Mathematics: CUET Mock Test - 10 - CUET MCQ

- The Assertion is correct because, in a binomial distribution, the probability of success (p) is indeed constant across trials.

- The Reason is false since, in a binomial distribution, the trials are independent, meaning the outcome of one trial does not affect the outcome of another.

- Therefore, the Reason does not explain the Assertion.

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= 

Concept:

Differential Equations by Variable Separable Method
If the coefficient of dx is the only function of x and coefficient of dy is only a function of y in the given differential equation then we can separate both dx and dy terms and integrate both separately.
∫f(x)dx = ∫g(y)dy

Calculation:
Given: xdy - ydx = 0
xdy = ydx
dy/y = dx/x
Integrating both sides, we get

ln y = ln x + ln c
Since ln x + ln y = ln (xy) will be:
⇒ ln(y) = ln cx
⇒ y = cx
Solution of the differential equation represents straight line passing through origin.

Given:

and

Concept:
Use concept of sum of matrix in which add elements of same places of both matrix.
determinant of matrix is expansion of matrix with respect to any one row or column.

Calculation:


A + B = 2(60 - 63) - 3(50 - 56) + 4(45 - 48)
A + B = - 6 + 18 - 12
A + B = 0
Hence the option (2) is correct.

Concept Used:

Any matrix M can be written as the sum of symmetric and skew-symmetric matrix i.e.,

A = S + K

Where S is skew-symmetric which is given by and K is the symmetric part which is given as .

Explanation:

Given Matrix

⇒

⇒

skew-symmetric part of the matrix is

Thus, the skew-symmetric part of the matrix is

⇒

⇒

Hence the option (4) is correct.

Given:

and such that A2 = B

Concept:

Two matrices are equal when all positional elements are equal for both matrices.

Calculation:

and

Then

Given A2 = B then

⇒ α² = 4 and α - 1 = 1

⇒ α = ± 2 and α = 2

Then α = 2

Hence the option (3) is correct.

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

-96x³ + 48x² - 12

• (A) Local minima: For a local minimum, the function has f'(x) = 0 and f''(x) > 0, which is represented by (III).
• (B) Local maxima: For a local maximum, the first derivative is zero and the second derivative is negative, i.e., f'(x) = 0, f''(x) < 0 (which corresponds to (II)).
• (C) Increasing function: A function is increasing if its derivative is positive, i.e., f'(x) > 0 (which corresponds to (I)).
• (D) Normal to the curve: The normal is the line perpendicular to the tangent at a point slope -1/f'(x) (which corresponds to (IV)).

Given that, y=e^2x+sin^-1⁡e^x

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

Replacing t with mx again we get,

f′(x) = 4x³ − 4x = 4x(x − 1)(x + 1).
The critical points of f are 0,−1,1 But f(0) = 5,f(1) = 4,f(−1) =  4,f(2) = 13.
So the range of f is [4,13] and grea, test value of f is at x=2

 are any three vectors then the correct expression for distributivity of scalar product over addition is :

If l, m, n are the direction cosines and a, b, c are the direction ratios of a line then , the directions cosines of the line are given by :

Midpoint of the line segment is

Parallel vector to the required line

Hence, the equation of the line is

Let E1,E2,E3andE4 be the events that the doctor comes by train, bus, scooter and car respectively. Then,
P(E1)=3/10,P(E2)=1/5,P(E3)=1/10andP(E4)=2/5.
Let E be the event that the doctor is late. Then,
P(E/E1) = probability that the doctor is late, given that he comes by train
=1/4.
P(E/E2)= probability that the doctor is late, given that he comes by bus
=1/3.
P(E/E3)= probability that the doctor is late, given that the comes by scooter
=1/12.
P(E/E4)= probability that the doctor is late, given that that he comes by car
=0.
Probability that he comes by train, given that he is late
P(E2/E)
[P(E2).P(E/E2)]/[P(E1).P(E/E1)+P(E2).P(E/E2)+P(E3).P(E/E3)+P(E4).P(E/E40)][by Bayes's theorem]
[(1/5×1/3)]/(3/10×1/4)+(1/5×1/3)+(1/10×1/12)+(2/5×0)
=(3/40×120/18)=4/9
Hence, the required probability is 4/9.

We know that P(A|B) = P(A∩B) / P(B). (By formula for conditional probability)
Which is equivalent to (3/13) / (7/13), hence the value of P(A|B) = 3/7.

Concept:

Calculation:

Multiplying and dividing by sin x

Put cot x = t

⇒ dt = - coesc2 x

Hence the integral becomes

∴ The value of the integral is .

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

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

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

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.

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

• (A) Area under a Line → (I) The integral used to calculate the area enclosed by a straight line on the x-y plane.
• (B) Area under a Circle → (II) The area under the curve of a circle, calculated by the definite integral of its equation.
• (C) Area between Curves → (III) The process of finding the area between two curves, where the integrals of both curves are calculated and the difference is taken.
• (D) Evaluation of Definite Integrals → (IV) The process of evaluating an integral with given limits to find the total area under the curve.

Thus, the correct answer is (1) (A) - (I), (B) - (II), (C) - (III), (D) - (IV).

• (A) Area under a Parabola → (I) The integral used to calculate the area enclosed by a parabola and a straight line.
• (B) Area between a Circle and a Line → (II) The area enclosed by a circle and a line, determined by the intersection points and the corresponding integrals.
• (C) Area between Two Parabolas → (III) The integral of the difference between the equations of two parabolas to find the enclosed area.
• (D) Area between a Parabola and an Ellipse → (IV) The area calculated by finding the difference between the parabola and the ellipse's equations over the given range.

Thus, the correct answer is (1) (A) - (I), (B) - (II), (C) - (III), (D) - (IV).