JEE Main Preparation Resources for Competitive Exams

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JEE Main MCQ & Objective Questions

The JEE Main exam is a crucial step for students aspiring to enter prestigious engineering colleges in India. It tests not only knowledge but also the ability to apply concepts effectively. Practicing MCQs and objective questions is essential for scoring better, as it helps in familiarizing students with the exam pattern and enhances their problem-solving skills. Engaging with practice questions allows students to identify important questions and strengthen their exam preparation.

What You Will Practise Here

Fundamental concepts of Physics, Chemistry, and Mathematics
Key formulas and their applications in problem-solving
Important definitions and theories relevant to JEE Main
Diagrams and graphical representations for better understanding
Numerical problems and their step-by-step solutions
Previous years' JEE Main questions for real exam experience
Time management strategies while solving MCQs

Exam Relevance

The topics covered in JEE Main are not only significant for the JEE exam but also appear in various CBSE and State Board examinations. Many concepts are shared with the NEET syllabus, making them relevant across multiple competitive exams. Common question patterns include conceptual applications, numerical problems, and theoretical questions that assess a student's understanding of core subjects.

Common Mistakes Students Make

Misinterpreting the question stem, leading to incorrect answers
Neglecting units in numerical problems, which can change the outcome
Overlooking negative marking and not managing time effectively
Relying too heavily on rote memorization instead of understanding concepts
Failing to review and analyze mistakes from practice tests

FAQs

Question: How can I improve my speed in solving JEE Main MCQ questions?
Answer: Regular practice with timed quizzes and focusing on shortcuts can significantly enhance your speed.

Question: Are the JEE Main objective questions similar to previous years' papers?
Answer: Yes, many questions are based on previous years' patterns, so practicing them can be beneficial.

Question: What is the best way to approach JEE Main practice questions?
Answer: Start with understanding the concepts, then attempt practice questions, and finally review your answers to learn from mistakes.

Now is the time to take charge of your preparation! Dive into solving JEE Main MCQs and practice questions to test your understanding and boost your confidence for the exam.

Chemistry Syllabus (JEE Main) Mathematics Syllabus (JEE Main) Physics Syllabus (JEE Main)

Q. For the function f(x) = 2x^3 - 9x^2 + 12x, find the inflection point.

A. (1, 1)
B. (2, 2)
C. (3, 3)
D. (4, 4)

Solution

f''(x) = 12x - 18. Setting f''(x) = 0 gives x = 1.5. The inflection point is (1.5, f(1.5)).

Correct Answer: B — (2, 2)

Q. For the function f(x) = 2x^3 - 9x^2 + 12x, find the intervals where the function is increasing.

A. (-∞, 1)
B. (1, 3)
C. (3, ∞)
D. (0, 3)

Solution

f'(x) = 6x^2 - 18x + 12. Setting f'(x) = 0 gives x = 1 and x = 3. Testing intervals shows f is increasing on (1, 3).

Correct Answer: B — (1, 3)

Q. For the function f(x) = 2x^3 - 9x^2 + 12x, find the local maxima.

A. (1, 5)
B. (2, 0)
C. (3, 0)
D. (0, 0)

Solution

f'(x) = 6x^2 - 18x + 12. Setting f'(x) = 0 gives x = 1 and x = 2. f(1) = 5 is a local maximum.

Correct Answer: A — (1, 5)

Q. For the function f(x) = 3x^2 - 12x + 7, find the coordinates of the vertex.

A. (2, -5)
B. (2, -1)
C. (3, -2)
D. (1, 1)

Solution

The vertex is at x = -b/(2a) = 12/(2*3) = 2. f(2) = 3(2^2) - 12(2) + 7 = -1. So, the vertex is (2, -1).

Correct Answer: B — (2, -1)

Q. For the function f(x) = 3x^3 - 12x^2 + 9, find the x-coordinates of the inflection points.

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

Solution

f''(x) = 18x - 24. Setting f''(x) = 0 gives x = 4/3. This is the inflection point.

