Mathematics: CUET Mock Test - 8 - CUET MCQ

Given:

A and B be two non zero square matrics

Calculations:

Matrix AB is defined means Columns is equal to the Rows of B

and BA is defined means Columns of B is equal to the Rows of A

Hence, Both matrices (A) and (B) have same order is Correct.

A. In the above matrix A = , we can see, the number of rows and columns are 2 respectively. Since the order of the matrix is 2 × 2, hence A is a square matrix.

B. The given 2 × 2 matrix

We first find the determinant of A.

Det A = (2 × 5) - (3 × -3) = 10 + 9 = 19

∴ |A| = 19

Since, |A| ≠ 0 ⇒ A-1 exists.

C. To know if a matrix is symmetric, find the transpose of that matrix. If the transpose of that matrix is equal to itself, it is a symmetric matrix. That is A = A^T

Here A = then A^T =

Here, A ≠ AT

Thus A is not a symmetric matrix.

D. We have already derived |A| = 19.

E. Null Matrix: If in a matrix all the elements are zero then it is called a null matrix. It is also called a zero matrix. Here we can see A is not null matrix.

Thus A, B, D is the correct answer.

The number of possible entries of 2 × 2 matrix is 4 Every entry has two choice, 0 or 1.

Thus, the total no. of choices is,

2 × 2 × 2 × 2 = 2⁴

= 16

Calculation:
y = (1/x )^x
take log on both sides
log y = x log(1/x)
log y = x(log 1 - log x)
log y = x(-log x)
Differentiate with respect to y
1/y × dy/dx = -( 1 + log x)
dy/dx = -(1/x)^x(1 + log x)again
Differentiate with respect to x .
d²y/dx² = -(dy/dx(1 + logx) +y(1/x)
d2y/dx2 = -(-(1/x)^x(1+ log x)² + (1/x)^x+1)
= 4 -1/e
Hence, option 2 is correct.

Concept:

For a function y = f(x):

• At the points of local maxima or minima, f'(x) = 0.
• In the regions where f(x) is increasing, f'(x) > 0.
• In the regions where f(x) is decreasing, f'(x) < 0.

Calculation:

For the function f(x) = x2 - 2x to be strictly increasing, f'(x) > 0.

For the function f(x) = x2 − 2x to be strictly decreasing, f'(x) < 0.

⇒ f'(x) = 2x - 2 < 0

⇒ x < 1

⇒ x ∈ (−∞, 1)

The matrix A is invertible if det(A) ≠ 0.
To determine when the matrix
 is invertible, we can analyze its determinant. A matrix is invertible if and only if its determinant is nonzero.
The determinant of the given matrix is given by:

Expanding along the first column:


Now, det(A) = (t - 1)(t - 2) ≠ 0
t ≠ 1, and t ≠ 2.
Thus, the matrix is invertible for all real numbers except 1 and 2.

Concept:

The method of finding Maxima and Minima of y = f(x)

• Find f’(x) and f”(x) for given function y = f(x)
• Equate f’(x) to zero to obtain stationary points x = a
• Calculate f”(x) at each stationary points x = a (i.e f”(a))
• The following three conditions are obtained:
• If f”(a) > 0 then f(x) has a minimum at x = a and minimum value will be f(a)
• If f”(a) < 0 then f(x) has a maximum at x = a and maximum value will be f(a)
• If f”(a) = 0 then f(x) may or may not have a maximum or a minimum at x = a

Calculation :
f(x) = x3 − 3x2 + 2x in [1, 2]
f(x)= x(x² - 3x +2)
f(x)= x(x-2)(x-1)
Critical Points : 0,1,2
f'(x) = 3x² - 6x +2
f''(x) =6x - 6
f''(0) = -6 < 0 ⇒ maximum
f''(1)= 0
f''(2) = 6 ⇒ minimum
Maxima at x = 0
maximum value of f(x)
⇒ f(0) = 0
Hence correct option is "4"

Concept Used:

Gaussian Elimination: The process of transforming a matrix into its echelon form by applying elementary row operations. These operations include adding one row to another, multiplying a row by a scalar, and swapping rows.

Explanation:

Given Matrix:

As there are only two linearly independent rows in the given matrix after row reduction. Thus, the rank of the matrix is 2.

Concept:

(i) A square matrix A is said to be an idempotent matrix if A² = A

(ii) Number of diagonal idempotent matrices of order n is 2ⁿ

Explanation:

The number of diagonal idempotent matrices of order 5 is 25 = 32

Option (2) is correct

Concept used:

dy/dx = 0, the value of x gives the minimum value when d2y/dx2 is greater than 0, and gives the maximum value when d2y/dx2 is less than 0

Calculation:

y = x2 - 4x ----(i)

Differentiating (i),

dy/dx = 0

⇒ 2x - 4 = 0 ----(ii)

⇒ x = 2

By double differentiating equation (ii),

d2y/dx2 = 2 > 0

That means the minimum value of equation (i) is at x = 2

The minimum value of equation (i)

⇒ 2² - 4 × 2

⇒ -4

∴ The minimum value is at x = 2.

