Q99 / 160 · AP EAMCET 2023 (17 May Shift 1) · Mathematics
If $\lim _{x \rightarrow 1} \frac{\sqrt[4]{x^{-3}}+a \sqrt[4]{x^5}}{x-1}=-2$, then the coefficient of $x$ in the expansion of $\left(\sqrt[4]{x^{-3}}+a \sqrt[4]{x^5}\right)^4$ is
- Correct answerA$6$
- B$-1$
- C$5$
- D$4$
Solution
$\begin{aligned}
& \text {} \lim _{x \rightarrow 1} \frac{\sqrt[4]{x^{-3}}+a \sqrt[4]{x^5}}{x-1}=-2 \\
& \Rightarrow \lim _{x \rightarrow 1} \frac{\sqrt[4]{x^{-3}}\left(1+a x^2\right)}{x-1}=-2
\end{aligned}$
This is possible if and if $a=1$
$\begin{aligned}
& \left(\sqrt[4]{x^{-3}}+a \sqrt[4]{x^5}\right)^4=\left(\sqrt[4]{x^{-3}}+\sqrt[4]{x^5}\right)^4 \\
& ={ }^4 C_0\left(\sqrt[4]{x^{-3}}\right)^0\left(\sqrt[4]{x^5}\right)^4+{ }^4 C_1\left(\sqrt[4]{x^{-3}}\right)^1\left(\sqrt[4]{x^5}\right)^3 \\
& +{ }^4 C_2\left(\sqrt[4]{x^{-3}}\right)^2\left(\sqrt[4]{x^5}\right)^2+{ }^4 C_3\left(\sqrt[4]{x^{-3}}\right)^3\left(\sqrt[4]{x^5}\right) \\
& +{ }^4 C\left(\sqrt[4]{x^{-3}}\right)^4\left(\sqrt[4]{x^5}\right)^0 \\
& \therefore \text { The coefficient of } x={ }^4 C_2=\frac{4 !}{2 ! 2 !}=\frac{4 \times 3}{2 \times 1}=6
\end{aligned}$
Hence, the correct answer is option: 'A'.