Correct Answer: B — 2

Q. For the function f(x) = 3x^3 - 12x^2 + 9x, the number of local maxima and minima is:

A. 0
B. 1
C. 2
D. 3

Solution

Finding f'(x) = 9x^2 - 24x + 9 and solving gives two critical points. The second derivative test confirms one maximum and one minimum.

Correct Answer: C — 2

Q. For the function f(x) = e^x - x^2, the point of inflection occurs at:

A. x = 0
B. x = 1
C. x = 2
D. x = -1

Solution

To find the point of inflection, we compute f''(x) = e^x - 2. Setting f''(x) = 0 gives e^x = 2, leading to x = ln(2). The closest integer is x = 1.

Correct Answer: B — x = 1

Q. For the function f(x) = ln(x), find the point where it is not differentiable.

A. x = 0
B. x = 1
C. x = -1
D. x = 2

Solution

f(x) = ln(x) is not defined for x ≤ 0, hence not differentiable at x = 0.

Correct Answer: A — x = 0

Q. For the function f(x) = sin(x) + cos(x), find the x-coordinate of the maximum point in the interval [0, 2π].

A. π/4
B. 3π/4
C. 5π/4
D. 7π/4

Solution

f'(x) = cos(x) - sin(x). Setting f'(x) = 0 gives tan(x) = 1, so x = π/4 + nπ. In [0, 2π], the maximum occurs at x = 3π/4.

Correct Answer: B — 3π/4

Q. For the function f(x) = x^2 + 2x + 1, what is f'(x)?

A. 2x + 1
B. 2x + 2
C. 2x
D. x + 1

Solution

f'(x) = 2x + 2.

Correct Answer: B — 2x + 2

Q. For the function f(x) = x^2 + 2x + 3, find the point where it is not differentiable.

A. x = -1
B. x = 0
C. x = 1
D. It is differentiable everywhere

Solution

The function is a polynomial and is differentiable everywhere.

Correct Answer: D — It is differentiable everywhere

Q. For the function f(x) = x^2 + kx + 1 to be differentiable at x = -1, what must k be?

A. -1
B. 0
C. 1
D. 2

Solution

Setting the derivative f'(-1) = 0 gives k = 1 for differentiability.

Correct Answer: C — 1

Q. For the function f(x) = x^2 - 2x + 1, find the slope of the tangent line at x = 1.

A. 0
B. 1
C. 2
D. 3

Solution

f'(x) = 2x - 2. Thus, f'(1) = 2(1) - 2 = 0.

Correct Answer: A — 0

Q. For the function f(x) = x^2 - 4x + 4, find the point where it is not differentiable.

A. x = 0
B. x = 2
C. x = 4
D. It is differentiable everywhere

Solution

As a polynomial, f(x) is differentiable everywhere, including at x = 2.

Correct Answer: D — It is differentiable everywhere

Q. For the function f(x) = x^2 - 4x + 5, find the minimum value.

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

Solution

The vertex occurs at x = 2. f(2) = 2^2 - 4*2 + 5 = 1, which is the minimum value.

Correct Answer: B — 2

Q. For the function f(x) = x^2 - 4x + 5, find the vertex.

A. (2, 1)
B. (2, 5)
C. (4, 1)
D. (4, 5)

Solution

The vertex is at x = -b/(2a) = 4/2 = 2. f(2) = 2^2 - 4(2) + 5 = 1, so the vertex is (2, 1).

Correct Answer: A — (2, 1)

Q. For the function f(x) = x^2 - 6x + 8, find the x-coordinate of the vertex.

A. 2
B. 3
C. 4
D. 5

Solution

The x-coordinate of the vertex is given by x = -b/(2a) = 6/(2*1) = 3.

Correct Answer: B — 3

Q. For the function f(x) = x^3 - 3x^2 + 2, find the points where it is not differentiable.

A. None
B. x = 0
C. x = 1
D. x = 2

Solution

As a polynomial, f(x) is differentiable everywhere, hence no points of non-differentiability.