Here, we have, 
Now, replacing C₃ = C₃ – 3C₁, we get,

Or, (x – 1)
Now, replacing R₁ = R₁ + 3R₃, we get,
Or, (x – 1) 
Or, (x – 1)[-1 {(11 – x)(5 – x) – 28}] = 0
Or, -(x – 1)(55 – 11x – 5x + x² – 28)
Or, (x – 1)(x² – 16x + 27) = 0
Thus, either x – 1 = 0 i.e. x = 1 or x² – 16x + 27 = 0
Therefore, solving x² – 16x + 27 = 0 further, we get,
x = 8 ± √37

The area of the triangle with vertices (2,3), (4,1), (5,0) is given by

Applying R₂→R₂-R₃

Expanding along R₂, we get
Δ=(1/2){-(-1)(3-0)+1(2-5)}
Δ=(1/2) (0-0)=0.

Consider the element -3 in Δ= 
The cofactor of the element -3 is given by
A₂₂=(-1)²⁺² M₂₂
M₂₂=  =1(5)-(-6)(-7)=5-42=-37
A₂₂=(-1)²⁺² (-37)=-37.

Given that, A= 
|A| = 
|A|=-cos⁡θ (cos⁡θ )-cotθ(-tan⁡θ)
|A|=-cos²⁡θ+1=sin²⁡θ.

The above matrix is a skew symmetric matrix and its order is odd
And we know that for any skew symmetric matrix with odd order has determinant = 0
Therefore, the value of the given determinant = 0

Let, D = 
Expanding D by the 1^st row we get,
D = – c 
= – c(0 – ab) + b(ac – 0)
= 2abc
Now, we have adjoint of D = D’

Or, D’ = 
Or, D’ = D²
Or, D’ = D² = (2abc)²

Let C(x,y) be a point on the line AB. Thus, the points A(5,1), B(4,0), C(x,y) are collinear. Hence, the area of the triangle formed by these points will be 0.
⇒ Δ = (1/2) 
Applying R₁→R₁-R₂

Expanding along R₁, we get
=(1/2) {1(0-y)-1(4-x)}=0
=(1/2){-y-4+x}=0
⇒ x-y = 4.

Consider the element 7 in the determinant Δ= 
The minor of the element 7 can be obtained by deleting R₂ and C₂
∴ M₂₂ = 2
Hence, the minor of the element 7 is 2.

Expanding along R₁, we get

Δ=5(24-24)-0+5(8-0)
Δ=0-0+40=40.

Here, C₁ and C₃ becomes equal when we put p = xⁿ
And R₁ and R₃ becomes equal when we put p = n + 1
And R1 and R3 becomes equal when we put p = n + 1

Given that the vertices are (3,2), (1,2), (5,k)
Therefore, the area of the triangle with vertices (3,2), (1,2), (5,k) is given by
Δ=(1/2) 
Applying R₁→R₁-R₂, we get
1/2 
Expanding along R₁, we get
(1/2) {2(2-k)-0+0} = 0
2-k = 0
k = 2 .

Consider the element 2 in the determinant Δ= 
The minor of the element 2 is given by
∴ M₂₂ = = 40-40 = 0
⇒ A₂₂ = (-1)²⁺² (0) = 0.

If a given system of equations has one or more solutions then the system is said to be consistent.

Consider y=(log2x)^sin3x
Applying log on both sides, we get
log⁡y=log(log2x)^sin3x
log⁡y=sin⁡3x log⁡(log⁡2x)
Differentiating with respect to x, we get

By using chain rule, we get

dy/dx =y(3 cos⁡3x log⁡(log⁡2x)+ 
∴ dy/dx=log⁡2x^sin⁡3x(3cos3xlog(log2x)+(sin3x/xlog2x))

Given that, y=9 log⁡t³

To solve:y=(8e^-x+2e^x)
Differentiating w.r.t x we get,
(dy/dx)= 8(-e^-x+2e^x)
∴ (dy/dx)= 2e^x - 8e^-x.

The rate of change of radius of the circle is dr/dt = 6 cm/s
The area of a circle is A=πr²
Differentiating w.r.t t we get,
(dA/dt = d/dt) (πr²) = 2πr (dr/dt) =2πr(6)=12πr.
dA/dt |_r=2=24π= 24×3.14=75.36 cm²/s

If a given system of equations has no solutions, then the system is said to be inconsistent.

Consider
Applying log on both sides, we get
log⁡y=log
log⁡y=log⁡4+  (∵log⁡ab  =log⁡a+log⁡b)
Differentiating both sides with respect to x, we get