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Saturday, 31 October

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JEE Main 2023 · MathsMultiple choiceSingle correctHardConceptual

JEE Main 6 April 2023, Shift 2, Maths Q9

Question 9 of 90 in this shift, Maths question 9 of 30, Section A.

\lim_{n\to\infty}\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right)\cdots\cdots\left(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\right)\right\}

is equal to

(1) $1$
(2) $0$ Official answer
(3) $\frac{1}{\sqrt{2}}$
(4) $\sqrt{2}$

Official answer

Option 2

NTA final key.

Chapter: Limits and Derivatives
Topic: No close match in our taxonomy
Idea tested: No close match in our taxonomy

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.