If the constant term in the expansion of (frac sqrt[5] 3 x +frac 2 x sqrt[3] 5 ) 12 , x neq 0 , is α × 2 8 × sqrt[5] 3 , then 25 α is equal to :

Mathematics · Binomial Theorem

JEE Main 2024 — 5 April, Shift 2 — Question 10

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

Option A:
639
Option B:
724
Option C:
693
Correct
Option D:
742

Answer: C

Step-by-step solution

$\quad \mathrm{T}_{\mathrm{r}+1}={ }^{12} \mathrm{C}_{\mathrm{r}}\left(\frac{3^{1 / 5}}{\mathrm{x}}\right)^{12-\mathrm{r}}\left(\frac{2 \mathrm{x}}{5^{1 / 3}}\right)^{\mathrm{r}}$

$\mathrm{T}_{\mathrm{r}+1}=\frac{{ }^{12} \mathrm{C}_{\mathrm{r}}(3)^{\frac{12-\mathrm{r}}{5}}(2)^{\mathrm{r}}(\mathrm{x})^{2 \mathrm{r}-12}}{(5)^{\mathrm{r} / 3}}$

$r=6$

$\mathrm{T}_{7}=\frac{{ }^{12} \mathrm{C}_{6}(3)^{6 / 5}(2)^{6}}{5^{2}}=\left(\frac{9 \times 11 \times 7}{25}\right) 2^{8} .3^{1 / 5}$

$25 \alpha=693$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 5 April, Shift 2
Subject: Mathematics
Chapter: Binomial Theorem
Topic: Binomial Coefficients