Mathematics: CUET Mock Test - 9 - Commerce MCQ

The given system of equations can be expressed in the form of AX=B,
⇒ X= A^-1 B

We know that, A^-1= (1/|A|) adj A
A^-1= (1/10) 
∴ X = A^-1 B= 

Consider y=2(tan⁡x)^cot⁡x
Applying log in both sides,
log⁡y=log⁡2(tan⁡x)^cot⁡x
log⁡y=log⁡2+log⁡(tan⁡x)^cot⁡x
log⁡y=log⁡2+cot⁡x log⁡(tan⁡x)
Differentiating both sides with respect to x, we get

dy/dx = 2(tan⁡x)^cot⁡x (−csc^2xlog(tanx)+cot^2x+1)
dy/dx = 2(tan⁡x)^cot⁡x (−csc^2xlog(tanx)+csc^2x)
dy/dx = 2(tan⁡x)^cot⁡x (csc²⁡x (1-log⁡(tan⁡x))
∴ dy/dx =2 csc^2⁡x.tan⁡x^cot⁡x (1-log⁡(tan⁡x))

Given that, y=4x⁴+2x
dy/dx = 16x³+2
d²y/dx² =48x²
48x²−96x³−12
=12(4x²-8x-1)

Consider y= 
y=5x³ (∴log⁡e^x=x)

∴ dy/dx = 5(3x²)=15x²

The marginal cost is given by the rate of change of revenue.
Hence, (dN(x)/dt) =0.18x²-0.02x+10.
∴ = 0.18(5)²-0.02(5)+10
= 4.5-0.1+10
= Rs. 14.4

Let the cost of apples per kg be x and the cost of oranges per kg be y

From the given information, we have 8x + 3y = 70 and 10x + 6y = 90

Solve the equations simultaneously to find x=3, y=2

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

 log⁡3+log⁡(cos⁡x)-x tan⁡x
dy/dx =y(log⁡(3 cos⁡x)-x tan⁡x)
dy/dx =(3 cos⁡x)^x (log⁡(3 cos⁡x)-x tan⁡x)

Given that, y=e^2x+sin^-1⁡e^x

Consider y=7 

Let the length be l, width be b and the area be A.
The Area is given by A=lb

Given that, dl/dt =4cm/s and dA/dt =8 cm/s
Substituting in the above equation, we get

Given that, l=4 cm and b=1 cm

If A is a singular matrix, then |A|=0
In this case, if (adj A) B≠O, then solution does not exist and the system of equations is called inconsistent.

Consider y= 
Applying log to both sides, we get
log⁡y=log 
log⁡y= 
log⁡y= (1/2) (log⁡(x+1)-log⁡(3x-1))
Differentiating with respect to x, we get

Given that, y=3x²+log⁡(4x)

Consider  log⁡x

Differentiating w.r.t x by using chain rule, we get

Given that, 
y=3x²-2x+7

4=6x-2
6x=6
⇒ x=1

The given system of equations can be expressed in the form of AX=B,
⇒X=A^-1 B
We know that, A^-1=1/|A| adj A

  

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

Given that, 

By using u.v rule, we get

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

Let the volume of cube be V.
V=x³

We have y = 2x/(x² + 1)
Differentiating y with respect to x, we get
dy/dx = d/dx(2x/(x² + 1))
= 2 * [(x² + 1)*1 – x * 2x]/(x² + 1)²
= 2 * [1 – x²]/(x² + 1)²
Thus, the slope of tangent to the curve at (0, 0) is,
[dy/dx]_{(0, 0)} = 2 * [1 – 0]/(0 + 1)²
Thus [dy/dx]_{(0, 0)} = 2.

Let y = √x. Let x=64 and Δx=0.3
Then, Δy= 

dy is approximately equal to Δy is equal to:

dy=0.3/16=0.01875
∴ The approximate value of  is 8+0.01875=8.01875

Using Integration by Substitution, Let xm=t
Differentiating w.r.t x, we get
mdx=dt

Replacing t with mx again we get,

∫8x³⁺¹dx
Using
∫8x³⁺¹dx=∫8x³dx+∫1dx

=2x⁴+x+C.

The given function  can also be written as  and is further used to solve integration by partial fractions numerical.

Let, Φ be the angle which the tangent to the curve y = f(x) at P makes with the positive direction of the x axis.
Then,
tanΦ = [f’(x)]_p = -1
= -tan(π/4)
So, it is clear that this can be written as,
= tan(π – π/4)
= tan(3π/4)
So, Φ = 3π/4
Therefore, the required angle which the tangent at P to the curve y = f(x) makes with positive direction of x axis is 3π/4.

Let y= Let x=49 and Δx=0.1
Then, Δy= 

dy is approximately equal to Δy is equal to

dy=0.1/14=0.00714
∴ The approximate value of  is 7+0.00714=7.00714

By using the method of integration by substitution,
Let x³=t
Differentiating w.r.t x, we get
3x² dx=dt
∫3x²(cosx³+8)dx=∫(cost+8)dt
∫(cost+8)dt=sint+8t
Replacing t with x³,we get
∫3x²(cosx³+8)dx=sinx³+8x³+C

To find ∫7x²−x³+2xdx
∫7x²−x³+2xdx=∫7x²dx−∫x³dx+2∫xdx
Using 

As it is not proper rational function, we divide numerator by denominator and get

Let 
So that, 5x–5 = A(x-3) + B(x-2)
Now, equating coefficients of x and constant on both sides, we get A + B = 5 and 3A + 2B = 5. Solving these equations, we get A=-5 and B=10.
Therefore, 

= x – 5log|x-2| + 10log|x-3|+C