JEE MATHEMATICS MCQ: MAXIMA & MINIMA

1 Hard

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Analyzing +4 | -1

a > b > 0

,then maximum value of

\frac{{ab\left( {{a^2} - {b^2}} \right)\sin x\cos x}}{{{a^2}{{\sin }^2}x + {b^2}{{\cos }^2}x}},\;x \in \left( {0,\;\pi /2} \right)

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Correct option: B

\frac{{{a^2} - {b^2}}}{2}

Option B (

\frac{{{a^2} - {b^2}}}{2}

) is correct because it represents the maximum value of the given expression, achieved when the denominator is minimized.

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