Q101 / 160 · AP EAMCET 2022 (08 Jul Shift 2) · Mathematics
Let $f(x)=\lim _{y \rightarrow \infty} y\left(x^{1 / y}-1\right)$, and $2022 f\left(\frac{1}{x}\right)+P f(x)=f\left(x^2\right)$, then $P=$
- A$2020$
- B$2021$
- C$2023$
- Correct answerD$2024$
Solution
Given, $f(x)=\lim _{y \rightarrow \infty} y\left(x^{1 / y}-1\right)$
and $2022 f\left(\frac{1}{x}\right)+p f(x)=f\left(x^2\right)$
Now, $f(x)=\lim _{y \rightarrow \infty} \frac{\left(x^{1 / y}-1\right)}{(1 / y)}$
On putting $y=1 / t$
$\begin{array}{r}\text { As } y \rightarrow \infty, \text { then } t \rightarrow 0 \\ f(x)=\lim _{t \rightarrow 0} \frac{x^t-1}{t}\end{array}$
Use L'Hospital's rule,
$\begin{aligned} & f(x)=\lim _{t \rightarrow 0} \frac{x^t(\ln x)-0}{(1)}=\ln (x) \\ & f\left(\frac{1}{x}\right)=\ln \left(\frac{1}{x}\right)=-\ln (x)\end{aligned}$
Now, $2022 f\left(\frac{1}{x}\right)+P f(x)=f\left(x^2\right)$
$\begin{aligned} & \Rightarrow \quad 2022[-\ln (x)]+P[\ln x]=\ln \left(x^2\right) \\ & \Rightarrow \quad-2022+P=2 \Rightarrow P=2024\end{aligned}$
Hence, the correct answer is option: 'D'.