JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Application +4 | -1

The minimum value of

\frac{12}{\pi}\left(\operatorname{cosec}^{-1}\left[3 x^2+\frac{5}{4}\right]+\sec ^{-1}\left[3 x^2+\frac{1}{4}\right]\right)

is equal to (where [.] denotes the greatest integer function)

View answer and explanation

Correct option: A

Option A (2) is correct because by analyzing the range of the inverse trigonometric functions and the greatest integer function, the minimum value is found to be 2.

More Questions from INVERSE TRIGONOMETRY

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The value of

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

is- The value of

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

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

is Find

\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

, then

x

is equal to If

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

then

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

is equal to

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

is equal to