JEE Main 2024 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 8 April 2024, Shift 2, Maths Q10

Question 10 of 90 in this shift, Maths question 10 of 30, Section A.

If the function

f(x) = 2x^3 - 9ax^2 + 12a^2x + 1

a > 0

has a local maximum at

x = \alpha

and a local minimum at

x = \alpha^2

, then

\alpha

and

\alpha^2

are the roots of the equation :

(1) $8x^2 - 6x + 1 = 0$
(2) $8x^2 + 6x - 1 = 0$
(3) $x^2 - 6x + 8 = 0$ Official answer
(4) $x^2 + 6x + 8 = 0$

Official answer

Option 3

NTA final key.

Chapter: Application of Derivatives
Topic: Maxima and Minima
Idea tested: Finding Parameters from Given Extrema

Same idea in other shifts

Asked 4× in all

2 Apr 2025, Shift 1 · Q8If the function $f(x)=2x^3-9ax^2+12a^2x+1$ , where $a>0$ , attains its local maximum and local minimum values at p and q, respectively, such…MediumSingle correct
4 Apr 2025, Shift 2 · Q17Let $a>0$ . If the function $f(x)=6x^3-45ax^2+108a^2x+1$ attains its local maximum and minimum values at the points $x_1$ and $x_2$ …MediumSingle correct
3 Aug 2021, Shift 2 · Q71If the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \frac{x^3}{3} - (2a+3b)x^2 + 24abx + 5$ has a local maxima at $-4$ and…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.