JEE Main 2019 · 10 Jan · Shift II

JEE Main 2019 Mathematics question on Binomial Theorem

The positive value of λ\lambda for which the coefficient of x2x^2 in the expression x2(x+λx2)10x^2\left(\sqrt{x} + \dfrac{\lambda}{x^2}\right)^{10} is 720, is
  1. A.
    3
  2. B.
    5\sqrt{5}
  3. C.
    222\sqrt{2}
  4. D.
    4
Show answer and solution

Correct option: D

Solution

The general term of (√x + λ/x²)10 is C(10,r)(√x)^(10-r)(λ/x²)r = C(10,r)λr · x^((10-r)/2 - 2r).

Multiplying by x² (from the x² outside), the power of x becomes (10-r)/2 - 2r + 2 = (14-5r)/2.

For coefficient of x², set (14-5r)/2 = 2 → 14-5r = 4 → r = 2.

The coefficient is C(10,2)·λ² = 45λ² = 720 → λ² = 16 → λ = 4.

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