JEE Main 2022 · MathsMultiple choiceSingle correctMediumCalculation

JEE Main 28 July 2022, Shift 1, Maths Q17

Question 17 of 90 in this shift, Maths question 17 of 30, Section A.

If the minimum value of

f(x)=\frac{5x^2}{2}+\frac{\alpha}{x^5},\ x>0

, is 14, then the value of

\alpha

is equal to :

(1) $32$
(2) $64$
(3) $128$ Official answer
(4) $256$

Official answer

Option 3

NTA final key (2022 Session 2).

Chapter: Application of Derivatives
Topic: Maxima and Minima
Idea tested: Optimization via AM-GM Inequality

Same idea in other shifts

Asked 5× in all

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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.