Mathematics: CUET Mock Test - 3 Free MCQ Practice Test with Solutions

Mathematics: CUET Mock Test - 3 - Question 1

Which of the following functions is not a solution for the differential equation y” + 9y = 0?

A.
y = 5 tan⁡3x
B.
y = 5 cos⁡3x
C.
y = cos⁡3x
D.
y = 6 cos⁡3x

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 1

Given the differential equation:
y'' + 9y = 0

We need to find which function satisfies this equation.

Step 1: Test each option by calculating y'' + 9y

Option A: y = 5 tan 3x

y = 5 tan 3x
y' = 5 * 3 sec² 3x = 15 sec² 3x
y'' = 15 * derivative of sec² 3x = 15 * 2 sec² 3x * tan 3x * 3 = 90 sec² 3x tan 3x (using chain rule)

Now, y'' + 9y = 90 sec² 3x tan 3x + 9 * 5 tan 3x = tan 3x (90 sec² 3x + 45) which is not zero for all x.

Option A does not satisfy the equation.

Option B: y = 5 cos 3x

First derivative: y' = 5 * (-3 sin 3x) = -15 sin 3x
Second derivative: y'' = -15 * 3 cos 3x = -45 cos 3x

Now, compute y'' + 9y:
y'' + 9y = -45 cos 3x + 9 * 5 cos 3x = -45 cos 3x + 45 cos 3x = 0

So, Option B satisfies the differential equation.

Option C: y = cos 3x

y' = -3 sin 3x
y'' = -9 cos 3x

Compute y'' + 9y:
-9 cos 3x + 9 cos 3x = 0

Option C satisfies the equation.

Option D: y = 6 cos 3x

y' = -18 sin 3x
y'' = -54 cos 3x

Compute y'' + 9y:
-54 cos 3x + 9 * 6 cos 3x = -54 cos 3x + 54 cos 3x = 0

Option D satisfies the equation.

Final answer:

Options B, C, and D satisfy the differential equation y'' + 9y = 0.

Among the given choices,
B: y = 5 cos 3x
C: y = cos 3x
D: y = 6 cos 3x

are solutions.

Only Option A is not a solution.

Mathematics: CUET Mock Test - 3 - Question 2

Which of the following differential equations given below has the solution y = log⁡x?

A.
B.
C.
D.

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 2

Consider the function y = log⁡x
Differentiating w.r.t x, we get

Differentiating (1) w.r.t x, we get

∴
=

Mathematics: CUET Mock Test - 3 - Question 3

Which of the following functions is a solution for the differential equation dy/dx -14x = 0?

A.
y = 7x²
B.
y = 7x³
C.
y = x⁷
D.
y = 14x

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

Consider the function y = 7x²
Differentiating w.r.t x, we get
dy/dx = 14x
⇒ dy/dx -14x = 0
Hence, the function y = 7x² is a solution for the differential equation dy/dx - 14x = 0

Mathematics: CUET Mock Test - 3 - Question 4

A.
B.
C.
D.

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 4

Mathematics: CUET Mock Test - 3 - Question 5

The area common to the circle x²+y² = 16 and the parabola y²= 6x is

A.
B.
C.
D.
none of these

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 5

Solving eqns. (i) and (ii), we get points of intersection (2, 2√3) and (2, -2√3)

Substituting these values of x in eq. (ii). Since both curves are symmetrical about r-axis.

Hence the required area

Mathematics: CUET Mock Test - 3 - Question 6

Given, f(x) = x³ – 12x² + 45x + 8. What is the maximum value of f(x)?

A.
61
B.
62
C.
63
D.
54

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 6

We have, f(x) = x³ – 12x² + 45x + 8 ……….(1)
Differentiating both sides of (1) with respect to x we
f’(x) = 3x² – 24x + 45
3x² – 24x + 45 = 0
Or x² – 8x + 15 = 0
Or (x – 3)(x – 5) = 0
Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5
Therefore, f’(x) = 0 for x = 3 and x = 5.
If h be a positive quantity, however small, then,
f’(3 – h) = 3*(3 – h – 3)(3 – h – 5) = 3h(h + 2) = positive.
f’(3 + h) = 3*(3 + h – 3)(3 + h – 5) = 3h(h – 2) = negative.
Clearly, f’(x) changes sign from positive on the left to negative on the right of the point x = 3.
So, f(x) has maximum at 3.
Putting, x = 3 in (1)
Thus, its maximum value is,
f(3) = 3³ – 12*3² + 45*3 + 8 = 62.

