If $I = \int_1^2 \frac{dx}{\sqrt{2x^3 - 9x^2 + 12x + 4}}$, then :

Option 1: $\frac{1}{9} < I^2 < \frac{1}{8}$

$\frac{1}{9} < I^2 < \frac{1}{8}$

$\frac{1}{16} < I^2 < \frac{1}{9}$

$\frac{1}{8} < I^2 < \frac{1}{4}$

$\frac{1}{6} < I^2 < \frac{1}{2}$

JEE Main 2020 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 8 January 2020, Shift 2, Maths Q61

Question 61 of 75 in this shift, Maths question 11 of 25, Section A.

I = \int_1^2 \frac{dx}{\sqrt{2x^3 - 9x^2 + 12x + 4}}

, then :

(1) $\frac{1}{9} < I^2 < \frac{1}{8}$ Official answer
(2) $\frac{1}{16} < I^2 < \frac{1}{9}$
(3) $\frac{1}{8} < I^2 < \frac{1}{4}$
(4) $\frac{1}{6} < I^2 < \frac{1}{2}$

Official answer

Option 1

NTA final key (Jan 2020).

Chapter: Integrals
Topic: Definite Integral Properties
Idea tested: Bounding and Estimation Property

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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.