) reflexive
(b) transitive
(c) symmetric
(d) None of these

R on the set {1, 2, 3} is defined by R = {(1, 1)} It is clear that R is reflexive.

Q.11: What is the principal value branch of tan–1 x ?
(a)
(b) [–1, 1]
(c) (–∞, ∞)
(d)

The tangent 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 tangent 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 tan-1 x , that is

Q.12: What is the derivative of ?
(a) xx (1 + log x)
(b) log xe
(c) ex log x
(d) xe (1 + log x)

Let y =  = xx

Q.13: Consider a square matrix The given matrix is an example of
(a) Diagonal matrix
(b) Identity matrix
(c) Scalar matrix
(d) Null matrix

A square matrix B = [bij] m × m is said to be a diagonal matrix if all its non diagonal elements are zero, that is a matrix B = [bij] m × m is said to be a diagonal matrix if bij = 0, when i ≠ j.

Q.14: If y  = x6 + logx2 then the value of dy/dx is
(a) 6x5 – 2x
(b) 6x5 + 2/x
(c) 6x5 - 2/x
(d) 6x6 + x
2

Q.15: If then |A αdj A| will be
(a) 95
(b) 0
(c) 950
(d) 957

∴ |A(αdj A)| = –33 × – 29 – 0
= 957

Q.16: Two numbers x and y related to each other as x – y = 3. What is the value of the numbers such that their product is minimum?
(a) x = 2.5, y = −0.5
(b) x = 1.5, y = −1.5
(c) x = 5.5, y = 2.5
(d) x = 1, y = −2

On applying Second Derivative test, At x = 1.5 f '' (x) > 0

Q.17: Consider the curve y =  x2/4. The Slope of the line parallel to tangent to the curve at x = 1 is
(a) 1/4
(b) 1/3
(c) - 1/2

(d) 1/2

y = x2/4

Parallel lines have same slopes
∴ Slope of tangent = 1/2

Q.18: The points on the curve at which the tangents are parallel to y-axis.
(a) (± 3, 0)
(b) (0, ± 3)
(c) (± 2, 0)
(d) (0, ± 2)

∵ Tangent line is parallel to the y-axis.
⇒ Slope of the normal is 0

Hence, Points at which tangent are Parallel to the y-axis are (±3, 0).

Q.19: Objective functions
Maximise  Z = 22x + 18y
Constraints : x + y ≤ 20
360x + 240y ≤ 5,760
Or 3x + 2y ≤ 48
x ≥ 0,   y ≥ 0

The maximum Z =
(a) 5,760
(b) 392
(c) 362
(d) 290

Vertices of feasible region are : A(0, 20), B(8, 12), C(16, 0) & O(0, 0)

Z(A) = 360, Z(B) = 392, Z(C) = 352

Max Z = 392

Q.20: Which of the following is true for the function f(x) = In x + 2 increasing.
(a) function is strictly increasing in (1,2).
(b) function is strictly decreasing in (1,2).
(c) function is constant in (1,2).
(d) None of the above

First derivative is always positive in the given interval.

### Section - B

Q.21: Let R be the relation in the set of integers Z given by R = {(a, b) : 3 divides a – b}. then R is:
(a) not reflexive
(b) reflexive
(c) symmetric
(d) transitive

By definition, a relation in Z is said to be reflexive if xRx, ∀ x ∈ Z.
So, α – α = 0 ⇒ 3 divides α – α ⇒ αRa.
Hence R is reflexive.

Q.22: If f(x) = |x|, then the function is continuous but not differentiable at:
(a) x = 0
(b) x ¹ 0
(c) x = 1
(d) x = 2

Given function is continuous but not differentiable at x = 0.

Q.23. A dietician wishes to mix together two kinds of food X and y in such a way that the mixture contains at least 15 units of carbohydrate, at least 17 units of protein and at most 6 units of fat. The nutrient contents of 1 kg food is given below:

1 kg of food X costs `16 and 1 kg of food Y cost `20. Find the least cost of the mixture which will produce the required diet.
(a) 100
(b) 128
(c) 110
(d) 99

Let Food X be p and Food Y be q.
Formulate the constraints as per statement.
The Constraints are : p + 2q ≥ 15; 3p + 2q ≥ 17; p + < 6; p ≥ 0; q ≥ 0
And objective function is Z = 16p + 20q
By solving the above equations, Corner points will be (1, 7), (1, 5), (3, 4), (5, 1) and (0, 6) After putting these points in Z, we get the least cost of the mixture, i.e., 100.

