Mathematics · 2019
JEE Main · 9 January 2019, Shift 2 · Q71
If x=3 tan t and y=3 sec t, then the value of (d^2 y)/(d x^2) at t =(π)/4, is:
If $\displaystyle x=3 \tan \mathrm{t}$ and $\displaystyle y=3 \sec \mathrm{t}$, then the value
of $\displaystyle \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}$ at $\displaystyle \mathrm{t}=\frac{\pi}{4}$, is :
Official answer
(2)
$\displaystyle \frac{1}{6 \sqrt{2}}$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.