Mathematics: CUET Mock Test - 8 - Commerce MCQ

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

Given that, y=tan²⁡⁡x+3 tan⁡x
dy/dx =2 tan⁡x sec²⁡⁡x+3 sec²⁡x=sec^2⁡⁡x (2 tan⁡x+3)
By using the u.v rule, we get
 (sec²⁡⁡x).(2 tan⁡x+3)+ (d/dx) (2 tan⁡x+3).sec²⁡⁡x
 =2 sec²⁡⁡x tan⁡x (2 tan⁡x+3)+sec^2⁡⁡x (2 sec⁡x tanx)
=2 sec²⁡x tan⁡x (2 tan⁡x+sec⁡x+3).

Consider y=8e^cos2x
Differentiating w.r.t x by using chain rule, we get

Let the edge of the cube be x. The rate of change of edge of the cube is given by dx/dt =7cm/s.
The area of the cube is A=6x²

By using the matrix method, the given equations can be expressed in the form of the equation AX=B, where

To find the value of x and y, we need to solve the matrix X
X = A^-1 B

Consider y=9^tan⁡3x
Applying log on both sides, we get
log⁡y=log⁡9^tan⁡3x
Differentiating both sides with respect to x, we get

 (∵ Using u.v = u′v + uv′)
(dy/dx) = y(3 sec²⁡⁡x.log⁡9+0)
(dy/dx) = 9^tan⁡3x (3 log⁡9 sec²⁡x)

Consider y=3 sin^-1⁡(e^2x)
dy/dx = d/dx (3 sin^-1⁡(e^2x))

Let the side of the square be x.
A = x², where A is the area of the square
Given that, 

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

By using u.v=u’ v+uv’, we get
(3x)log(cos3x)+(d/dx)(log(cos3x)).3x
(dy/dx)=y(3 log⁡(cos⁡3x) +(cos3x).3x)
(dy/dx)=y(3 log⁡(cos⁡3x) +
(dy/dx)=y(3 log⁡(cos⁡3x) +
(dy/dx) = y(3 log⁡(cos⁡3x) – 9x tan⁡3x)
(dy/dx) = (cos⁡3x)^3x (3 log⁡(cos⁡3x) – 9x tan⁡3x)

Given that, y=2e^2x-3 log⁡(2x-3)

Consider y=(log⁡(log⁡(x⁵)))