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AP EAPCET 08-Jul-2022 Shift 2 Paper

Section: Mathematics

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Question : 61 of 160

Marks: +1, -0

Let

f(x)=\lim\limits_{y\rightarrow\infty}y\left(x^{1/y}-1\right)

2022 f\left(\frac{1}{x}\right)+P f(x)=f(x^2)

P=

2020
2021
2023
2024

Solution: 👈: Video Solution

Given,

f(x)=\lim\limits_{y\rightarrow\infty}y\left(x^{1/y}-1\right)

and

2022 f\left(\frac{1}{x}\right)+p f(x)=f(x^2)

Now,

f(x)=\lim\limits_{y\rightarrow\infty}\frac{x^{1/y}-1}{1/y}

On putting

y=\frac{1}{t}

As

y\rightarrow\infty

, then

t\rightarrow0

f(x)=\lim\limits_{t\rightarrow0}\frac{x^{t}-1}{t}

Use

L'

Hospital's rule,

f(x)=\lim\limits_{t\rightarrow0}\left(x^{t}\ln x-0\right)=

ln

(x)

f\left(\frac{1}{x}\right)=\ln\left(\frac{1}{x}\right)=-\ln(x)

Now,

2022 f\left(\frac{1}{x}\right)+P f(x)=f(x^2)

\Rightarrow 2022[-\ln(x)]+P[\ln x]=\ln(x^2)

\Rightarrow -2022+P=2 \Rightarrow P=2024

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