JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Application +4 | -1

f(x)=\mathrm{e}^{\operatorname{cosecos}^{-1} x^2+\tan ^{-1} \cot ^{-1} x^2}

then minimum value of

f(x)

View answer and explanation

Correct option: B

e^2

Option B (

e^2

) is correct because the minimum value of the function is obtained when

x^2 + \frac{1}{x^2}

is minimized, which occurs at

x^2 = 1

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\cos ^{-1}(\cos 12)-\sin ^{-1}(\sin 12)

\sin \left(2 \tan ^{-1}(1 / 3)\right)+

\cos \left(\tan ^{-1} 2 \sqrt{2}\right)

\cos ^ {-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right\}=

\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1

x

{x^2} + {y^2} + {z^2} = {r^2},

{\tan ^{ - 1}}\left( {\frac{{xy}}{{zr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{yz}}{{xr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{xz}}{{yr}}} \right)\;

2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)