Q60 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics
If $T_4$ represents the 4th term in the expansion of $\left(5 x+\frac{7}{x}\right)^{-3 / 2}$ and $x \notin\left[-\sqrt{\frac{7}{5}}, \sqrt{\frac{7}{5}}\right]$, then $\left(x^7 \sqrt{5 x}\right) T_4=$
- A$\frac{7^4}{2^5 5^3}$
- B$-\frac{7^4}{2^5 5^3}$
- Correct answerC$-\frac{7^4}{2^4 5^3}$
- D$\frac{7^4}{2^4 5^3}$
Solution
We know expansion of
$\begin{aligned}
& (1+x)^n=1+n x+\frac{n(n-1) x^2}{2!}+\ldots \ldots \\
& \left(5 x+\frac{7}{x}\right)^{-3 / 2}=5^{-3 / 2} x^{-3 / 2}\left(1+\frac{7}{5 x^2}\right)^{-3 / 2}
\end{aligned}$
The expansion of $\left(1+\frac{7}{5 x^2}\right)^{-3 / 2}$ $\begin{array}{r} =1+\frac{-3}{2} \times\left(\frac{7}{5 x^2}\right)+\frac{\left(\frac{-3}{2}\right) \times\left(\frac{-3}{2}-1\right)\left(\frac{7}{5 x^2}\right)^2}{2!} \\ +\frac{\left(\frac{-3}{2}\right)\left(\frac{-3}{2}-1\right)\left(\frac{-3}{2}-2\right)\left(\frac{7}{5 x^2}\right)^3}{3!} \end{array}$ $\begin{aligned} & T_4=\frac{(5)^{\frac{-3}{2}} \times(x)^{\frac{-3}{2}} \times\left(\frac{-3}{2}\right) \times\left(\frac{-5}{2}\right) \times\left(\frac{-7}{2}\right) \times\left(\frac{7}{5 x^2}\right)^3}{3!} \\ & T_4=\frac{-(5)^{\frac{-7}{2}} \times(7)^4 \times(x)^{\frac{-15}{2}}}{2^4} \therefore x^7 \sqrt{5 x} \times T_4=-\frac{7^4}{2^4 \times 5^3} . \end{aligned}$
The expansion of $\left(1+\frac{7}{5 x^2}\right)^{-3 / 2}$ $\begin{array}{r} =1+\frac{-3}{2} \times\left(\frac{7}{5 x^2}\right)+\frac{\left(\frac{-3}{2}\right) \times\left(\frac{-3}{2}-1\right)\left(\frac{7}{5 x^2}\right)^2}{2!} \\ +\frac{\left(\frac{-3}{2}\right)\left(\frac{-3}{2}-1\right)\left(\frac{-3}{2}-2\right)\left(\frac{7}{5 x^2}\right)^3}{3!} \end{array}$ $\begin{aligned} & T_4=\frac{(5)^{\frac{-3}{2}} \times(x)^{\frac{-3}{2}} \times\left(\frac{-3}{2}\right) \times\left(\frac{-5}{2}\right) \times\left(\frac{-7}{2}\right) \times\left(\frac{7}{5 x^2}\right)^3}{3!} \\ & T_4=\frac{-(5)^{\frac{-7}{2}} \times(7)^4 \times(x)^{\frac{-15}{2}}}{2^4} \therefore x^7 \sqrt{5 x} \times T_4=-\frac{7^4}{2^4 \times 5^3} . \end{aligned}$
Hence, the correct answer is option: 'C'.