Mathematics: CUET Mock Test - 3 - Question 7

The Poisson distribution comes under which probability distribution?

A.
Continuous probability distribution
B.
Sine probability distribution
C.
Discrete probability distribution
D.
Mutual probability distribution

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 7

Poisson distribution shows the number of times an event is likely to occur within a specified time. It is used only for independent events that occur at a constant rate within a given interval of time.

Mathematics: CUET Mock Test - 3 - Question 8

What is the formula for the Poisson distribution probability?

A.
P(x; μ) = (e^-μ) (μ^x) / x!
B.
P(x; μ) = (e^-x) (μ^x) / x!
C.
P(x; μ) = (e^-μ) (μ) / x!
D.
P(x; μ) = (e^-μ) (μ^x) / x

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 8

Poisson distribution shows the number of times an event is likely to occur within a specified time. The Poisson distribution probability formula is P(x; μ) = (e^-μ) (μ^x) / x!

Mathematics: CUET Mock Test - 3 - Question 9

If A is any square matrix, then

A.
A+A^t is symmetric
B.
A−A^t is symmetric
C.
A+A^t is skew-symmetric
D.
none of these.

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 9

For any square matrix A, both (A + A^T) is symmetric and (A - A^T) is skew-symmetric. Therefore, options A and B are true.

(A + A^T) is symmetric.
(A - A^T) is skew-symmetric.

Mathematics: CUET Mock Test - 3 - Question 10

The number of arbitrary constants in a particular solution of a fourth order differential equation is ______

A.
1
B.
0
C.
4
D.
3

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

The number of arbitrary constants for a particular solution of n^th order differential equation is always zero.

Mathematics: CUET Mock Test - 3 - Question 11

How many arbitrary constants will be there in the general solution of a second order differential equation?

A.
3
B.
4
C.
2
D.
1

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 11

The number of arbitrary constants in a general solution of a nth order differential equation is n.
Therefore, the number of arbitrary constants in the general solution of a second order D.E is 2.

Mathematics: CUET Mock Test - 3 - Question 12

If the normal to the ellipse x² + 3y² = 12 at the point be inclined at 60° to the major axis, then at what angle does the line joining the curve to the point is inclined to the same axis?

A.
90°
B.
45°
C.
60°
D.
30°

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 12

Given, x² + 3y² = 12 Or x²/12 + y²/4 = 1
Differentiating both sides of (1) with respect to y we get,
2x*(dx/dy) + 3*2y = 0
Or dx/dy = -3y/x
Suppose the normal to the ellipse (1) at the point P(√12cosθ, 2sinθ) makes an angle 60° with the major axis. Then, the slope of the normal at P is tan60°
Or -[dx/dy]_P = tan60°
Or -(-(3*2sinθ)/√12cosθ) = √3
Or √3tanθ = √3
Or tanθ = 1
Now the centre of the ellipse (1) is C(0, 0)
Therefore, the slope of the line CP is,
(2sinθ – 0)/(√12cosθ – 0) = (1/√3)tanθ = 1/√3 [as, tanθ = 1]
Therefore, the line CP is inclined at 30° to the major axis.

Mathematics: CUET Mock Test - 3 - Question 13

What will be the value of angle between the curves x² - y² = 2a² and x² + y² = 4a²?

