Mathematics: CUET Mock Test - 10 - Commerce MCQ

# Mathematics: CUET Mock Test - 10 - Commerce MCQ

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## 30 Questions MCQ Test - Mathematics: CUET Mock Test - 10

Mathematics: CUET Mock Test - 10 for Commerce 2024 is part of Commerce preparation. The Mathematics: CUET Mock Test - 10 questions and answers have been prepared according to the Commerce exam syllabus.The Mathematics: CUET Mock Test - 10 MCQs are made for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Mathematics: CUET Mock Test - 10 below.
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Mathematics: CUET Mock Test - 10 - Question 1

### What will be the average rate of change of the function [y = 16 – x2] between x = 3 and x = 4?

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 1

Let, y = f(x) = 16 – x2
If x changes from 3 to 4, then, δx = 4 – 3 = 1
Again f(4) = 16 – 42 = 0
And f(3) = 16 – 32 = 7
Therefore, δy = f(4) – f(3) = 0 – 7 = -7
Hence, the average rate of change of the function between x = 3 and x = 4 is:
δy/δx = -7/1 = -7.

Mathematics: CUET Mock Test - 10 - Question 2

### What will be the approximate change in the surface area of a cube of side xm caused by increasing the side by 2%.

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 2

Let the edge of the cube be x. Given that dx or Δx is equal to 0.02x(2%).
The surface area of the cube is A=6x2
Differentiating w.r.t x, we get
dA/dx =12x
dA= (dA/dx)Δx=12x(0.02x)=0.24x2
Hence, the approximate change in volume is 0.24x2.

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Mathematics: CUET Mock Test - 10 - Question 3

### Find

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 3

Let −cot−1x=t
Differentiating w.r.t x, we get

=et
Replacing t with -cot-1x, we get

Mathematics: CUET Mock Test - 10 - Question 4

Find the integral of .

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 4

To find

Mathematics: CUET Mock Test - 10 - Question 5

equals ______

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 5

We know that
By simplifying it we get,
Now equating the coefficients we get A = 0, B = 0, C=1.

Therefore after integrating we get log|x|– (1/2)log(x2+1)  + C.

Mathematics: CUET Mock Test - 10 - Question 6

equals ___

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 6

By simplifying, it we get
By solving the equations, we get, A+B=0 and 3A-3B=1
By solving these 2 equations, we get values of A=1/6 and B=-1/6.
Now by putting values in the equation and integrating it we get value,

Mathematics: CUET Mock Test - 10 - Question 7

What will be the average rate of change of the function [y = 16 – x2] at x = 4?

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 7

Let, y = f(x) = 16 – x2
dy/dx = -2x
Now, [dy/dx]x = 4 = [-2x]x = 4
So, [dy/dx]x = 4 = -8

Mathematics: CUET Mock Test - 10 - Question 8

Find the approximate value of f(4.04), where f(x)=7x3+6x2-4x+3.

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 8

Let x=4 and Δx=0.04
Then, f(x+Δx)=7(x+Δx)3+7(x+Δx)2-4(x+Δx)+3
Δy=f(x+Δx)-f(x)
∴f(x+Δx)=Δy+f(x)
Δy=f'(x)Δx
⇒ f(x+Δx)=f(x)+f’ (x)Δx
Here, f'(x)=21x2+12x-4
f(4.04)=(7(4)3+6(4)2-4(4)+3)+(21(4)2+12(4)-4)(0.04)
f(4.04)=(7(4)3+6(4)2-4(4)+3)+(21(4)2+12(4)-4)(0.04)
f(4.04)=531+380(0.04)=546.2

Mathematics: CUET Mock Test - 10 - Question 9

Find the integral of .

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 9

Let x5+9=t
Differentiating w.r.t x, we get
5x4 dx=dt

Replacing t with x5+9, we get

Mathematics: CUET Mock Test - 10 - Question 10

Find ∫3cosx+(1/x)dx .

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 10

To find

Mathematics: CUET Mock Test - 10 - Question 11

What will be the value of the co-ordinate whose position of a particle moving along the parabola y2 = 4x at which the rate at of increase of the abscissa is twice the rate of increase of the ordinate?

