If the constant term in the expansion of \(\left(\frac{\sqrt[5]{3}}{x}+\frac{2x}{\sqrt[3]{5}}\right)^{12}, x\neq 0\), is \(\alpha \times 2^{8} \times \sqrt[5]{3}\), then \(25\alpha\) is equal to

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Binomial Theorem

If the constant term in the expansion of

(\frac{5 3}{x} + \frac{2 x}{3 5})^{12}, x \neq = 0

, is

α \times 2^{8} \times 53

, then

25 α

is equal to

Answer: (D)

General term:

T_{r + 1} = (r 12) (\frac{3 ^{1/5}}{x})^{12 - r} (\frac{2 x}{5 ^{1/3}})^{r} = (r 12) 3^{(12 - r) /5} 5^{- r /3} 2^{r} x^{2 r - 12}

.
Constant term needs

2 r - 12 = 0 ⟹ r = 6

T_{7} = (6 12) 3^{6/5} 5^{- 2} 2^{6} = 924 \times 3 \times 3^{1/5} \times \frac{64}{25}

Matching this to

α \times 2^{8} \times 3^{1/5} = 256 α \times 3^{1/5}

α = \frac{924 \times 3 \times 64}{25 \times 256} = \frac{924 \times 3}{25 \times 4} = \frac{2772}{100} = 27.72

25 α = 25 \times 27.72 = 693

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Single correct option