Mathematics · 2024
JEE Main · 6 April 2024, Shift 2 · Q12
If ∫ 1/(a^2 sin^2 x+ b^2 cos^2 x) d x=1/12 tan^-1(3 tan x)+ constant, then the maximum value of a sin x+b cos x, is:
If $\displaystyle \int \frac{1}{\mathrm{a}^2 \sin ^2 x+\mathrm{b}^2 \cos ^2 x} \mathrm{~d} x=\frac{1}{12} \tan ^{-1}(3 \tan x)+$ constant, then the maximum value of $\displaystyle a \sin x+b \cos x$, is :
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle \sqrt{40}$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.