Correct Answer: A — None

Q. For the function f(x) = x^3 - 3x^2 + 4, find the points where it is not differentiable.

A. None
B. x = 0
C. x = 1
D. x = 2

Solution

The function is a polynomial and is differentiable everywhere, hence there are no points where it is not differentiable.

Correct Answer: A — None

Q. For the function f(x) = x^3 - 3x^2 + 4, find the value of x where f is not differentiable.

A. 0
B. 1
C. 2
D. 3

Solution

The function is a polynomial and is differentiable everywhere, so there is no such x.

Correct Answer: A — 0

Q. For the function f(x) = x^3 - 3x^2 + 4, find the x-coordinate of the point where f is differentiable.

A. -1
B. 0
C. 1
D. 2

Solution

f(x) is a polynomial and is differentiable everywhere. The x-coordinate can be any real number.

Correct Answer: C — 1

Q. For the function f(x) = x^3 - 6x^2 + 9x, find the critical points.

A. x = 0, 3
B. x = 1, 2
C. x = 2, 3
D. x = 3, 4

Solution

First, find f'(x) = 3x^2 - 12x + 9. Setting f'(x) = 0 gives (x - 3)(x - 1) = 0, so critical points are x = 1 and x = 3.

Correct Answer: A — x = 0, 3

Q. For the function f(x) = x^3 - 6x^2 + 9x, find the intervals where the function is increasing.

A. (-∞, 0)
B. (0, 3)
C. (3, ∞)
D. (0, 6)

Solution

f'(x) = 3x^2 - 12x + 9. The critical points are x = 1 and x = 3. The function is increasing on (1, 3) and (3, ∞).

Correct Answer: B — (0, 3)

Q. For the function f(x) = x^4 - 8x^2 + 16, find the coordinates of the inflection point.

A. (0, 16)
B. (2, 0)
C. (4, 0)
D. (2, 4)

Solution

Find f''(x) = 12x^2 - 16. Setting f''(x) = 0 gives x^2 = 4, so x = ±2. f(2) = 0, thus the inflection point is (2, 0).

Correct Answer: B — (2, 0)

Q. For the function f(x) = x^4 - 8x^2 + 16, find the intervals where the function is increasing.

A. (-∞, -2)
B. (-2, 2)
C. (2, ∞)
D. (-2, ∞)

Solution

f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x(x^2 - 4) = 0, so x = -2, 0, 2. Test intervals: f' is positive in (-2, ∞).

Correct Answer: D — (-2, ∞)

Q. For the function f(x) = { x^2 + 1, x < 0; 2x + b, x = 0; 3 - x, x > 0 to be continuous at x = 0, what is b?

A. -1
B. 0
C. 1
D. 2

Solution

Setting the left limit (0 + 1 = 1) equal to the right limit (3 - 0 = 3), we find b = 1.

Correct Answer: B — 0

Q. For the function f(x) = { x^2, x < 1; 3, x = 1; 2x, x > 1 }, what is the value of f(1)?

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

Solution

By definition, f(1) = 3, as given in the piecewise function.

Correct Answer: C — 3

Q. For the function f(x) = { x^2, x < 1; kx + 1, x >= 1 }, find k such that f is differentiable at x = 1.

A. 0
B. 1
C. 2
D. 3

Solution

Setting f(1-) = f(1+) and f'(1-) = f'(1+) gives k = 2 for differentiability.

Correct Answer: B — 1

Q. For the function f(x) = { x^2, x < 3; 9, x = 3; 3x, x > 3 } to be continuous at x = 3, the value of f(3) must be:

A. 6
B. 9
C. 3
D. 12

Solution

For continuity, f(3) must equal the limit as x approaches 3, which is 9.

Correct Answer: B — 9

Q. For the function f(x) = |x - 2| + |x + 3|, find the point where it is not differentiable.

A. -3
B. 2
C. 0
D. 1

Solution

The function is not differentiable at x = -3 and x = 2, but the first point of interest is -3.

Correct Answer: A — -3

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