JEE Main Preparation Resources for Competitive Exams

My Learning Cart

JEE Main

Download Q&A

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. Evaluate ∫_0^1 (x^3 - 3x^2 + 3x - 1) dx.

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

Solution

The integral evaluates to [x^4/4 - x^3 + (3/2)x^2 - x] from 0 to 1 = 0.

Correct Answer: A — 0

Q. Evaluate ∫_0^1 (x^4 - 2x^2 + 1) dx.

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

Solution

∫_0^1 (x^4 - 2x^2 + 1) dx = [x^5/5 - (2/3)x^3 + x] from 0 to 1 = (1/5 - 2/3 + 1) = 1/15.

Correct Answer: B — 1

Q. Evaluate ∫_0^1 (x^4) dx.

A. 1/5
B. 1/4
C. 1/3
D. 1/2

Solution

The integral evaluates to [x^5/5] from 0 to 1 = 1/5.

Correct Answer: A — 1/5

Q. Evaluate ∫_0^π/2 cos^2(x) dx.

A. π/4
B. π/2
C. 1
D. 0

Solution

∫_0^π/2 cos^2(x) dx = π/4.

Correct Answer: A — π/4

Q. Evaluate ∫_0^π/2 sin^2(x) dx.

A. π/4
B. π/2
C. π/3
D. π/6

Solution

Using the identity sin^2(x) = (1 - cos(2x))/2, the integral evaluates to π/4.

Correct Answer: A — π/4

Q. Evaluate ∫_1^2 (3x^2 - 4) dx.

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

Solution

The integral evaluates to [x^3 - 4x] from 1 to 2 = (8 - 8) - (1 - 4) = 3.

Correct Answer: A — 1

Q. Evaluate ∫_1^2 (3x^2 - 4x + 1) dx.

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

Solution

∫_1^2 (3x^2 - 4x + 1) dx = [x^3 - 2x^2 + x] from 1 to 2 = (8 - 8 + 2) - (1 - 2 + 1) = 1.

Correct Answer: B — 1

Q. Evaluate ∫_1^3 (2x + 1) dx.

A. 8
B. 10
C. 12
D. 14

Solution

The integral evaluates to [x^2 + x] from 1 to 3 = (9 + 3) - (1 + 1) = 10.

Correct Answer: B — 10

Q. Evaluate: sin^(-1)(0) + cos^(-1)(0).

A. 0
B. π/2
C. π
D. 1

Solution

sin^(-1)(0) = 0 and cos^(-1)(0) = π/2, thus the sum is 0 + π/2 = π/2.

Correct Answer: B — π/2

Q. Evaluate: sin^(-1)(1) + cos^(-1)(0).

A. π/2
B. π
C. 0
D. 1

Solution

sin^(-1)(1) = π/2 and cos^(-1)(0) = π/2. Therefore, π/2 + π/2 = π.

Correct Answer: A — π/2

Q. Find the 10th term of the sequence defined by a_n = 3n + 2.

A. 32
B. 30
C. 28
D. 34

Solution

a_10 = 3(10) + 2 = 30 + 2 = 32.

Correct Answer: A — 32

Q. Find the 10th term of the sequence defined by a_n = 3n^2 + 2n.

A. 320
B. 302
C. 290
D. 310

Solution

a_10 = 3(10^2) + 2(10) = 300 + 20 = 320.

Correct Answer: B — 302

Q. Find the angle between the lines represented by the equation 2x^2 - 3xy + y^2 = 0.

A. 30 degrees
B. 45 degrees
C. 60 degrees
D. 90 degrees

Solution

The angle between the lines can be found using the formula tan(θ) = |(m1 - m2) / (1 + m1*m2)|, where m1 and m2 are the slopes of the lines. The slopes can be found from the equation.

Correct Answer: B — 45 degrees

Q. Find the angle between the lines y = 2x + 1 and y = -0.5x + 3.

A. 60 degrees
B. 45 degrees
C. 90 degrees
D. 30 degrees

Solution

The slopes are m1 = 2 and m2 = -0.5. The angle θ is given by tan(θ) = |(m1 - m2) / (1 + m1*m2)| = |(2 + 0.5) / (1 - 1)|, which is undefined, indicating 90 degrees.

Correct Answer: A — 60 degrees

Q. Find the angle between the vectors (1, 0, 0) and (0, 1, 0).

A. 0 degrees
B. 90 degrees
C. 45 degrees
D. 180 degrees

Solution

The angle θ = cos⁻¹((u · v) / (|u| |v|)) = cos⁻¹(0) = 90 degrees.

Correct Answer: B — 90 degrees

Q. Find the angle between the vectors A = (1, 2, 2) and B = (2, 0, 2).

A. 0°
B. 45°
C. 60°
D. 90°

Solution

cos(θ) = (A · B) / (|A| |B|). A · B = 1*2 + 2*0 + 2*2 = 6. |A| = √(1^2 + 2^2 + 2^2) = 3, |B| = √(2^2 + 0^2 + 2^2) = 2√2. cos(θ) = 6 / (3 * 2√2) = 1/√2, θ = 45°.