A.
π/2
B.
π/4
C.
π/6
D.
π/3

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 13

x² – y² = 2a² ……….(1) and x² + y² = 4a² ……….(2)
Adding (1) and (2) we get, 2x² = 6a²
Again, (2) – (1) gives,
2y² = 2a²
Therefore, 2x² * 2y² = 6a² * 2a²
4x²y² = 12a²
Or x²y² = 3a⁴
Or 2xy = ±2√3
Differentiating both side of (1) and (2) with respect to x we get,
2x – 2y(dy/dx) = 0
Or dy/dx = x/y
And 2x + 2y(dy/dx) = 0
Or dy/dx = -x/y
Let (x, y) be the point of intersection of the curves(1) and (2) and m₁ and m₂ be the slopes of the tangents to the curves (1) and (2) respectively at the point (x, y); then,
m₁ = x/y and m₂ = -x/y
Now the angle between the curves (1) and (2) means the angle between the tangents to the curve at their point of intersection.
Therefore, if θ is the required angle between the curves (1) and (2), then
tanθ = |(m₁ – m₂)/(1 + m₁m₂)|
Putting the value of m₁, m₂ in the above equation we get,
tanθ = |2xy/(y² – x²)|
As, 2xy = ±2√3a² and x² – y² = 2a²
tanθ = |±2√3a²/-2a²|
Or tanθ = √3
Thus, θ = π/3.

Mathematics: CUET Mock Test - 3 - Question 14

The term Bernoulli trials is termed after which swiss mathematician?

A.
Jacob Bernoulli
B.
Albert Einstein
C.
Johann Gutenberg
D.
Archimedes

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 14

Bernoulli trials is termed after swiss mathematician Jacob Bernoulli. Bernoulli trials is also called a Dichotomous experiment and is repeated n times. If in each trial the probability of success is constant, then such trials are called Bernoulli trials.

Mathematics: CUET Mock Test - 3 - Question 15

Bag 1 contains 4 white and 6 black balls while another Bag 2 contains 4 white and 3 black balls. One ball is drawn at random from one of the bags and it is found to be black. Find the probability that it was drawn from Bag 1.

A.
12/13
B.
5/12
C.
7/11
D.
7/12

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 15

Let E1 = event of choosing the bag 1, E2 = event of choosing the bag 2.
Let A be event of drawing a black ball.
P(E1) = P(E2) = 1/2.
Also, P(A|E1) = P(drawing a black ball from Bag 1) = 6/10 = 3/5.
P(A|E2) = P(drawing a black ball from Bag 2) = 3/7.
By using Bayes’ theorem, the probability of drawing a black ball from bag 1 out of two bags is-:
P(E1 | A) = P(E1)P(A | E1)/( P(E1)P(A│E1)+P(E2)P(A | E2))
= (1/2 × 3/5) / ((1/2 × 3/7)) + (1/2 × 3/5)) = 7/12.

Mathematics: CUET Mock Test - 3 - Question 16

Find the angle between 2x + 3y – 2z + 4 = 0 and (2, 1, 1).

A.
38.2
B.
19.64
C.
89.21
D.
29.34

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 16

Angle between a plane and a line sin θ =
sinθ = 0.49
θ = sin-1(0.49)
θ = 29.34

Mathematics: CUET Mock Test - 3 - Question 17

Match List-I with List-II:

Choose the correct answer from the options given below:

A.
(A) - (I), (B) - (IV), (C) - (III), (D) - (II)
B.
(A) - (II), (B) - (I), (C) - (IV), (D) - (III)
C.
(A) - (IV), (B) - (II), (C) - (III), (D) - (I)
D.
(A) - (I), (B) - (II), (C) - (III), (D) - (IV)

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 17

(A) Area of the curve y = x²: The area under the curve y = x² from 0 to 1 is 1/3 (which is (I)).
(B) Area of the curve y = x³: The area under the curve y = x³ from 0 to 1 is 1/4 (which is (II)).
(C) Area of the curve y = 1/x: The area under the curve y = 1/x from 1 to ∞ is 1 (which is (III)).
(D) Area of the curve y = x: The area under the curve y = x from 0 to 1 is 1/2 (which is (IV)).

Mathematics: CUET Mock Test - 3 - Question 18

Match List-I with List-II:

Choose the correct answer from the options given below:

A.
(A) - (III), (B) - (II), (C) - (I), (D) - (IV)
B.
(A) - (I), (B) - (IV), (C) - (III), (D) - (II)
C.
(A) - (II), (B) - (I), (C) - (IV), (D) - (III)
D.
(A) - (IV), (B) - (III), (C) - (II), (D) - (I)

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 18

(A) Scalar (Dot) Product → (III) A scalar quantity obtained by multiplying two vectors and the cosine of the angle between them.
(B) Projection of a Vector → (II) The component of one vector along the direction of another.
(C) Vector (Cross) Product → (I) A vector quantity obtained by multiplying two vectors and the sine of the angle between them.
(D) Scalar Triple Product → (IV) A determinant-based operation used to find the volume of a parallelepiped formed by three vectors.

