JEE Main 27 July 2022, Shift 2, Maths Q26

Question 26 of 90 in this shift, Maths question 26 of 30, Section B.

For the curve

C : (x^2 + y^2 - 3) + (x^2 - y^2 - 1)^5 = 0

, the value of

3y' - y^3 y''

, at the point

(\alpha, \alpha)

\alpha > 0

, on C, is equal to __________.

Official answer

NTA final key.

Chapter: Continuity and Differentiability
Topic: Chain Rule
Idea tested: Implicit Chain Rule

Same idea in other shifts

Asked 2× in all

4 Sep 2020, Shift 1 · Q59If $(a + \sqrt{2}\, b\cos x)(a - \sqrt{2}\, b\cos y) = a^2 - b^2$ , where $a > b > 0$ , then $\frac{dx}{dy}$ at…MediumSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.