Mathematics: CUET Mock Test - 1 - CUET MCQ

Probability of 3 students,
P(A) = 1/2, P(Ā) = 1/2

P(B) = 1/3, P(B̄) = 2/3

P(C) = 1/4, P(C̄) = 3/4

So, probability that no one solves the question is = 1/2 × 2/3 × 3/4 = 1/4
⇒ P(None) = 1/4

Then, the probability to solve the question is = 1 – 1/4 = 3/4

Let H₁H₂H₃H₄ B be the horses in which B is the winning horse.
Required probability

Any person can apply for any house.
Required probability

To calculate the degree or the order of a differential equation, the powers of derivatives should be an integer.
On squaring both sides, we get a differential equation with the integral power of derivatives.

⇒ Order (the highest derivative) = 2
⇒ Degree (the power of highest degree) = 2

Answer: D
Explanation: Given that, y=9 log⁡t³

∫(-3 to 3) (x+1)dx

=  ∫(-3 to -1) (x+1)dx +  ∫(-1 to 3) (x+1) dx 

= [x² + x](-3 to -1) + [x² + x](-1 to 3)

= [½ - 1 - (9/2 - 3)] + [9/2 + 3 - (½ - 1)]

= -[-4 + 2] + [4 + 4]

= -[-2] + [8]

= 10

We want to find the indefinite integral of |x| with respect to x:

∫ |x| dx

1. Consider x ≥ 0:

   • Here, |x| = x.
   • So, ∫ |x| dx = ∫ x dx = (x²)/2 + C₁

2. Consider x < 0:

   • Here, |x| = -x.
   • So, ∫ |x| dx = ∫ (-x) dx = - (x²)/2 + C₂

To combine these results into a single expression, we note that for an indefinite integral, constants C₁ and C₂ can be absorbed into a single arbitrary constant C. A concise way to write this is:

∫ |x| dx = (x · |x|)/2 + C

• For x ≥ 0, this becomes x·x/2 = x²/2.
• For x < 0, this becomes x·(-x)/2 = -x²/2.
• Combining both the results , we get x|x|/2 + C

Objective function is Z = - x + 2 y ……………………(1).
The given constraints are : x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.

Corner points Z =  - x + 2y

Here , the open half plane has points in common with the feasible region .
Therefore , Z has no maximum value.

A vector has both magnitude as well as direction.

Distributive law is given by : 

Answer: B

Solution:

2^3x3 = 2⁹ = 512.

The number of elements in a 3 X 3 matrix is the product 3 X 3=9.

Each element can either be a 0 or a 1.

Given this, the total possible matrices that can be selected is 2⁹=512

Answer: c) 2.67
Explanation: First, find k: ΣP(X) = 1 → k(1² + 2² + 3²) = k(1 + 4 + 9) = 14k = 1 → k = 1/14.
P(X = 1) = 1/14, P(X = 2) = 4/14, P(X = 3) = 9/14.
E(X) = Σ[x * P(X)] = 1 * (1/14) + 2 * (4/14) + 3 * (9/14) = 1/14 + 8/14 + 27/14 = 36/14 ≈ 2.6

Answer: b) 0.134
Explanation: P(X ≥ 5) = 1 - P(X ≤ 4). Use binomial cumulative probability:
P(X ≤ 4) = Σ[C(15, k) * (0.2)^k * (0.8)^(15-k)] for k = 0 to 4 ≈ 0.866.
P(X ≥ 5) = 1 - 0.866 = 0.134.

Let R be the feasible region (convex polygon) for a linear programming problem and let Z = ax + by be the objective function. When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities,then , optimal value must occur at a corner point (vertex) of the feasible region.

Let number of packages of screws A produced = x 
And number of packages of screws B produced = y 
Therefore , the above L.P.P. is given as : 
Maximise , Z = 7x +10y , subject to the constraints : 4x +6y ≤ 240 and. 6x +3y ≤ 240 i.e. 2x +3y ≤ 120 and 2x +y ≤ 80 , x, y ≥ 0.

i.e 30 packages of screws A and 20 packages of screws B; Maximum profit = Rs 410.

Here , maximize Z = 11x + 7y , subject to the constraints :2x + y ≤ 6, x ≤ 2, x ≥ 0, y ≥ 0.

y = Ae^2x + Be^-2x 
dy/dx = 2Ae^2x – 2Be^-2x
d²y/dx² = 4Ae^2x + 4Be^-2x
= 4*y
d²y/dx² – 4y = 0

Given equation is : 

(dy/dx)² + 1/(dy/dx) = 1

((dy/dx)³ + 1)/(dy/dx) = 1

(dy/dx)³ +1 = dy/dx

So, final equation is

(dy/dx)³ - dy/dx + 1 = 0

So, degree = 3

Clearly the range is (−∞,−5)∪[0,1]∪(5,∞)