Q.24. What is the domain of the cosec–1x ?
(a) [– ∞, ∞]
(b) (– ∞, ∞)
(c) (– 1, 1)
(d) (– ∞, – 1] ∪ [1, ∞)

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. Now for the inverse of a function the domain becomes range and the range becomes domain. Thus the range of cosec function, that is, (– ∞, – 1] ∪ [1, ∞) becomes the domain of inverse function.

Q.25. For a given matrix both the minor and cofactor of the element α31 is.

(a) 40

(b) 42

(c) 45

(d) 35

Q.26: What is the derivative of ?
(a) xx (1 + log x)
(b) log xe
(c) ex log x
(d) xe (1 + log x)

Let y =  = xx

Q.27: Given a set A = {1, 2, 3} and R1 = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}; R2 = {(2, 2), (1, 3), (3, 1)}; and R3 = {(1, 3), (3, 3)}. Which of the following relations are equivalence relation?
(a) All of them
(b) R1, R3
(c) None of them
(d) R
1

R1 is reflexive but neither symmetric nor transitive. R2 is symmetric but neither reflexive nor transitive. R2 is transitive but neither reflexive nor symmetric.

Q.28: If A = then A2 is:
(a)
(b)
(c)
(d)

Q.29: If  y = e9x, then:
(a) y2 = 9e9x
(b) y2 = 81e
(c) y2 = 81e9x
(d) y= 81e
x

We have: e9x
This expression can be differentiated using the "chain rule".
Let u = 9x ⇒ u ' = 9 and v = eu ⇒ v ' = eu:

We can now replace u with 9x:

Q.30: The principal value of is:
(a) π/3
(b) - π/3
(c) π/2
(d) -π/2

The principal value of means that we need to find an angle in the principal branch of the function where the cosec function is equal to -2/√3. Hence the required value is - π/3.

Q.31: If y = Ae8x + Be–8x, then d2y/dx2 is:
(a) 64y
(b) 8y
(c) –64y
(d) 40y

Q.32: If then the value of x is:
(a) 95/3
(b) 97/3
(c)
(d)

Q.33: In which of the following, the constants of a linear function has to be maximized or minimized according to a set of given conditions
(a) Constraints
(b) Objective functions
(c) Decision variables
(d) Feasible solutions

The objective function is a function in the LPP which is to be optimized. The LPP objective function either has maximum value or minimum value or has no solutions.

Q.34: If the curve y = –x3 + 4x2 + 8x – 32, then d2y/dxis

(a) –2(3x + 4)
(b) 2(3x + 4)
(c) –2(3x – 4)
(d) 2(3x – 4)

Q.35: Let then |AB| is :

(a) 4600

(b) 5600

(c) 82000
(d) –82000

Q.36: The principal value of cot−1 (√3) is:
(a) π/6
(b) - π/6
(c) π/3
(d) - π/3

The principal value of cot−1(√3) means that we need to find an angle in the principal branch of the function where the cotangent function is equal to √3. Hence the required value is π/6.

Q.37: Let W be the set of words in the English dictionary. A relation R is defined on W as R = {(x, y) ∈ W × W such that x and y have at least one letter in common}. Then R is:
(a) Symmetric but not reflexive
(b) Both Symmetric and reflexive
(c) Reflexive but not symmetric
(d) Symmetric and transitive

For any word x ∈ W, x and x have at least one (all) letter in common Therefore, (x, x) ∈ R,
⇒ x ∈ W so R is reflexive.

Symmetric : Let (x, y) ∈ R, xy ∈ w
⇒ x and y have at least one letter in common
⇒ y and x have at least one letter in common
⇒ (y, x) ∈ R So, R is symmetric.