Correct Answer: C — 60°

Q. Find the angle between the vectors A = (1, 2, 2) and B = (2, 1, 1).

A. 60°
B. 45°
C. 30°
D. 90°

Solution

cos(θ) = (A · B) / (|A| |B|). A · B = 1*2 + 2*1 + 2*1 = 6; |A| = √(1^2 + 2^2 + 2^2) = 3; |B| = √(2^2 + 1^2 + 1^2) = √6. Thus, cos(θ) = 6 / (3√6) = 1/√6, θ = 45°.

Correct Answer: B — 45°

Q. Find the angle between the vectors A = (3, -2, 1) and B = (1, 1, 1) if A · B = |A||B|cos(θ).

A. 60°
B. 45°
C. 90°
D. 30°

Solution

A · B = 3*1 + (-2)*1 + 1*1 = 3 - 2 + 1 = 2. |A| = √(3^2 + (-2)^2 + 1^2) = √14, |B| = √3. cos(θ) = 2/(√14 * √3). θ = 60°.

Correct Answer: A — 60°

Q. Find the angle between the vectors A = (3, -2, 1) and B = (1, 1, 1).

A. 60°
B. 45°
C. 90°
D. 30°

Solution

cos(θ) = (A · B) / (|A| |B|). A · B = 3*1 + (-2)*1 + 1*1 = 2. |A| = √(3^2 + (-2)^2 + 1^2) = √14, |B| = √3. θ = cos^(-1)(2/(√14 * √3)).

Correct Answer: A — 60°

Q. Find the area between the curves y = x^2 and y = 4 from x = -2 to x = 2.

A. 8/3
B. 16/3
C. 8
D. 4

Solution

The area between the curves is given by ∫(from -2 to 2) (4 - x^2) dx = [4x - x^3/3] from -2 to 2 = 16/3.

Correct Answer: B — 16/3

Q. Find the area between the curves y = x^2 and y = 4 from x = 0 to x = 2.

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

Solution

The area between the curves y = x^2 and y = 4 is given by ∫(from 0 to 2) (4 - x^2) dx = [4x - x^3/3] from 0 to 2 = (8 - 8/3) = 4/3.

Correct Answer: A — 4

Q. Find the area between the curves y = x^3 and y = x from x = 0 to x = 1.

A. 1/4
B. 1/3
C. 1/2
D. 1/6

Solution

The area between the curves is given by ∫(from 0 to 1) (x - x^3) dx = [x^2/2 - x^4/4] from 0 to 1 = (1/2 - 1/4) = 1/4.

Correct Answer: B — 1/3

Q. Find the area of the triangle formed by the points (0, 0), (4, 0), and (0, 3).

A. 6
B. 12
C. 8
D. 10

Solution

Area = 1/2 * base * height = 1/2 * 4 * 3 = 6.

Correct Answer: A — 6

Q. Find the area of the triangle formed by the points A(1, 2, 3), B(4, 5, 6), and C(7, 8, 9) using the vector product.

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

Solution

Area = 0.5 * |AB × AC| = 0, as points are collinear.

Correct Answer: A — 0

Q. Find the area of the triangle with vertices (0,0), (4,0), (0,3).

A. 6
B. 12
C. 8
D. 10

Solution

Area = 0.5 * base * height = 0.5 * 4 * 3 = 6.

Correct Answer: A — 6

Q. Find the area of the triangle with vertices (0,0), (4,0), and (4,3).

A. 6
B. 12
C. 8
D. 10

Solution

Area = 1/2 * base * height = 1/2 * 4 * 3 = 6.

Correct Answer: B — 12

Q. Find the area of the triangle with vertices at (0,0), (4,0), and (0,3).

A. 6
B. 12
C. 8
D. 10

Solution

Area = 1/2 * base * height = 1/2 * 4 * 3 = 6.

Correct Answer: A — 6

Q. Find the area under the curve y = e^x from x = 0 to x = 1.

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

Solution

The area is given by the integral from 0 to 1 of e^x dx. This evaluates to [e^x] from 0 to 1 = e - 1.

Correct Answer: A — e - 1

Q. Find the area under the curve y = x^2 + 2x from x = 0 to x = 3.

A. 9
B. 12
C. 15
D. 18

Solution

The area under the curve is given by ∫(from 0 to 3) (x^2 + 2x) dx = [x^3/3 + x^2] from 0 to 3 = (27/3 + 9) = 18.

Correct Answer: C — 15

Q. Find the area under the curve y = x^2 from x = 0 to x = 2.

A. 2
B. 4
C. 8/3
D. 3

Solution

The area under the curve y = x^2 from 0 to 2 is given by the integral ∫(from 0 to 2) x^2 dx = [x^3/3] from 0 to 2 = (2^3/3) - (0^3/3) = 8/3.

Correct Answer: C — 8/3

Showing 1831 to 1860 of 10700 (357 Pages)

School

College

Degree

Competitve

Skills

Soulshift Feedback ×

On a scale of 0–10, how likely are you to recommend The Soulshift Academy?

Not likely Very likely