Mathematics: CUET Mock Test - 3 - Commerce MCQ

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

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

∴ 
= 

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

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.

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

f(x) = x -[x] is derivable at all x ∈ R – I , and f ‘(x) = 1 for all x ∈ R – I .

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 

Let, y = sin(x + α)/sin(x + β)
Then,
dy/dx = [cos(x + α)sin(x + β) – sin(x + α)cos(x + β)]/sin²(x + β)
= sin(x+β – x-α)/sin²(x + β)
Or sin(β – α)/sin²(x + β)
So, for minimum or maximum value of x we have,
dy/dx = 0
Or sin(β – α)/sin²(x + β) = 0
Or sin(β – α) = 0 ……….(1)
Clearly, equation (1) is independent of x; hence, we cannot have a real value of x as root of equation (1).
Therefore, y has neither a maximum or minimum value.

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.

Given, y² = 3x ……….(1) and y = 2x + 4 ……….(2)
Differentiating both sides of (1) with respect to y we get,
2y = 3(dx/dy)
Or dx/dy = 2y/3
Let P (x₁, y₁) be any point on the parabola (1). Then the slope of the normal to the parabola (1) at point P is
-[dx/dy]_P = -2y₁/3
If the normal at the point P to the parabola (1) be perpendicular to the line (2) then we must have,
-2y₁/3*2 = -1
Since the slope of the line (2) is 2
Or y₁ = 3/4
Since the point P(x₁, y₁) lies on (1) hence,
y₁² = 3x₁
As, y₁ = 3/4, so, x₁ = 3/16
Therefore, the required equation of the normal is
y – y₁ = -(2y₁)/3*(x – x₁)
Putting the value of x₁ and y₁ in the above equation we get,
16x + 32y = 27
And the coordinates of the foot of the normal are (x₁, y₁) = (3/16, 3/4)

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.

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.

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.

Bernoulli trial has only two possible outcomes and is mutually exclusive. Those two outcomes are ‘success’ and ‘failure’. So, it is also called as a ‘yes’ or ‘no’ question.

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.

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!

Bayes theorem is the method in which the calculated probabilities are revised with values of new probabilities, whereas Updation theorem, Revision theorem and Dependent theorem are not related to the concept of probability.

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.

Concept:

Marginal revenue is the rate of change total revenue with respect to the number of items sold at an instant.

MR =

Calculation:

Given, Total revenue R(x) = 4x2 + 10x + 3

∴ Marginal revenue, MR =

=

= 8x + 10

⇒ MR(at x = 20) = 8(20) + 10 = 160 + 10 = 170

∴ The marginal revenue on selling 20 such units is Rs. 170.

Concept:

1) The critical point of a function is the point where its first derivative is 0.

2) A function has minima if its second derivative at a critical point is greater than 0.

Calculation:

The given equation of the curve is x2 – 8x + 17.

Let f(x) = x2 – 8x + 17

∴ f'(x) = 2x - 8

For critical point, put f'(x) = 0

⇒ 2x - 8 = 0

⇒ x = 4

f''(x) = 2 > 0 hence f(x) has minima at x = 4.

∴ Minimum value of f(x) = f(4)

f(4) = 42 - 8 × 4 + 17 = 1

∴ The minimum value of x2 – 8x + 17 is 1.

The correct answer is option 3.

μ = = =
Hence, 9μ + 4 =
Therefore, the required value is 10.

Random Variable: A random variable is a variable whose value is determined by the outcome of a random process or experiment. In other words, it is a mathematical function that assigns a numerical value to each possible outcome of a random experiment.
The mean of a random variable is a measure of its central tendency and represents the average value of the variable over all possible outcomes. Mathematically, the mean is defined as the sum of the products of each possible value of the variable and its corresponding probability, divided by the total number of possible outcomes. The mean is also sometimes referred to as the expected value of the random variable.

The correct answer is Rs. 1
Key Points

Concept:

The Squeeze Theorem (The Sandwich Theorem): is used on a function where it will be almost impossible to differentiate.

• The squeeze theorem states that if we define functions such that h(x) ≤ f(x) ≤ g(x) and if , then .

Calculation:

We know that -1 ≤ sin θ ≤ 1.

⇒ -1 ≤ ≤ 1

Since, e^x is a strictly increasing function for all real values of x, we can say that:

⇒ e^-1 ≤ ≤ e¹

Also, since x² ≥ 0, we can say that:

⇒ x²e-1 ≤ ≤ x²e1

⇒

So, we can consider h(x) = , f(x) = and g(x) = x²e.

Now, .

And .

Since, , we must have .

Hence, .

Concept:

exists if

f(x) is Continuous at x = a ⇔

Calculation:

1. This statement is false, as the limit at the given point should be equal to the existence of the limit.

2. If f(x) is a continuous function at a point, then it is not necessary that the function 1/ f(x) will be continuous.

Take, f (x) = x, which is continuous at a point,

For 1/f (x) = 1/x, which will not be continuous at the same point x = 0

So, this statement is not true.

Hence, option (4) is correct.

Problem:

We are given a piecewise function and need to find the value of k that makes the function f(x) continuous at x = 0.

Function Definition:

• For x < 0:
  f(x) = kx / |x|

• For x ≥ 0:
  f(x) = 3/2

Step 1: Understand the condition for continuity at x = 0

For a function to be continuous at x = 0, three conditions must be met:

1. f(0) must be defined.

2. The left-hand limit (as x approaches 0 from the left) must exist.

3. The right-hand limit (as x approaches 0 from the right) must exist.

4. Left-hand limit = Right-hand limit = f(0)

Step 2: Evaluate f(0)

Since f(x) = 3/2 for x ≥ 0, and 0 falls in this range, we have:

f(0) = 3/2

Step 3: Find the Left-Hand Limit (LHL)

For x < 0, |x| = –x (because x is negative in this region), so:

f(x) = kx / |x| = kx / (–x) = –k

So, as x approaches 0 from the left:

LHL = –k

Step 4: Find the Right-Hand Limit (RHL)

For x > 0, f(x) = 3/2. So, as x approaches 0 from the right:

RHL = 3/2

Step 5: Apply continuity condition

Set the LHL and RHL equal to each other and equal to f(0):

–k = 3/2

Solving for k:

k = –3/2

Final Answer:

The value of k that makes the function continuous at x = 0 is:

–3/2

Concept:

• A function f(x) is said to be continuous at a point x = a, in its domain if exists or its graph is a single unbroken curve.
• f(x) is Continuous at x = a ⇔

Calculation:

Given:

Let’s check continuity at x = 0

∴ It is not continuous at x = 0