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AP EAPCET 25-Aug-2021 Shift 2 Paper

Section: Mathematics

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Question : 3 of 160

Marks: +1, -0

Define

f: R \rightarrow R

f(x)=\cos ({\tan}^{-1}(\sin ({\tan}^{-1} x)))

\lim\limits_{x\to\infty}(f o f) x=

$\frac{3}{2\sqrt{3}}$
$\frac{\sqrt{2}}{3}$
$\sqrt{\frac{2}{3}}$
$\frac{2}{3\sqrt{3}}$

Solution: 👈: Video Solution

Given,

f: R \rightarrow R

by

f(x)=\cos [\tan^{-1}\{\sin (\tan^{-1} x)\}]

To find,

\lim\limits_{x\to\infty}(f o f)(x)

Let us find

(f o f)(x)

first,

∴

(fo f)(x)=f(f(x))

=f(\cos (\tan^{-1}(\sin (\tan^{-1} x))))

=f\left(\cos \left(\tan^{-1}\left(\sin \left(\sin^{-1} \frac{x}{\sqrt{1+x^{2}}}\right)\right)\right)\right)

\{\because \tan^{-1} x = \sin^{-1} \frac{x}{\sqrt{1+x^{2}}}\}

=f\left(\cos^{1}\left(\tan^{-1} \frac{x}{\sqrt{1+x^{2}}}\right)\right)

\{\because \tan^{-1} x = \cos^{-1} \frac{1}{\sqrt{1+x^{2}}}\}

=f\left(\cos \left(\cos^{-1} \frac{1}{\sqrt{1+\left(\frac{x}{\sqrt{1+x^{2}}}\right)^{2}}}\right)\right)

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

∴

(f o f)(x)=\frac{\sqrt{1+\left(\sqrt{\frac{1+x^{2}}{1+2x^{2}}}\right)^{2}}}{1+2\left(\sqrt{\frac{1+x^{2}}{1+2x^{2}}}\right)^{2}}

where

f(x)=\sqrt{\frac{1+x^{2}}{1+2x^{2}}}

=\sqrt{\frac{1+2x^{2}+1+x^{2}}{1+2x^{2}+2+2x^{2}}}=\sqrt{\frac{2+3x^{2}}{3+4x^{2}}}

Let

x=\frac{1}{y}

, then

y\to 0

, when

x\to\infty

∴

\lim\limits_{x\to\infty}\sqrt{\frac{2+3x^{2}}{3+4x^{2}}}=\lim\limits_{y\to 0}\sqrt{\frac{2+\frac{3}{y^{2}}}{3+\frac{4}{y^{2}}}}

=\lim\limits_{y\to 0}\sqrt{\frac{2y^{2}+3}{3y^{2}+4}}=\sqrt{\frac{3}{4}}

=\frac{\sqrt{3}}{2}

or

\frac{3}{2\sqrt{3}}

∴

\lim\limits_{x\to\infty}(f o f)(x)=\frac{3}{2\sqrt{3}}

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