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

Mathematics: CUET Mock Test - 3 - Question 19

Maximize Z = 3x + 4y subject to 2x + y ≤ 10, x + 2y ≤ 10, x ≥ 2, y ≥ 2. At which point is the optimal solution achieved?

A.
(2,2)
B.
(3,4)
C.
(10/3, 10/3)
D.
(5/4,2)

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 19

Step 1: Identify the feasible region

The inequalities define a polygon in the xy-plane bounded by:

2x + y ≤ 10
x + 2y ≤ 10
x ≥ 2
y ≥ 2

Step 2: Find the corner points (vertices) of the feasible region

The optimal solution of a linear programming problem lies at one of the vertices of the feasible region.
We will find the intersection points of the constraints to get the vertices.

Vertex A: Intersection of x = 2 and y = 2

Point: (2, 2)

Check constraints:

2(2) + 2 = 4 + 2 = 6 ≤ 10 ✓
2 + 2(2) = 2 + 4 = 6 ≤ 10 ✓

Vertex B: Intersection of x = 2 and 2x + y = 10

Put x = 2 in 2x + y = 10:
2(2) + y = 10 → 4 + y = 10 → y = 6

So, point (2, 6).

Check x + 2y ≤ 10:
2 + 2(6) = 2 + 12 = 14 which is NOT ≤ 10

So, (2, 6) is not feasible.

Vertex C: Intersection of y = 2 and 2x + y = 10

Put y = 2 in 2x + y = 10:
2x + 2 = 10 → 2x = 8 → x = 4

Point (4, 2).

Check x + 2y ≤ 10:
4 + 2(2) = 4 + 4 = 8 ≤ 10

Also, x ≥ 2 and y ≥ 2 are satisfied.

So (4, 2) is feasible.

Vertex D: Intersection of x = 2 and x + 2y = 10

Put x = 2 in x + 2y = 10:
2 + 2y = 10 → 2y = 8 → y = 4

Point (2, 4).

Check 2x + y ≤ 10:
2(2) + 4 = 4 + 4 = 8 ≤ 10

Feasible.

Vertex E: Intersection of y = 2 and x + 2y = 10

Already found as point (4, 2), same as C.

Vertex F: Intersection of 2x + y = 10 and x + 2y = 10

Solve:
2x + y = 10
x + 2y = 10

From the first equation:
y = 10 - 2x

Substitute in second:
x + 2(10 - 2x) = 10
x + 20 - 4x = 10
-3x = -10
x = 10/3 ≈ 3.33

Then
y = 10 - 2*(10/3) = 10 - 20/3 = (30 - 20)/3 = 10/3 ≈ 3.33

Check x ≥ 2, y ≥ 2: both satisfied.

So point (10/3, 10/3) ≈ (3.33, 3.33) is feasible.

Step 3: Evaluate objective function Z = 3x + 4y at all feasible vertices

Point	Z = 3x + 4y
A (2, 2)	32 + 42 = 6 + 8 = 14
C (4, 2)	34 + 42 = 12 + 8 = 20
D (2, 4)	32 + 44 = 6 + 16 = 22
F (10/3, 10/3)	3(10/3) + 4(10/3) = 10 + 40/3 = 70/3 ≈ 23.33

Step 4: Conclusion

The maximum value of Z is approximately 23.33 at the point (10/3, 10/3) ≈ (3.33, 3.33).

Mathematics: CUET Mock Test - 3 - Question 20

The void relation (a subset of A x A) on a non empty set A is:

A.
Anti symmetric
B.
Transitive
C.
Reflexive
D.
Transitive and symmetric

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 20

The relation { } ⊂ A x A on a is surely not reflexive. However, neither symmetry nor transitivity is contradicted. So { } is a transitive and symmetry relation on A.

Mathematics: CUET Mock Test - 3 - Question 21

The domain of the function f = {(1, 3), (3, 5), (2, 6)} is

A.
{1, 3, 2}
B.
1, 3 and 2
C.
3, 5 and 6
D.
{3, 5, 6}

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 21

The domain in ordered pair (x,y) is represented by x-coordinate. Therefore, the domain of the given function is given by : {1, 3, 2}.

