Mathematics: CUET Mock Test - 9

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

The given situation can be represented by a system of equations as:
8x+3y=70
10x+5y=90
Consider, A= 
It can be expressed in the form of AX=B
⇒ X=A^-1 B
We know that, A^-1= 
 
∴ x=3, y=2.
i.e. The cost of apples is Rs 3 per kg and the cost of oranges is Rs 2 per kg.

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

Let, y = f(x) = √(x² + 2)
So, f(x) = (x² + 2)^1/2
On differentiating it we get,
f’(x) = d/dx[(x² + 2)^1/2]
f’(x) = 1/2 * 1/√(x² + 2) * 2x
So f’(x) = x/√(x² + 2)
So the differential equation is:
dy = f’(x)dx
Hence, dy = x/√(x² + 2) dx

Let x=5 and Δx=0.03
Then, f(x+Δx)=4(x+Δx)²-7(x+Δx)+2
Δy=f(x+Δx)-f(x)
∴f(x+Δx)=Δy+f(x)
Δy=f’ (x)Δx
⇒ f(x+Δx)=f(x)+f’ (x)Δx
f(5.03)=(4(5)²-7(5)+2)+(8(5)-7)(0.03) (∵ f’ (x)=8x-7)
f(5.03)=(100-35+2)+(40-7)(0.03)
f(5.03)=67+33(0.03)
f(5.03)=67+0.99=67.99

Let x²=t
Differentiating w.r.t x, we get
2x dx=dt
∫6x(x²+6)dx=3∫(t+6)dt
3∫(t+6)dt=3 
Replacing t with x²
∫6x(x²+6)dx= +18x²+C

To find ∫ 2 sin⁡2x+3 dx
∫2sin2x+3dx=∫2sin2xdx+∫3dx
∫2sin2x+3dx=2∫sin2xdx+3∫dx
∫2sin2x+3dx=  + 3x
∴∫2 sin⁡2x+3 dx = -cos⁡2x+3x+C

It is a proper rational function. Therefore,

Where real numbers are determined, 1 = A(x+2) + B(x+1), Equating coefficients of x and the constant term, we get A+B = 0 and 2A+B = 1.
Solving it we get A=1, and B=-1.
Thus, it simplifies to, 
= log|x+1| – log|x+2| + C

Let, y = f(x) = log(x² + 4)
So f(x) = log(x² + 4)
On differentiating it we get,
f’(x) = d/dx[log(x² + 4)]
So f’(x) = 2x/(x² + 4)
So the differential equation is:
dy = f’(x)dx
Hence, dy = 2x/(x² + 4) dx

Let y=√x. Let x=9 and Δx=2
Then, Δy= 

dy is approximately equal to Δy is equal to

dy=2/6=0.34
∴ The approximate value of √11 is 3+0.34=3.34.

Let log⁡x=t
Differentiating w.r.t x, we get

Replacing t with log⁡x, we get

To find 

An improper integration factor can be reduced to proper fraction by division, i.e., if the numerator and denominator have same degree, then they must be divided in order to reduce it to proper fraction.