Mathematics: CUET Mock Test - 10

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.

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.

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

=e^t
Replacing t with -cot^-1x, we get

To find 

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.

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, 

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

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

Replacing t with x⁵+9, we get

To find 

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).

Let y=(x)^1/3. Let x=125 and Δx=2
Then, Δy=(x+Δx)^1/3-x^1/3
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

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


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

To find ∫(2+x) x √x dx

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.

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.

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.

Let 1+x⁴=t
4x³ dx=dt

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

To find: 
 

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, 

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)

Let y=x^1/4. Let x=81 and Δx=1
Then, Δy=(x+Δx)^1/4-x^1/4
Δy=82^1/4-81^1/4
82^1/4=Δy+3
dy is approximately equal to Δy is equal to
dy = (dy/dx)Δx

∴ The approximate value of 82^1/4 is 3+0.00925=3.00925

Let x³=t
3x² dx=dt
x² dx=dt/3

Replacing t with x³, we get

To find ∫sin2x+e^3x−cos3xdx
∫sin2x+e^3x−cos3xdx=∫sin2xdx+∫e^3xdx−∫cos3xdx
∫sin2x+e^3x−cos3xdx=

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.

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

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

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

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

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

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.

Let the height on the wall be x and laser touches the ground at distance y from the wall. The length of the ladder is 20ft.
By Pythagoras theorem:
x² + y² = 400
Differentiating with respect to t:
2x(dx/dt) + 2y(dy/dt) = 0
Dividing throughout by 2:
x (dx/dt) + y (dy/dt) = 0
Now, dx/dt = -2 ft /s. negative because downwards
x(-2) + y (dy/dt) = 0 ………..(1)
When lower end is 12 ft from wall, let us find x:
x² + 12² = 400
x² = 400 – 144= 256
x = 16
x(2) + y (dy/dt) = 0 from (1)
16(-2) + 12 (dy/dt) = 0
-32 + 12(dy/dt) = 0
dy/dt = (32/12) = (8/3)
Thus, lower end moves on a horizontal floor when it is 12 ft from the wall at the rate of 8/3 ft/s
Now, assume that the ladder makes an angle θ with the horizontal plane at time t.
If, m be the slope of the ladder at time t, then,
m = tanθ = x/y
Thus, dm/dt = d/dt(x/y) = [y(dx/dt) – x(dx/dt)]/y²
Therefore, the rate of change of slope of ladder is,
[dm/dt]_{y = 2} = [12*(-2) – 16*(8/3)]/(12)²
Now, putting the value of x = 16, when y = 12 and dx/dt = -2, dy/dt = 8/3
We get, [dm/dt]_{y = 12} = [12(-2) – 16(8/3)]/(12)² = -25/54

We have, x = t³ + 6t² – 15t + 18
Let, ‘v’ be the velocity of the particle and ‘a’ be the acceleration of the particle at the end of t seconds. Then,
v = dx/dt = d/dt(t³ + 6t² – 15t + 18)
So, v = 3t² + 12t – 15
Therefore, a = dv/dt = d/dt(3t² + 12t – 15)
So, a = 6t + 12
Thus, acceleration of the particle at the end of 2 seconds is, [dv/dt]_{t = 2} = 6(2) + 12 = 24cm/sec².

Let ‘r’ be the radius and ‘h’ be the height of the water level at time t.
Then the volume of water level ‘V’ at time t is given by,
V = 1/3 (πr²h) ……….(1)
Given the radius of the base of the cone is = (OA)’ = 27/2 cm and its height = (OB)’ = 45cm.
Again at time t, the radius of the water level = r = (CD)’ and its height = h = (CB)’
Clearly, the triangle OAB and CBD are similar.
Therefore, (CD)’/(CB)’ = (OA)’/(OB)’
Or, r/h = (22/7)/45 = 3/10
or, r = 3h/10
Thus, from (1) we get,
V = 1/3 π (3h/10)² h = (3π/100)h³
Thus, dV/dt = (3π/100)*3h²(dh/dt)
= (9πh²/100)(dh/dt)
When, h = 30cm, then, 11 = (9π/100) (30)² (dh/dt) [as for all the values of t we have, dV/dt = 11]
Or, dh/dt = (11/(9π*9)) = 0.043
Thus, the rising rate of rising is 0.043 cm/minute.

We have, x = t³ + 6t² – 15t + 18
The particle stops when dx/dt = 0
And, dx/dt = d/dt(t³ + 6t² – 15t + 18)
3t² + 12t – 15 = 0
⇒ t² + 4t –5 = 0
⇒ (t – 1)(t + 5)= 0
Thus, t = 1 or t = -5
Hence, the particle stops at the end of 1 second.

We have, x = t³ + 6t² – 15t + 18
Let, v be the velocity of the particle at the end of t seconds. Then, v = dx/dt = d/dt(t³ + 6t² – 15t + 18)
So, v = 3t² + 12t – 15
Thus, velocity of the particle at the end of 2 seconds is, [dx/dt]_{t = 2} = 3(2)² + 12(2) – 15 = 21cm/sec.

If L₁ and L₂ have the direction ratios
a₁,b₁,c₁ and a₂,b₂,c₂  respectively then the angle between the lines is given by

The angle between two lines is given by the equation


0 = p+1
p = -1

We know that, if two lines are perpendicular to each other then, a₁a₂ + b₁b₂ + c₁c₂ = 0
i.e.4(p)+(-2)2+4(-12)=0
4p-4-48=0
4p=52
p = 52/4 = 13.

Given that the equations of the lines are

∴ the angle between the two lines is given by

If two lines L₁ and L₂ are having direction cosines l₁,m₁,n₁ and l₂,m₂,n₂  respectively, then the angle between the lines is given by cos⁡θ=|l₁l₂+m₁m₂+n₁n₂|