JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

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Analyzing +4 | -1

The value of

'a'

for which

a{x^2} + {\sin ^{ - 1}}\left( {{x^2} - 2x + 2} \right) + {\cos ^{ - 1}}\left( {{x^2} - 2x + 2} \right) = 0

has a real solution, is

View answer and explanation

Correct option: C

- \frac{\pi }{2}

Option C (

- \frac{\pi }{2}

) is correct because substituting x=1 into the equation and solving for 'a' yields a = -π/2.

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\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)