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 11

Here, y2 = 4x ….(1)
Let, (x, y) be the position of the particle moving along the parabola (1) at time t.
Now, differentiating both sides of (1) with respect to t, we get:
2y(dy/dt) = 4(dx/dt)
Or, y(dy/dt) = 2(dy/dt) ……….(2)
By question, dx/dt = 2 * dy/dt ……….(3)
From (2) and (3) we get, y(dy/dt) = 2 * 2 dy/dt
Or, y = 4
Putting y = 4 in (1) we get, 42 = 4x
So, x = 4
Thus, the co-ordinate of the particle is (4, 4).

Mathematics: CUET Mock Test - 10 - Question 12

Find the approximate value of (127)1/3.

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 12

Let y=(x)1/3. Let x=125 and Δx=2
Then, Δy=(x+Δx)1/3-x1/3
Δy=(127)1/3-(125)1/3
(127)1/3=Δy+5
dy is approximately equal to Δy is equal to

dy=2/75=0.0267
∴ The approximate value of (127)1/3 is 5+0.0267=5.0267

Mathematics: CUET Mock Test - 10 - Question 13

Find

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 13

Let √x = t
Differentiating w.r.t x,we get

=12(-cos⁡t)=-12 cos⁡t
Replacing t with √x, we get

Mathematics: CUET Mock Test - 10 - Question 14

Find ∫(2+x)x √x dx .

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 14

To find ∫(2+x) x √x dx

Mathematics: CUET Mock Test - 10 - Question 15

Which form of rational function  represents?

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 15

It is a form of the given partial fraction  which can also be written as  and is further used to solve integration by partial fractions numerical.

Mathematics: CUET Mock Test - 10 - Question 16

The time rate of change of the radius of a sphere is 1/2π. When it’s radius is 5cm, what will be the rate of change of the surface of the sphere with time?

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 16

Let, r be the radius and s be the area of the surface of the sphere at time t.
By question, dr/dt = 1/2π
Now, s = 4πr2;
Thus, ds/dt = 4π * 2r(dr/dt) = 8πr(dr/dt)
When r = 5cm and dr/dt = 1/2π
Then ds/dt = 8π*5*(1/2π) = 20
Thus, correct answer is 20 sq cm.

Mathematics: CUET Mock Test - 10 - Question 17

Find the approximate change in the volume of cube of side xm caused by increasing the side by 6%.

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 17

We know that the volume V of a cube is given by V=x3
Differentiating w.r.t x, we get
dV/dx = 3x2
dV = (dV/dx)Δx=3x2 Δx
dV=3x2 (6x/100)=0.18x3
Therefore, the approximate change in volume is 0.18x3.

Mathematics: CUET Mock Test - 10 - Question 18

Find

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 18

Let 1+x4=t
4x3 dx=dt

=5 log⁡t
Replacing t with 1+x4, we get
= 5log(1+x4)+C

Mathematics: CUET Mock Test - 10 - Question 19

Find ∫7x8−4e2x− (2/x2)dx .

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 19

To find:

Mathematics: CUET Mock Test - 10 - Question 20

equals ______

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 20

Now equating, (x2+x+1) = A (x2+1) + (Bx+C) (x+2)
After equating and solving for coefficient we get values,  now putting these values in the equation we get,

Hence it comes,

Mathematics: CUET Mock Test - 10 - Question 21

A solid cube changes its volume such that its shape remains unchanged. For such a cube of unit volume, what will be the value of rate of change of volume?

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 21

Let x be the length of a side of the cube.
If v be the volume and s the area of any face of the cube, then
v = x3 and s = x2
Thus, dv/dt = dx3/dt = 3x2 (dx/dt)
And ds/dt = dx2/dt = 2x(dx/dt)
Now, (dv/dt)/(ds/dt) = 3x/2
Or, dv/dt = (3x/2)(ds/dt)
Now, for a cube of unit volume we have,
v = 1
⇒ x = 1 [as, x is real]
Therefore, for a cube of unit volume [i.e. for x = 1], we get,
dv/dt = (3/2)(ds/dt)
Thus the rate of change of volume = 3/2*(rate of change of area of any face of the cube)

Mathematics: CUET Mock Test - 10 - Question 22

Find the approximate value of (82)1/4.

