Mathematics · 2024
JEE Main · 29 January 2024, Shift 2 · Q9
Let y= log_e((1-x^2)/(1+x^2)),-1<x<1. Then at x=1/2, the value of 225(y^′-y^′ ′) is equal to
Let $\displaystyle y=\log _e\left(\frac{1-x^2}{1+x^2}\right),-1<x<1$. Then at $\displaystyle x=\frac{1}{2}$, the value of $\displaystyle 225\left(y^{\prime}-y^{\prime \prime}\right)$ is equal to
Official answer
From NTA’s final answer key for this paper.
(4)
$\displaystyle 736$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.