JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Analyzing +4 | -1

a,\;b

are positive quantities and if

{a_1} = \frac{{a + b}}{2},\;{b_1} = \sqrt {{a_1}b} ,\;{a_2} = \frac{{{a_1} + {b_1}}}{2},\;{b_2} = \sqrt {{a_2}{b_1}}

and so on, then

View answer and explanation

Correct option: B

{b_\infty } = \frac{{\sqrt {{b^2} - {a^2}} }}{{{{\cos }^{ - 1}}\left( {\frac{a}{b}} \right)}}

Option D is incorrect. The correct answer is

{b_\infty } = \frac{{\sqrt {{{{b^2} - {a^2}}}} }}{{{{\cos }^{ - 1}}\left( {\frac{a}{b}} \right)}}

which is not listed, and the provided solution derives this result using trigonometric substitution and limit evaluation.

More Questions from INVERSE TRIGONOMETRY

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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)