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Integral: JEE Main Previous Year Questions (2021-2026)
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Page 1 JEE Main Previous Year Questions (2021-2026): Indefinite Integrals (January 2026) Q1: Let be such that f(0) = -26 + 24 log?(2). If f(1) = a + b log?(3), where a, b ? Z, then a + b is equal to: A: -5 B: -11 C: -18 D: -26 Answer: B Explanation: Given f(0) = -26 + 24 ln 2 and f(1) = a + b ln 3, a, b ? Z Use substitution Let x = t 6 = ? dx = 6t 5 dt x 2/3 = (t 6) 2/3 = t 4 x 1/2 = (t 6) 1/2 = t³ Substituting these into the integral: manipulate Now, integrate term by term: f(x) = 3t² - 12t + 24 ln|t + 2| + C Back-substitute t = x 1/6 : f(x) = 3x 1/3 - 12x 1/6 + 24 ln|x 1/6 + 2| + C Finding the Constant C Page 2 JEE Main Previous Year Questions (2021-2026): Indefinite Integrals (January 2026) Q1: Let be such that f(0) = -26 + 24 log?(2). If f(1) = a + b log?(3), where a, b ? Z, then a + b is equal to: A: -5 B: -11 C: -18 D: -26 Answer: B Explanation: Given f(0) = -26 + 24 ln 2 and f(1) = a + b ln 3, a, b ? Z Use substitution Let x = t 6 = ? dx = 6t 5 dt x 2/3 = (t 6) 2/3 = t 4 x 1/2 = (t 6) 1/2 = t³ Substituting these into the integral: manipulate Now, integrate term by term: f(x) = 3t² - 12t + 24 ln|t + 2| + C Back-substitute t = x 1/6 : f(x) = 3x 1/3 - 12x 1/6 + 24 ln|x 1/6 + 2| + C Finding the Constant C We are given the initial condition f(0) = -26 + 24 ln(2) : f(0) = 3(0) - 12(0) + 24 ln(0 + 2) + C -26 + 24 ln(2) = 24 ln(2) + C = ? C = -26 So, the complete function is: f(x) = 3x 1/3 - 12x 1/6 + 24 ln(x 1/6 + 2) - 26 Solving for a and b Now we evaluate f(1): f(1) = 3(1) 1/3 - 12(1) 1/6 + 24 ln(1 1/6 + 2) - 26 f(1) = 3 - 12 + 24 ln(3) - 26 f(1) = -35 + 24 ln(3) Comparing this to the given form f(1) = a + b ln(3): a = -35 and b = 24 Then a + b = -35 + 24 = -11 Q2: where C is the constant of integration, then is equal to A: 1/v3 (26 - v3) B: 4/v3 (8 - v6) C: 1/v3 (26 + v3) D: 2/v3 (4 + v6) Answer: B Explanation: The given integral is: Page 3 JEE Main Previous Year Questions (2021-2026): Indefinite Integrals (January 2026) Q1: Let be such that f(0) = -26 + 24 log?(2). If f(1) = a + b log?(3), where a, b ? Z, then a + b is equal to: A: -5 B: -11 C: -18 D: -26 Answer: B Explanation: Given f(0) = -26 + 24 ln 2 and f(1) = a + b ln 3, a, b ? Z Use substitution Let x = t 6 = ? dx = 6t 5 dt x 2/3 = (t 6) 2/3 = t 4 x 1/2 = (t 6) 1/2 = t³ Substituting these into the integral: manipulate Now, integrate term by term: f(x) = 3t² - 12t + 24 ln|t + 2| + C Back-substitute t = x 1/6 : f(x) = 3x 1/3 - 12x 1/6 + 24 ln|x 1/6 + 2| + C Finding the Constant C We are given the initial condition f(0) = -26 + 24 ln(2) : f(0) = 3(0) - 12(0) + 24 ln(0 + 2) + C -26 + 24 ln(2) = 24 ln(2) + C = ? C = -26 So, the complete function is: f(x) = 3x 1/3 - 12x 1/6 + 24 ln(x 1/6 + 2) - 26 Solving for a and b Now we evaluate f(1): f(1) = 3(1) 1/3 - 12(1) 1/6 + 24 ln(1 1/6 + 2) - 26 f(1) = 3 - 12 + 24 ln(3) - 26 f(1) = -35 + 24 ln(3) Comparing this to the given form f(1) = a + b ln(3): a = -35 and b = 24 Then a + b = -35 + 24 = -11 Q2: where C is the constant of integration, then is equal to A: 1/v3 (26 - v3) B: 4/v3 (8 - v6) C: 1/v3 (26 + v3) D: 2/v3 (4 + v6) Answer: B Explanation: The given integral is: We can rewrite the integrand by splitting the terms: Page 4 JEE Main Previous Year Questions (2021-2026): Indefinite Integrals (January 2026) Q1: Let be such that f(0) = -26 + 24 log?(2). If f(1) = a + b log?(3), where a, b ? Z, then a + b is equal to: A: -5 B: -11 C: -18 D: -26 Answer: B Explanation: Given f(0) = -26 + 24 ln 2 and f(1) = a + b ln 3, a, b ? Z Use substitution Let x = t 6 = ? dx = 6t 5 dt x 2/3 = (t 6) 2/3 = t 4 x 1/2 = (t 6) 1/2 = t³ Substituting these into the integral: manipulate Now, integrate term by term: f(x) = 3t² - 12t + 24 ln|t + 2| + C Back-substitute t = x 1/6 : f(x) = 3x 1/3 - 12x 1/6 + 24 ln|x 1/6 + 2| + C Finding the Constant C We are given the initial condition f(0) = -26 + 24 ln(2) : f(0) = 3(0) - 12(0) + 24 ln(0 + 2) + C -26 + 24 ln(2) = 24 ln(2) + C = ? C = -26 So, the complete function is: f(x) = 3x 1/3 - 12x 