Mathematics: CUET Mock Test - 10 - CUET MCQ

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.

Calculations:

-(1 - e) + (e¹ - 1)

-1 + e + e - 1

2e - 2

2 (e - 1)

Hence, the Correct Answer is 4.

Concept:
P (A ∪ B) = P (A) + P (B) - P (A ∩ B) ----(1)

Calculation:
Given:
P(A) = 5/12 , P(B) = 3/12 and P (A ∪ B) = 7/12
putting given values in eqn (1),
⇒ (7/12) = (5/12) + (3/12) - P (A ∩ B)

The correct answer is 159/80
Key Points

As we know that ∑P ​=1
Therefore, 0.1 + 2k + k + k + 2k = 1 ⇒ 6k = 0.9 ⇒ k = 0.15
Mean of the given data (m)=∑XP = 0 × 1 + 1 × 2 × 0.15 + 2 × 0.15 + 3 × 0.15 + 4 × 2 × 0.15 = 0 + 0.3 + 0.3 + 0.45 + 1.2 = 2.25
Hence, m2= (2.25)²= 5.0625
The variance of the given data, var(X) =∑X²P − m²
∑X2P = 0² × 0.1 + 1² × 2 × 0.15 + 2² × 0.15 + 3² × 0.15 + 4² × 2 × 0.15 = 0.3 + 0.6 + 1.35 + 4.8 = 7.05
∑X2P − m2 = 7.05 − 5.0625 = 1.9875 = 159/80
Hence, var(X) = 159/80 h

Calculations:
2x + 3y = 28
x + y = 10

At (0, 9.3) value of z = 4x + 2y is 18.6.
At (10, 0) value of z = 4x + 2y is 40.
At (2, 8) value of z = 4x + 2y is 24.
At (0, 0) value of z = 4x + 2y is 0.
Hence, the Correct Option is option no 2.

Formula Used:

2 sin x cos x =sin 2x

Calculation:

Let . . .(1)

Now, Put sin x - cos x = t

⇒​ ( sin x + cos x ) dx = dt

and (sin x - cos x)² = t²

⇒ 1 - 2 sin x cos x = t2

⇒ 1 - sin 2x = t2

⇒ 1 - t2 = sin 2x

Substituting all the values in (1)

Given:

Probability of winning a game p(x) = 0.7

Formula:

Probability of losing a game p(y) = Probability of an event - Probability of winning a game

Variance = n(p(x)) × (1 - p(x))

Calculation:

Let the probability of an event (n) = 1

Probability of losing a game p(y) = Probability of an event - Probability of winning a game

⇒ 1 - 0.7

⇒ 0.3

Variance = n(p(x)) × (1 - p(x))

⇒ 1(0.7) × (1 - 0.7)

⇒ 0.7 × 0.3

⇒ 0.21

The Correct Answer is 0.21.

Solution

⇒

Dividing and multiplying the term by its conjugate.

⇒ × dx

⇒ ...(1 - sin²x = cos²x)

⇒ =

Since, (1/cos²x = sec²x and sin/cos²x = tan x × sec x)

⇒ ∫(sec²x - tan x sec x) dx

⇒ ∫sec²x dx - ∫tan x sec x dx

⇒ tan x - sec x + c

The correct option is 2.

Concept:

Feasible Region:

The feasible region is the set of all feasible solutions that satisfy all the constraints of the linear programming problem. It is represented by shaded regions in the graph of the problem.

Convex set:

A convex set is a set of points in which any two points can be connected by a straight line that lies entirely within the set.

Calculation:

The feasible set in a linear programming problem is always a convex set.

Since, the constraints of the problem are linear, also the intersection of any two linear constraints is always a line or a plane, which is a convex set.

Also, the feasible set is always bounded because the objective function of a linear programming problem is always a linear function.

∴ The feasible region of LPP is always a convex set.

The correct answer is option 4.

Concept:

• The standard deviation (SD) measures the amount of variability, or dispersion, from the individual data values to the mean, while the standard error of the mean (SEM) or mean deviation measures how far the sample mean (average) of the data is likely to be from the true population mean.
• The SEM is always smaller than the SD.

• In a symmetrical distribution, mean deviation equals 4/5 of standard deviation.

Calculation:

Since the distribution is symmetrical,

⇒ Mean deviation = 4/5 of Standard deviation

⇒ 12 = (4/5) × Standard deviation

⇒ Standard deviation = 15

Hence, the value of the standard deviation is 15.

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)