Mathematics · 2016
JEE Main · 10 April 2016, Shift 1 · Q71
Let a, b ∈ R,(a ≠ 0). If the function f defined as f(x)= {(2 x^2)/a, 0 ≤ x<1; a, 1 ≤ x<√ 2; (2 b^2-4 b)/x^3, √ 2 ≤ x<∞} is continuous in the interval…
Let $\displaystyle a, b \in \mathbf{R},(a \neq 0)$. If the function $\displaystyle f$ defined as
$$f(x)= \begin{cases}\frac{2 x^2}{a}, & 0 \leq x<1 \\ a, & 1 \leq x<\sqrt{2} \\ \frac{2 b^2-4 b}{x^3}, & \sqrt{2} \leq x<\infty\end{cases}
$$
is continuous in the interval $\displaystyle [0, \infty)$, then an ordered pair $\displaystyle (a, b)$ is :
Official answer
(1)
$\displaystyle (\sqrt{2}, 1-\sqrt{3})$
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JEE Main 2016 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.