Mathematics: CUET Mock Test - 3 - Question 22

Let R be the relation on N defined as x R y if x + 2 y = 8. The domain of R is

A.
{2, 4, 8}
B.
{1, 2, 3, 4}
C.
{2, 4, 6, 8}
D.
{2, 4, 6}

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 22

As x R y if x + 2y = 8, therefore, domain of the relation R is given by x = 8 – 2y ∈ N.
When y = 1,
⇒ x = 6 ,when y = 2,
⇒ x = 4, when y = 3,
⇒ x = 2.
therefore domain is {2, 4, 6}.

Mathematics: CUET Mock Test - 3 - Question 23

The principal value of
is.

A.
2π/5
B.
-2π/5
C.
3π/5
D.
-3π/5

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 23

tan^-1 (tan 3π/5)
This can be written as:
tan^-1 (tan 3π/5) = tan^-1 (tan[π – 2π/5])
= tan^-1 (- tan 2π/5) {since tan(π – x) = -tan x}
= –tan^-1 (tan 2π/2)
= –2π/5

Mathematics: CUET Mock Test - 3 - Question 24

Find the value of tan^-1⁡(1/3) + tan^-1⁡(1/5) + tan^-1⁡(1/7).

A.
tan^-1⁡(4/7)
B.
tan^-1⁡(9/7)
C.
tan^-1⁡(7/9)
D.
tan^-1⁡1

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 24

Mathematics: CUET Mock Test - 3 - Question 25

If P₁(x₁, y₁, z₁) and P₂(x₂, y₂, z₂) are any two points, then the vector joining P1 and P2is the vector P1P2. Magnitude of the vector

A.
B.
C.
D.

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 25

If P₁(x₁, y₁, z₁) and P₂(x₂, y₂, z₂) are any two points, then the vector joining P1 and P2is the vector P1P2, then ;

Mathematics: CUET Mock Test - 3 - Question 26

Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (– 5, 7).

A.
B.
C.
D.

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 26

The scalar and vector components of the vector with initial point (2, 1) and terminal point (– 5, 7) is given by : (- 5 – 2) i.e. – 7 and (7 – 1) i.e. 6. Therefore, the scalar components are – 7 and 6 .,and vector components are

Mathematics: CUET Mock Test - 3 - Question 27

Find the shortest distance between the lines and

A.
B.
C.
D.

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 27

Find the shortest distance between the lines

On comparing them with :

we get :

Mathematics: CUET Mock Test - 3 - Question 28

Determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0

A.
B.
C.
D.

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 28

Mathematics: CUET Mock Test - 3 - Question 29

Matrix A when multiplied with Matrix C gives the Identity matrix I, what is C?

A.
Identity matrix
B.
Inverse of A
C.
Square of A
D.
Transpose of A

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 29

Any square matrix when multiplied with its inverse gives the identity matrix. Note that non square matrices are not invertible.

*Answer can only contain numeric values

Mathematics: CUET Mock Test - 3 - Question 30

Let for any three distinct consecutive terms a, b, c of an A.P, the lines ax + by + c = 0 be concurrent at the point P and Q(α, β) be a point such that the system of equations
x + y + z = 6,
2x + 5y + αz = β and
x + 2y + 3z = 4,
has infinitely many solutions. Then (PQ)² is equal to ____.

Detailed Solution for Mathematics: CUET Mock Test - 3 - Question 30

∵ a, b, c and in A.P
⇒ 2b = a + c ⇒ a - 2b + c = 0
∴ ax + by + c passes through fixed point (1, -2)
∴ P = (1, -2)
For infinite solution,
D = D₁ = D₂ = D₃ = 0

⇒ α = 8

∴ Q = (8, 6)
∴ Q² = 113

Mathematics: CUET Mock Test - 3 - CUET MCQ

30 Questions MCQ Test - Mathematics: CUET Mock Test - 3

Which of the following functions is not a solution for the differential equation y” + 9y = 0?

Which of the following differential equations given below has the solution y = log⁡x?

Which of the following functions is a solution for the differential equation dy/dx -14x = 0?

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