JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Analyzing +4 | -1

y = {\tan ^{ - 1}}\left( {\frac{{{{\log }_e}\left( {e/{x^2}} \right)}}{{{{\log }_e}\left( {e{x^2}} \right)}}} \right) + {\tan ^{ - 1}}\left( {\frac{{3 + 2{{\log }_e}x}}{{1 - 6{{\log }_e}x}}} \right),{\text{then}}\,\,\frac{{{d^2}y}}{{d{x^2}}}\,\,{\text{is}}

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Correct option: C

Option D ({0}) is correct because the expression inside the square root must be non-negative, which implies that {x} must be 0.

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

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