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


50 Questions MCQ Test Sample Papers for Class 12 Medical and Non-Medical | Test: Class 12 Mathematics: CBSE Sample Question Paper- Term I (2021-22)


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QUESTION: 1

What is the principal value branch of cot–1x ?

Solution:

The cotangent 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 cotangent function this interval is considered as (0, π).

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 cot–1 x that is (0, π).

QUESTION: 2

What is the continuity of [x] at x = 3.5 ?

Solution:

L.H. L=R.H. L = f(x)

QUESTION: 3

Choose the correct option for the given matrix.

Solution:

A square matrix is called lower triangular if all the entries above the main diagonal are zero. Similarly, a square matrix is called upper triangular if all the entries below the main diagonal is zero.
Hence, given matrix is lower triangular matrix.

QUESTION: 4

If a matrix has 4 rows and 3 columns then how many elements will be there in this matrix?

Solution:

Matrix is represented by m × n
Where m = no. of rows & n = no. of column
And number of total elements = mn
If a matrix has 4 rows and 3 columns hen the number of elements in the matrix is 12.

QUESTION: 5

What is the equation of the normal to the curve y = 3x2 – 7x + 5 at (0, 5)?

Solution:

Tangent and normal are perpendicular to each other


∴ Slope of normal = 1/7
Equation of normal is

QUESTION: 6

If A is a square matrix of order 3, such that A(adj A) = 10I, then |adj A| is equal to

Solution:

Consider the equation
A(adj A) = |A| I
Here, A(adj A) = 10 I
Then, |A| = 10
Since, |adj A| = |A|n–1
Where n is order of matrix
Here, = |A|3–1 = 102 = 100

QUESTION: 7

R be the relation in the set N given by R = {(a, b) : a = b –2, b > 6}. Then, the correct option is:

Solution:

Take b =8, ⇒ a = 6.
Hence, (6, 8) ∈ R

QUESTION: 8

Evaluate the determinant of the matrix 

Solution:

The determinant of a square matrix of order 2 is given by the difference of the product of diagonal elements and the product of the offdiagonal elements.

QUESTION: 9

Which of the function is not differentiable everywhere in R?

Solution:

|x| is not differentiable at x=0

QUESTION: 10

All the trigonometric functions have inverse functions irrespective of the domain.

Solution:

For the inverse of a function to exist the function should be bijective which none of the trigonometric function is as they are periodic functions.

QUESTION: 11

Let R be the relation in the set {5, 6, 7, 8} given by R = {(5, 6), (6, 6), (5, 5), (8, 8), (5, 7), (7, 7), (7, 6)}.
Choose the correct answer:

Solution:

Let R be the relation in the set {1, 2,3, 4} is given by:
R = {(5,6), (6,6), (5,5), (8,8), (5,7), (7,7), (7,6)}
(a) (5,5), (6,6), (7,7), (8,8) ∈ R Therefore, R is reflexive.
(b) (5,6) ∈ R but (6,5) ∈ R. Therefore, R is not symmetric.
(c) If (5, 7) ∈ R and (7, 6) ∈ R then (5, 6) ∈ R. Therefore, R is transitive.

QUESTION: 12

What is the equation of the tangent to the curve y – 5x2 = 0 and parallel to line 5x + y + 1 = 0 ?

Solution:

Given equation of the curve is y – 5x2 = 0


Slope of the line 5x + y + 1 = 0 is – 5 slope of the tangent = slope of the parallel line
i.e., 10x = – 5


QUESTION: 13

if  then A2−5A + 7I2 is equal to :

Solution:

Since, all the elements of matrix are zero. So, given matrix is null/zero matrix.

QUESTION: 14

If y = log(cosex) then dy/dx will be:

Solution:


QUESTION: 15

For a square matrix A = [aij] the quantity calculated for any element aij in A as the determinant of thesquare sub-matrix of order (n – 1) obtained by leaving the ith row and jth column of A is known as:

Solution:

The minor of an element aij in A is calculated as the determinant of the square sub-matrix of order (n-1) obtained by leaving the ith row and jth column of A.

QUESTION: 16

Which of the following is satisfied by the given function y = sin10x−cos10x ?

Solution:


QUESTION: 17

If matrix A = [2 3 5], then the value of A.A' is:

Solution:

A = [2 3 5]

QUESTION: 18

The function y = 5x2 – 32x has a local minimum in the interval (0,10).

Solution:

let f(x) = 5x2 – 32x
f '(x) = 10x – 32
10x – 32 = 0
f"(x) for x = 3.2 and f'(x)<0 for x >3.2 .