1/6 + 24 ln(x 1/6 + 2) - 26 Solving for a and b Now we evaluate f(1): f(1) = 3(1) 1/3 - 12(1) 1/6 + 24 ln(1 1/6 + 2) - 26 f(1) = 3 - 12 + 24 ln(3) - 26 f(1) = -35 + 24 ln(3) Comparing this to the given form f(1) = a + b ln(3): a = -35 and b = 24 Then a + b = -35 + 24 = -11 Q2: where C is the constant of integration, then is equal to A: 1/v3 (26 - v3) B: 4/v3 (8 - v6) C: 1/v3 (26 + v3) D: 2/v3 (4 + v6) Answer: B Explanation: The given integral is: We can rewrite the integrand by splitting the terms: Q3: A: 1 + v2 B: -1 - 2v2 C: -1 - v2 D: -1 + v2 Answer: C Explanation: Page 5 JEE Main Previous Year Questions (2021-2026): Indefinite Integrals (January 2026) Q1: Let be such that f(0) = -26 + 24 log?(2). If f(1) = a + b log?(3), where a, b ? Z, then a + b is equal to: A: -5 B: -11 C: -18 D: -26 Answer: B Explanation: Given f(0) = -26 + 24 ln 2 and f(1) = a + b ln 3, a, b ? Z Use substitution Let x = t 6 = ? dx = 6t 5 dt x 2/3 = (t 6) 2/3 = t 4 x 1/2 = (t 6) 1/2 = t³ Substituting these into the integral: manipulate Now, integrate term by term: f(x) = 3t² - 12t + 24 ln|t + 2| + C Back-substitute t = x 1/6 : f(x) = 3x 1/3 - 12x 1/6 + 24 ln|x 1/6 + 2| + C Finding the Constant C We are given the initial condition f(0) = -26 + 24 ln(2) : f(0) = 3(0) - 12(0) + 24 ln(0 + 2) + C -26 + 24 ln(2) = 24 ln(2) + C = ? C = -26 So, the complete function is: f(x) = 3x 1/3 - 12x 1/6 + 24 ln(x 1/6 + 2) - 26 Solving for a and b Now we evaluate f(1): f(1) = 3(1) 1/3 - 12(1) 1/6 + 24 ln(1 1/6 + 2) - 26 f(1) = 3 - 12 + 24 ln(3) - 26 f(1) = -35 + 24 ln(3) Comparing this to the given form f(1) = a + b ln(3): a = -35 and b = 24 Then a + b = -35 + 24 = -11 Q2: where C is the constant of integration, then is equal to A: 1/v3 (26 - v3) B: 4/v3 (8 - v6) C: 1/v3 (26 + v3) D: 2/v3 (4 + v6) Answer: B Explanation: The given integral is: We can rewrite the integrand by splitting the terms: Q3: A: 1 + v2 B: -1 - 2v2 C: -1 - v2 D: -1 + v2 Answer: C Explanation: Q4: A: 30 B: 29 C: 28Read More
FAQs on Integral: JEE Main Previous Year Questions (2021-2026)
| 1. What is the significance of integrals in JEE Main examinations? |  |
Ans. Integrals are significant in JEE Main examinations as they are essential for understanding concepts in calculus, which is a critical component of the syllabus. They are used to calculate areas under curves, volumes of solids, and in solving problems related to motion, physics, and various applications in engineering and technology.
| 2. How are definite and indefinite integrals different in the context of JEE? | |
Ans. Definite integrals have specific upper and lower limits and yield a numerical value representing the area under a curve between those limits. In contrast, indefinite integrals represent a family of functions and include a constant of integration, as they do not have specified limits. Both types are important for solving different kinds of problems in JEE.
| 3. What are some common techniques for solving integrals that JEE aspirants should know? | |
Ans. Common techniques for solving integrals include substitution, integration by parts, partial fractions, and trigonometric substitutions. Mastering these techniques allows students to tackle a wide range of integral problems efficiently, which is crucial for success in the examination.
| 4. Can you provide an example of a typical integral problem that might appear in the JEE? | |
Ans. A typical integral problem in the JEE may ask to evaluate the integral ∫(x² + 3x + 2) dx. The solution involves applying the power rule of integration, resulting in (1/3)x³ + (3/2)x² + 2x + C, where C is the constant of integration.
| 5. Why is it important to practice previous year questions related to integrals for JEE preparation? | |
Ans. Practising previous year questions related to integrals is important for JEE preparation because it helps students understand the exam pattern, identify frequently tested concepts, and improve problem-solving speed and accuracy. It also builds familiarity with the types of questions that may appear in the examination, thereby boosting confidence.
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