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

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).

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

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.

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.

Tangent and normal are perpendicular to each other


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

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

Take b =8, ⇒ a = 6.
Hence, (6, 8) ∈ R

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.

|x| is not differentiable at x=0

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.

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.

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

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

The minor of an element a_ij in A is calculated as the determinant of the square sub-matrix of order (n-1) obtained by leaving the i^th row and j^th column of A.

A = [2 3 5]

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 .

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.

On applying Second Derivative test,

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.

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

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

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

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

So, f(x) always increases.

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.

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

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

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.