QUESTION: 19

Z = 30x1 + 30x2, subject to x1 ≥ 0, x2 ≥ 0, x1 + 3x2 ≤ 6, 4x1 + 8x2 ≥ 16, x1 + x2 ≤ 4. The minimum value of Z occurs at

Solution:

Corner Points are (0, 2), (2, 1), (4, 0) and (3, 1).
The objective function for (x, y) is 30x1 + 30x2.
After putting the corner points in the objective function, we get the minimum value of Z.

QUESTION: 20

A rectangle of maximum area is inscribed in a circle of radius R.Then which of the following is true?

Solution:

On applying Second Derivative test,

QUESTION: 21

Let T be the set of all triangles in a plane with R is a relation in T given by R = {(T1, T2) : T1 is congruent to T2}. Then R is

Solution:

For reflexive, T1 is congruent to T1
⇒ (T1,T1) Î R For symmetric, (T1,T2) ∈ R ⇒ T1 is congruent to T2 ⇒ T2 is congruent to T1 ⇒ (T2,T1) ∈ R. Hence it is symmetric.
For transitive, (T1, T2) ∈ R ⇒ T1 is congruent to T2 and (T2,T3) ∈ R ⇒ T2 is congruent to T3 which implies T1 is congruent to T3 ⇒ (T1,T3) ∈ R. Hence, it is transitive.
Hence, R is an equivalence relation.

QUESTION: 22

Consider the function 3x4 +20x3 −36x2 +44 in the interval [−5,10] , function is maximum at

Solution:

Let f(x) = 3x4 + 20x3 – 36x2 + 44
f '(x) = 12x3 + 60x2 – 72x
f'(x) = 0
12x3 + 60x2 – 72x = 0
12x (x2 + 5x – 6) = 0
12x (x – 3) (x – 2) = 0
x = 0, 2, 3
f''(x) = 36x2 + 120x – 72
f''(0) = –72 < 0
f''(2) = 312 > 0
f''(3) = 612 > 0
∴ function has a maxima at x = 0

QUESTION: 23

The graph of x ≤ 2 and y ≥ 2 will be situated in quadrants.

Solution:

 

QUESTION: 24

What is the value of l in the function  provided function is continuous at x = 1?

Solution:

For a function to be continuous L.H. L=R.H. L= f (x)

QUESTION: 25

If A is a 3 × 3 matrix such that |A| = 8, then |3A| equals

Solution:

Here |A| = 8
Then |3A| = 33|A| = 27 × 8 = 216

QUESTION: 26

The function f(x) = tan x – x

Solution:

We have
f(x) = tan x – x
On differentiating with respect to x, we
get

So, f(x) always increases.

QUESTION: 27

What is the domain of the cot–1 x ?

Solution:

The cot 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 cot function, that is, (-∞, ∞) becomes the domain of inverse function.

QUESTION: 28

If A is skew symmetric matrix of order 3, then the value of |A| is

Solution:

Determinant value of skew symmetric matrix is always '0'.

QUESTION: 29

The function f(x) = 4sin3x – 6 sin2x + 12 sinx + 100 is strictly

Solution:

Given that, f(x) = 4sin3x – 6 sin2x + 12 sinx 100
On differentiating with respect to x, we get




QUESTION: 30

If A = {1, 2}, B = {3,4, 5} and f = {(1, 3), (2, 5)} is a function from A to B, then f(x) is

Solution:

Given, A = {1, 2}, B = {3,4, 5} and f : A → B is defined as f = {(1, 3), (2, 5)} i.e., f(1) = 3, f(2) = 5.
We can see that the images of distinct elements of A under f are distinct. So, f is one-one.

QUESTION: 31

The function f(x) = e|x| is

Solution:

Given that,
f (x) = e|x|

The functions ex and |x| are continuous functions for all real value of x. Since ex is differentiable everywhere but |x| is non-differentiable at x = 0.
Thus, the given functions f(x) = e|x| is continuous everywhere but not differentiable at x = 0.

QUESTION: 32

The matrix 

Solution:


So, the given matrix is a symmetric matrix. [Since, in a square matrix A, if A’ = A, then A is called symmetric matrix.]

QUESTION: 33

Z = 7x + y, subject to 5x + y £ 5, x + y ³ 3, x ³ 0, y ≥ 0. Maximum value of Z occurs at

Solution:

QUESTION: 34

Which of the following functions is decreasing on 

Solution:

In the given interval 
f(x) = cos x
On differentiating with respect to x, we get f´(x) = – sin x
which gives

Hence, f(x) = cosx is decreasing in 

QUESTION: 35

If A and B are symmetric matrices of same order, then AB – BA is a:

Solution:

A and B are symmetric matrices.
⇒ A = A’ and B = B’
Now, (AB – BA)’ = (AB)’ – (BA)’ ...(i)
⇒ (AB – BA)’=B’A’ – A’B’ [By reversal law]
⇒ (AB – BA)’ = BA – AB [From Eq. (i)]
⇒ (AB – BA)’ = –(AB – BA)
⇒ (AB – BA) is a skew-symmetric matrix.

