Mathematics · 2020
JEE Main · 5 September 2020, Shift 1 · Q59
If the function f(x)= {k_1(x-π)^2-1, x ≤ π; k_2 cos x, x>π} is twice differentiable, then the ordered pair ( k_1, k_2) is equal to:
If the function
$\displaystyle f(x)=\left\{\begin{array}{ll}\mathrm{k}_1(x-\pi)^2-1, & x \leq \pi \\ \mathrm{k}_2 \cos x, & x>\pi\end{array}\right.$ is twice differentiable, then the ordered pair $\displaystyle \left(\mathrm{k}_1, \mathrm{k}_2\right)$ is equal to :
Official answer
(1)
$\displaystyle \left(\frac{1}{2}, 1\right)$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.