Mathematics · 2019
JEE Main · 9 January 2019, Shift 2 · Q75
If ∫_0^π / 3 (tan θ)/(√(2 k sec θ)) d θ=1-1/(√ 2),( k >0), then the value of k is:
If $\displaystyle \int_0^{\pi / 3} \frac{\tan \theta}{\sqrt{2 \mathrm{k} \sec \theta}} \mathrm{d} \theta=1-\frac{1}{\sqrt{2}},(\mathrm{k}>0)$, then the
value of $\displaystyle k$ is :
Official answer
(1)
$\displaystyle 2$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.