NIMCET 2021 Question Paper with solutions

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Question : 2 of 120

Marks: +1, -0

The function

f(x)={x}/{1+x \tan x} 0 \≤ x \≤ {\π}/{2}

is maximum when

$x=\sec x$
$x=\tan x$
$x=\cos x$
None of the above

Solution:

y={x}/{1+x \tan x}

y={1}/{{1}/{x}+\tan x}

f(x)={1}/{x}+\tan x

f'(x)={-1}/{x^{2}}+\sec ^{2} x

For critical point

f'(x)=0

{-1}/{x^{2}}=\sec ^{2} x

x=\±\cos x

now

f"(x)={2}/{x^{3}}+2 \sec ^{2} x \tan x

f"(-\cos x)=-v e

f"(\cos x)=+v e

(x)={1}/{x}+\tan x

is minimum at

x=\cos x

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