Learn

Saturday, 31 October

JEE Main 2022 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 30 June 2022, Shift 1, Maths Q6

Question 6 of 90 in this shift, Maths question 6 of 30, Section A.

For two positive real numbers a and b such that 1a2+1b3=4\frac{1}{a^2} + \frac{1}{b^3} = 4, the minimum value of the constant term in the expansion of (ax18+bx−112)10\left(ax^{\frac{1}{8}} + bx^{-\frac{1}{12}}\right)^{10} is :
  1. (1)1052\frac{105}{2}
  2. (2)1054\frac{105}{4}
  3. (3)1058\frac{105}{8}Official answer
  4. (4)10516\frac{105}{16}

Official answer

Option 3

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.