Mathematics · 2021
JEE Main · 4 August 2021, Shift 1 · Q70
The equation of the tangent to the curve y=∫_x^2^x^3 (d t)/(√(1+t^2)) at the point whose abscissa is 1, is
The equation of the tangent to the curve $\displaystyle y=\int_{x^2}^{x^3} \frac{d t}{\sqrt{1+t^2}}$ at the point whose abscissa is $\displaystyle 1$, is
Official answer
(4)
$\displaystyle x-\sqrt{2} y-1=0$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.