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JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Binomial Theorem

If the constant term in the binomial expansion of

(x - \frac{k}{x ^{2}})^{10}

is 405, then

∣ k ∣

equals:

(A)9
(B)1
(C)3✓ Correct
(D)2

Answer: (C)

Solution

The general term is

T_{r + 1} = (r 10) (x)^{10 - r} (- \frac{k}{x ^{2}})^{r} = (r 10) (- k)^{r} x^{5 - \frac{5 r}{2}}

. The constant term needs the exponent of

x

to vanish,

5 - \frac{5 r}{2} = 0

, so

r = 2

. That term is

(2 10) k^{2} = 45 k^{2} = 405

, giving

k^{2} = 9

and

∣ k ∣ = 3

.

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Single correct option

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