ExamHabit
Practise free

JEE Main 2019 · 10 April · Shift II

JEE Main 2019 Mathematics question on Binomial Theorem

The smallest natural number nn, such that the coefficient of xx in the expansion of (x2+1x3)n\left(x^2 + \dfrac{1}{x^3}\right)^n is nC23^nC_{23}, is
  1. A.
    35
  2. B.
    23
  3. C.
    58
  4. D.
    38
Show answer and solution

Correct option: D

Solution

The general term is C(n,r) · (x²)^(n-r) · (1/x³)r = C(n,r) · x^(2n-5r).

For the coefficient of x¹: 2n - 5r = 1, so r = (2n-1)/5.

The coefficient is C(n,r) = C(n,23), so r = 23 (or r = n-23).

Case 1: r = 23 → 2n - 5(23) = 1 → 2n = 116 → n = 58. Check: r=23 < n=58 ✓

Case 2: r = n-23 → 2n - 5(n-23) = 1 → -3n + 115 = 1 → n = 38. Check: r = 15 < 38 ✓

Both n=38 and n=58 work, but the smallest natural number is n = 38.

[JEE Main 2019 · 10 April · Shift II]

Practise Binomial Theorem the way it is asked

A timed set of real JEE Main Binomial Theorem questions, graded as the exam grades. No sign-up.

Start a Binomial Theorem test

More: all JEE Main Binomial Theorem questions · JEE Main 2019 · Mathematics chapter list