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Saturday, 31 October

JEE Main 2021 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 26 February 2021, Shift 1, Maths Q71

Question 71 of 90 in this shift, Maths question 11 of 30, Section A.

The maximum value of the term independent of 't' in the expansion of (tx15+(1−x)110t)10\left(tx^{\frac{1}{5}}+\frac{(1-x)^{\frac{1}{10}}}{t}\right)^{10} where x∈(0,1)x\in(0,1) is :
  1. (1)10!3(5!)2\frac{10!}{\sqrt{3}(5!)^2}
  2. (2)2.10!33(5!)2\frac{2.10!}{3\sqrt{3}(5!)^2}Official answer
  3. (3)2.10!3(5!)2\frac{2.10!}{3(5!)^2}
  4. (4)10!3(5!)2\frac{10!}{3(5!)^2}

Official answer

Option 2

NTA final key.

Chapter
Binomial Theorem
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.