Q.38: Let two matrices be A = [2] and B = then A + B:
(a) 32
(b) 40
(c) 42
(d) A + B = not possible

Matrix A is of Order |X| and B is of order 2×2. Hence, as per property of matrices, Addition can’t be operated on the given two matrices of different order.

Q.39: d/dx(sin 3x) at x = π/3 is:

(a) 0

(b) 1

(c) 2

(d) –3

Q.40: If  = 0, then x is:
(a) 0
(b) 6
(c) –6
(d) –2

Comparing the elements of both sides,
x – 6 = 0
x = 6.

### Section - C

Q.41: Z = 20x1 + 20x2, subject to x1 ≥ 0, x2 ≥ 2, x1 + 2x2 ≥ 8, 3x1 + 2x2 ≥ 15, 5x1 + 2x2 ≥ 20.

The minimum value of Z occurs at :
(a) (8, 0)
(b)
(c) (0, 10)
(d)

Q.42: Maximum slope of the curve y = –x3 + 3x2 + 9x – 27  is:
(a) 0
(b) 12
(c) 16
(d) 32

Given that,

So, the maximum slope of given curve is at x = 1.

Q.43: The curve y = x1/5 has at (0, 0)
(a) a vertical tangent (parallel to y-axis)
(b) a horizontal tangent (parallel to x-axis)
(c) an oblique tangent
(d) no tangent

Given that, y =x1/5 On differentiating with respect to x, we get

So, the curve y = x1/5 has a vertical tangent at (0, 0), which is parallel to y-axis.

Q.44: The corner points of the feasible region determined by the system of linear constraints are (0, 0), (0, 40), (20, 40), (60, 20), (60, 0). The objective function is Z = 4x + 3y. Compare the quantity in Column A and Column B

(a) The quantity in column A is greater.
(b) The quantity in column B is greater.
(c) The two quantities are equal.
(d) The relationship cannot be determined on the basis of the information supplied.

Q.45: Is the value of the determinant dependent on the value of q.
(a) Yes
(b) No
(c) None of the above
(d) All of the above

The determinant comes out to be –x3 after calculation which is independent of θ.

CASE-STUDY

Neha wants to prepare a handmade gift box for her sister's birthday at home. For making lower part of box; the takes a square piece of card board of side 10 cm.

Based on the given information, answer the following questions.

Q.46: If x  cm be the length of the square card board which is to be cut off from corners of the square piece of side 10 cm, the possible value of x will be given by the interval.
(a) [0, 10]
(b) (0, 20)
(c) (0, 5)
(d) None of these

Since, side of square is of length 10 cm therefore x ∈ (0, 5)

Q.47: Volume of the open box formed by folding up the cutting corner can be expressed as
(a) V = x(10 – 2x)2
(b) V = (x – 10)(10 – 2x)2
(c) V = x(10 – 2x)(10 + 2x)
(d) V = (x – 5)(10 – 2x)
2

Height of open box = x cm
Length of open box = 10 – 2x
and width of open box = 10 – 2x
∴ Volume of the open box
= x(10 – 2x)(10 – 2x)
= x(10 – 2x)2

Q.48: The values of x for which dV/dx = 0 are,

(a) 5/3, - 2

(b) 5, 5/3
(c) 5-, 3/2
(d) 5, 3

V = x(10 – 2x)2

+ (10 – 2x)2
= (10 – 2x)[–4x + 10 – 2x]
= (10 – 2x)(10 – 6x)
Now dV/dx = 0

⇒ 10 – 2x = 0 or 10 – 6x = 0
⇒ x = 5 or 5/3.

Q.49: Neha is intrested in maximising the volume of the box. So, what should be the side of the square to be cut off so that the volume of the box is maximum.
(a) 4 cm
(b) 3 cm
(c) 5 cm
(d) 5/3 cm

∴ Volume of the box will be maximum
When x = 5/3 cm.

Q.50: The maximum volume of the box is
(a) 1000/9 cm3
(b) 2000/9 cm3
(c) 2000/27 cm3
(d) 1000/27 cm
3

V = x(10 – 2x)2

∴ Maximum volume

The document Class 12 Mathematics: CBSE Sample Question Paper- Term I (2021-22)- 3 | 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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