JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Application +4 | -1

- \frac{1}{2} \le x \le \frac{1}{2}

, then

{\cos ^{ - 1}}\left( {4{x^3} - 3x} \right)

equals

View answer and explanation

Correct option: B

2\pi - 3{\cos ^{ - 1}}x

Option D (None of these) is correct because the correct expression is

3{\cos ^{ - 1}}x

when

0 \leq x \leq 1

, and

2\pi - 3{\cos ^{ - 1}}x

when

-1 \leq x \leq -1/2

. Since the given range is

-1/2 \le x \le 1/2

, the expression is

3\cos^{-1}x

only when

1/2 \le x \le 1

More Questions from INVERSE TRIGONOMETRY

Practice All Chapter MCQs →

\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)