Mathematics · 2019
JEE Main · 10 January 2019, Shift 2 · Q76
If ∫_0^x f( t ) dt =x^2+∫_x^1 t^2 f( t ) dt, then f^′(1/2) is:
If $\displaystyle \int_0^x f(\mathrm{t}) \mathrm{dt}=x^2+\int_x^1 \mathrm{t}^2 f(\mathrm{t}) \mathrm{dt}$, then
$\displaystyle f^{\prime}\left(\frac{1}{2}\right)$ is :
Official answer
(3)
$\displaystyle \frac{24}{25}$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.