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JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Binomial Theorem

If the term without

x

in the expansion of

(x^{\frac{2}{3}} + \frac{α}{x ^{3}})^{22}

is 7315, then

∣ α ∣

is equal to ______.

Answer: 1

Solution

The general term is

T_{r + 1} = (r 22) (x^{2/3})^{22 - r} (\frac{α}{x ^{3}})^{r} = (r 22) α^{r} x^{\frac{44 - 11 r}{3}}

. The term free of

x

needs

44 - 11 r = 0

, so

r = 4

, giving the term

(4 22) α^{4}

. Since

(4 22) = \frac{22 \cdot 21 \cdot 20 \cdot 19}{24} = 7315

, the condition

7315 α^{4} = 7315

gives

α^{4} = 1

, so

∣ α ∣ = 1

.

JEE Main · 2023 · 01 Feb 2023 · Shift 2 · Integer answer

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