Q38 / 75 · JEE Main 2025 (24 Jan Shift 1) · Mathematics
Let $f: R-\{0\} \rightarrow R$ be a function such that $f(x)-6 f\left(\frac{1}{x}\right)=\frac{35}{3 x}-\frac{5}{2}$.
If the $\lim _{x \rightarrow 0}\left(\frac{1}{\alpha x}+f(x)\right)=\beta ; \alpha, \beta \in R$, then $\alpha+2 \beta$ is equal to
- A$5$
- B$3$
- Correct answerC$4$
- D$6$
Solution
$\begin{aligned}
& f(x)-6 f\left(\frac{1}{x}\right)=\frac{35}{3 x}-\frac{5}{2}...(1) \\
& 6\left(f\left(\frac{1}{x}\right)-6 f(x)=\frac{35 x}{3}-\frac{5}{2}\right) \\
& 6 f\left(\frac{1}{x}\right)-36 f(x)=\frac{210 x}{3}-\frac{30}{2}...(2) \\
& (1)+(2) \\
& -35 f(x)=\frac{35}{3}\left[\frac{1}{x}+6 x\right]-\frac{5}{2}(1+6) \\
& -f(x)=\frac{1}{3}\left(\frac{1}{x}+6 x\right)-\frac{1}{2} \\
& f(x)=-\frac{1}{3 x}-2 x+\frac{1}{2} \\
& \lim _{x \rightarrow 0}\left[\frac{1}{\alpha x}-\frac{1}{3 x}-2 x+\frac{1}{2}\right]=\beta \\
& \Rightarrow \alpha=3 \\
& \beta+2 \beta=3+2 \times \frac{1}{2}=4
\end{aligned}$
Hence, the correct answer is option: 'C'.