Second Order Derivative

The correct answer is 8 f^{'} (0) = 4 f (x) + 0 \int x (x - t) f^{'} (t) d t = (e^{2 x} + e^{- 2 x}) cos 2 x + \frac{2}{a} x Put x = 0 : f (0) = 2 f^{'} (x) + x (f^{'} (x)) + 0 \int x f^{'} (t) d t - x f^{'} (x) = (e^{2 x} + e^{- 2 x}) (- 2 sin 2 x) + cos 2 x (2 e^{2 x} - 2 e^{- 2 x}) + \frac{2}{a} \Rightarrow f^{'} (x) + f (x) - 2 = (e^{2 x} + e^{- 2 x}) (- 2 sin 2 x) + cos 2 x (2 e^{2 x} - 2 e^{- 2 x}) + \frac{2}{a} Put x = 0 4 + 2 - 2 = 0 + (2 - 2) + 2/ a \Rightarrow a = \frac{1}{2} (2 a + 1)^{5} a^{2} = 2^{5} \cdot \frac{1}{2 ^{2}} = 8

About Second Order Derivative - JEE-MAIN

Second Order Derivative is a vital chapter for JEE-MAIN aspirants. Mastering the concepts covered in this chapter is essential for securing a top rank.

By rigorously practicing the previous year questions associated with this chapter, you can identify high-yield topics, understand the examiner's perspective, and boost your confidence during the actual exam.

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Analyzing PYQs for this specific chapter reveals the most frequently tested concepts and the typical complexity of questions, allowing you to tailor your study plan efficiently.

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Review the topic breakdown to see which sub-topics within Second Order Derivative carry the most weight. Then, tackle the questions iteratively to solidify your understanding.

High-Yield Trend

Chapter Questions
1 MCQs

Official Solution

About Second Order Derivative - JEE-MAIN

Frequently Asked Questions

Why focus on Second Order Derivative PYQs?

How to best use this analysis?

Second Order Derivative

High-Yield Trend

Chapter Questions 1 MCQs

Official Solution

About Second Order Derivative - JEE-MAIN

Frequently Asked Questions

Why focus on Second Order Derivative PYQs?

How to best use this analysis?

Chapter Questions
1 MCQs