JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Medium

1 of 1

Application +4 | -1

{\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \frac{{\;\;\;\;\;3\pi }}{2}\;\mathrm{and}\;f\left( 1 \right) = 2

f\left( {p + q} \right) = f\left( p \right).f\left( q \right),\forall \;p,\;q \in R,\;

then

{x^{f\left( 1 \right)}} + {y^{f\left( 2 \right)}} + {z^{f\left( 3 \right)}} - \frac{{\left( {x + y + z} \right)}}{{{x^{f\left( 1 \right)}} + {y^{f\left( 2 \right)}} + {z^{f\left( 3 \right)}}}}\;

is equal to

View answer and explanation

Correct option: C

Option C (2) is correct because given the conditions, x=y=z=1, f(1)=2, f(2)=4, f(3)=8, which simplifies the expression to 3 - 1 = 2.

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)