JEE Main 2024 · MathsNumerical answerNumerical valueHardCalculation

JEE Main 1 February 2024, Shift 2, Maths Q25

Question 25 of 90 in this shift, Maths question 25 of 30, Section B.

y=\frac{(\sqrt{x}+1)(x^2-\sqrt{x})}{x\sqrt{x}+x+\sqrt{x}}+\frac{1}{15}(3\cos^2x-5)\cos^3x

, then

96\,y'\left(\frac{\pi}{6}\right)

is equal to ________.

Official answer

105

NTA final key.

Chapter: Continuity and Differentiability
Topic: Chain Rule
Idea tested: Chain Rule for Trigonometric Powers and Arguments

Same topic in other shifts

All Chain Rule questions

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5 Apr 2024, Shift 2 · Q20If $y(\theta) = \frac{2\cos\theta + \cos 2\theta}{\cos 3\theta + 4\cos 2\theta + 5\cos\theta + 2}$ , then at $\theta = \frac{\pi}{2}$ ,…HardSingle correct
9 Apr 2024, Shift 2 · Q8If $\log_e y=3\sin^{-1}x$ , then $(1-x^2)y''-xy'$ at $x=\frac{1}{2}$ is equal toMediumSingle correct
29 Jan 2024, Shift 1 · Q9Suppose $f(x)=\frac{(2^x+2^{-x})\tan x\sqrt{\tan^{-1}(x^2-x+1)}}{(7x^2+3x+1)^3}$ . Then the value of $f'(0)$ is equal toHardSingle 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.