QUESTION: 36

The principal value of 

Solution:

The principal value of  means that we need to find an angle in the principal branch of the function where the cosine function is equal to √3/2 . Hence the required value
is π/6

QUESTION: 37

The domain of sin–1 x is :

Solution:

The domain of sin-1 x is [-1, 1].

QUESTION: 38

if  then the value of x is

Solution:


On expanding along R1
2(x – 9x) – 3(x – 4x)+ 2(9x – 4x) + 3 = 0
2(–8x) – 3(–3x) + 2(5x) + 3 = 0
– 16x + 9x + 10x + 3 = 0
3x + 3 = 0
3x = –3
x = - 3/3
x = –1

QUESTION: 39

The contentment obtained after eating x units of a new dish at a trial function is given by the functionf(x) = x3 + 6x2 + 5x + 3. The marginal contentment when 3 units of dish are consumed is ____________.

Solution:


At x = 3,
Marginal contentment
= 3x (3)2 + 12 x 3 + 5
= 27+ 36 +5
68 units.

QUESTION: 40

The matrix 

Solution:

We know that, in a square matrix, if bij = 0 when i ≠ j then it is said to be a diagonal matrix. Here, b12, b13…. ≠ 0
so the given matrix is not a diagonal matrix.
Now,

So, the given matrix is a skew-symmetric matrix, since we know that in a square matrix B, if B’ = − B, then it is called skew-symmetric matrix.

QUESTION: 41

A dealer in rural area wishes to purchase a number of sewing machines. He has only ₹5,760 to invest and
has space for at most 20 items for storage. An electronic sewing machine cost him ₹360 and a manually
operated sewing machine ₹240. He can sell an electronic sewing machine at a profit of ₹22 and a manually
operated machine at a profit of ₹18. Assume that the electronic sewing machines he can sell is x and that
of manually operated machines is y.

The objective function is _______.

Solution:
QUESTION: 42

The maximum value of sin x. cos x is

Solution:

Let us assume that,
f(x) = sin x cos x
Now, we know that



QUESTION: 43

The maximum value of (1/x)is:

Solution:



Hence, the maximum value of

QUESTION: 44

From the figure, which point is not belongs to feasible region

Solution:
QUESTION: 45

Which of the given values of x and y make the following pair of matrices equal

Solution:

It is given that

Equating the corresponding elements, we get
3x + 7 = 0


We find that on comparing the corresponding elements of the two matrices, we get two different values of x, which is not possible.
Hence, it is not possible to find the values of x and y for which the given matrices are equal.

QUESTION: 46

Case - Study

The Relation between the height of the plant (y in cm) with respect to exposure to sunlight is governed
by the following equation y = 4x -1/2x2 . where x is the number of days exposed to sunlight.

The rate of growth of the plant with respect to sunlight is ______ .

Solution:

∴ rate of growth of the plant with respect
to sunlight

QUESTION: 47

Case - Study

The Relation between the height of the plant (y in cm) with respect to exposure to sunlight is governed
by the following equation y = 4x -1/2x2 . where x is the number of days exposed to sunlight.

What is the number of days it will take for the plant to grow to the maximum height?

Solution:

The number of days it will take for the plant to grow to the maximum height,

QUESTION: 48

Case - Study

The Relation between the height of the plant (y in cm) with respect to exposure to sunlight is governed
by the following equation y = 4x -1/2x2 . where x is the number of days exposed to sunlight.

What is the maximum height of the plant?

Solution:

We have, number of days for maximum height of plant = 4 Days
∴ Maximum height of plant

QUESTION: 49

Case - Study

The Relation between the height of the plant (y in cm) with respect to exposure to sunlight is governed
by the following equation y = 4x -1/2x2 . where x is the number of days exposed to sunlight.

What will be the height of the plant after 2 days?

Solution:

Height of plant after 2 days

QUESTION: 50

Case - Study

The Relation between the height of the plant (y in cm) with respect to exposure to sunlight is governed
by the following equation y = 4x -1/2x2 . where x is the number of days exposed to sunlight.

If the height of the plant is 7/2 cm, the number of days it has been exposed to the sunlight is _____ .

Solution:



We will take x = 1, because it will take 4 days for the plant to grow to the maximum height i.e. 8 cm and 7/2 cm is not maximum height so, it will take less than 4 days. i.e., 1 Day.

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