Mathematics · 2021
JEE Main · 18 March 2021, Shift 1 · Q70
If f(x)= {1/(|x|);|x| ≥ 1; a x^2+ b;|x|<1} is differentiable at every point of the domain, then the values of a and b are respectively:
If $\displaystyle f(x)=\left\{\begin{array}{ll}\frac{1}{|x|} \quad ;|x| \geqslant 1 \\ \mathrm{a} x^2+\mathrm{b} ;|x|<1\end{array}\right.$ is differentiable at every point of the domain, then the values of $\displaystyle \mathbf{a}$ and $\displaystyle \mathbf{b}$ are respectively :
Official answer
(2)
$\displaystyle -\frac{1}{2}, \frac{3}{2}$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.