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 22

Let y=x1/4. Let x=81 and Δx=1
Then, Δy=(x+Δx)1/4-x1/4
Δy=821/4-811/4
821/4=Δy+3
dy is approximately equal to Δy is equal to
dy = (dy/dx)Δx

∴ The approximate value of 821/4 is 3+0.00925=3.00925

Mathematics: CUET Mock Test - 10 - Question 23

Integrate

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 23

Let x3=t
3x2 dx=dt
x2 dx=dt/3

Replacing t with x3, we get

Mathematics: CUET Mock Test - 10 - Question 24

Find the integral ∫sin2x+e3x−cos3xdx.

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 24

To find ∫sin2x+e3x−cos3xdx
∫sin2x+e3x−cos3xdx=∫sin2xdx+∫e3xdx−∫cos3xdx
∫sin2x+e3x−cos3xdx=

Mathematics: CUET Mock Test - 10 - Question 25

Identify the type of the equation (x+1)2.

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 25

As it represents the identity (b+a)2 it satisfies the identity (b+a)2 = (a2 + b2 +2ab) and is not linear, cubic or an imaginary equation so the correct option is Identity Equation.

Mathematics: CUET Mock Test - 10 - Question 26

A 5 ft long man walks away from the foot of a 12(½) ft high lamp post at the rate of 3 mph. What will be the rate at which the shadow increases?

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 26

Let, AB be the lamp-post whose foot is A, and B is the source of light, and given (AB)’ = 12(½) ft.
Let MN denote the position of the man at time t where (MN)’ = 5ft.
Join BN and produce it to meet AM(produced) at P.
Then the length of man’s shadow= (MP)’
Assume, (AM)’ = x and (MP)’ = y. Then, (PA)’ = (AM)’ + (MP)’ = x + y
And dx/dt = velocity of the man = 3
Clearly, triangles APB and MPN are similar.
Thus, (PM)’/(MN)’ = (PA)’/(AB)’
Or, y/5 = (x + y)/12(½)
Or, (25/2)y = 5x + 5y
Or, 3y = 2x
Or, y = (2/3)x
Thus, dy/dt = (2/3)(dx/dt)
As, dx/dt = 2,
= 2/3*3 = 2mph

Mathematics: CUET Mock Test - 10 - Question 27

Find the approximate error in the volume of the sphere if the radius of the sphere is measured to be 6cm with an error of 0.07cm.

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 27

Let x be the radius of the sphere.
Then, x=6cm and Δx=0.07cm
The volume of a sphere is given by V= (4/3)πx3

dV=4×π×62×0.07
dV=10.08π cm3

Mathematics: CUET Mock Test - 10 - Question 28

Find

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 28

Let cos-1⁡x=t
Differentiating w.r.t x, we get

= t2/2
Replacing t with cos-1x,we get

Mathematics: CUET Mock Test - 10 - Question 29

Find the integral of (ax2+b)2.

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 29

To find (ax2+b)2
∫(ax2+b)2dx=∫(a2x4+b2+2ax2b)dx
∫(ax2+b)2dx=∫a2x4dx+∫b2dx+2∫ax2bdx
∫(ax2+b)2dx=a2∫x4dx+b2∫dx+2ab∫x2dx

Mathematics: CUET Mock Test - 10 - Question 30

For the given equation (x+2) (x+4) = x2 + 6x + 8, how many values of x satisfies this equation?

Detailed Solution for Mathematics: CUET Mock Test - 10 - Question 30

If we solve the L.H.S. (Left Hand Side) of the equation, we get the following value.
(x+2) (x+4) = x2 + 4x + 2x + 8 = x2 + 6x + 8.
This value is same as the R.H.S. (Right Hand Side).
So, all the values of x satisfy the equality.

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