COMEDK UGET 2025 Calculus Practice Questions with Solutions

COMEDK UGET 2025 Calculus Practice Questions with Solutions: If you are planning to appear in COMEDK entrance exam 2025, you must focus on Calculus to score well in the Mathematics section. COMEDK UGET 2025 exam features a significant number of questions from Calculus. Therefore, you should solve COMEDK UGET 2025 Calculus practice questions with solutions to have comprehensive knowledge and excellent preparation. Calculus covers a weightage of around 25% and you can expect as many as 15 questions from this chapter. Limits and Derivatives are the most important topic in Calculus centred around which the maximum number of questions are asked. In this article, we have provided a detailed Calculus syllabus weightage and some of the most important and expected COMEDK UGET 2025 Calculus practice questions with solutions.

Also Check - Do or Die Chapters for COMEDK UGET 2025 Mathematics

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COMEDK UGET 2025 Calculus Important Topics

The COMEDK UGET 2025 Mathematics syllabus features all the important topics based on which COMEDK question paper 2025 will be prepared. In the following table, we have provided the detailed Calculus topics for the entrance exam.

COMEDK UGET 2025 Calculus Expected Weightage

Calculus weightage will help you know the expected number of questions that will be asked in COMEDK UGET 2025 question paper. As per the trends from the past few years, the expected weightage of Calculus in COMEDK UGET 2025 is around 25%. You can check the detailed COMEDK UGET 2025 Calculus weightage below:

COMEDK UGET 2025 Calculus Practice Questions with Solutions

COMEDK Calculus practice questions will help you gain a better understanding of concepts and fundamentals. In the following section, we have listed down some of the most important COMEDK UGET 2025 Calculus practice questions with solutions:

Q1. Which of the following function is injective?

a. f(x) = |x +2|, x ∊ [ –2, ∞)

b. f(x) = x ² + 2, x ∊ (- ∞, ∞)

c. f(x) = 4x ² + 3x - 5, x ∊ (- ∞, ∞)

d. f(x) = (x - 4) (x - 5), x ∊ (- ∞, ∞)

Ans. a. f(x) = |x +2|, x ∊ [ –2, ∞)

Solution: Let's examine each option to check for injectivity. A function is injective (one-to-one) if different inputs always produce different outputs. In other words, for any two numbers a and b in the domain, if f(a) = f(b) then a = b.

Below is a breakdown of each option:

Option a: f(x) = |x +2|, x ∊ [ –2, ∞)

For x ≥ -2, the expression x + 2 is non-negative, so the absolute value function simplifies:

f(x) |x + 2|  = x + 2

The function now is a linear function, f(x) = x + 2, which is strictly increasing over the domain [-2, ∞)

Because an increasing linear function never repeats the same output for different inputs, this function is injective. Thus option a is correct.

Q2. equals to:

a. -3/2

b. 1/2a ^3/2

c. 1/2

d. 2a ^-3/2

Ans. b. 1/2a ^3/2

Solution: The given limit is in the form of 0/0 as x →0. . To solve this, we will use the rationalization method.

Multiplying both numerator and denominator by the conjugate of the numerator, we get:

Now, we can directly substitute x = 0 into the expression, to get:

1/√[a(a+0)(√a+0 + √a)]

Simplifying the equation,

1/2a ^3/2

Q3. If Sin y = x(cos(a+y)), then find dy/dx when x = 0:

a. 1

b. Sec a

c. Cos a

d. -1

Ans. c. Cos a

Solutions: If Sin y = x(cos(a+y)), then find dy/dx when x = 0

Differentiating both sides of the equation with respect to x, we get

Cos ydy/dx = cos (a+y) + x(-sin(a+y)dy/dx

Cos y dy/dx + x sin (a+y)dy/dx = cos(a+y)

dy/dx(cos y + x sin(a+y)) = cos(a+y)

dy/dx = cos(a+y)/[cos y+xsin(a+y)]

When x = 0, we have

dy/dx = cos(a+y)/cos y

Since Sin y = xcos (a+y), when x = 0, we have sin y = 0. This implies y = 0 (since we are looking for the derivative at x =0)

Therefore, x = 0.

dy/dx = cos(a+0)/cos0 = Cos a.

Q4. The rate of change of the volume of a sphere with respect to its surface area S is?

a. 1/2√S/π

b. √S/π

c. 2/3√S/π

d. 1/4√S/π

Ans. d. 1/4√S/π

Solution: To find the rate of change of the volume of a sphere with respect to its surface area, we first need to express both the volume and the surface area in terms of the radius of the sphere.

The volume V of a sphere is given by the formula:

V = 4/3 πr ³

The surface area S of a sphere is given by the formula:

S = 4πr ²

We need to find the rate of change of V with respect to S, which is expressed as dV/dS. To do this, we use the chain rule:

dV/dS = dV/dr.dr/dS

First, we find dV/dr,

dV/dr = d/dr(4/3πr ³ ) = 4πr ²

Next, we find dS/dr,

dS/dr = d/dr(4πr ² ) = 8πr

Now, we need to find dr/dS. Since dS/dr = 8πr

we can write: dr/dS = 1/8πr

Finally, we substitute, dV/dr and dr/dS back into the chain rule expression:

dV/dS = (4πr ² ).(1/8πr) = r/2

We know from the surface area formula that S = 4πr ² . Solving for r in terms of S, we get:

r ² = S/4π = √S/4π

Substituting this back into dV/dS, we get:

dV/dS = 2/s√S/4π = 1/2.1/2√S/π = 1/4√S/π

Q5. The area enclosed by the pair of lines x = 0, the line x - 4 = 0 and y + 5 = 0 is

a. 10 square unit

b. 20 square unit

c. 50 square unit

d. 5/4 square unit

Ans. b. 20 square unit

Solution: We have,

xy = 0, x - 4 = 0, y + 5 = 0

x = 0, y = 0, x = 4, y = -5

So, area enclosed = Area of rectangle of OABC

AB X BC

5 X 4 = 20 square unit.

You should also revise Calculus equations and formulas from COMEDK 2025 Mathematics formula sheet to tackle the questions easily and promptly, thus saving valable time in the exam hall.

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