Mathematics · 2019
JEE Main · 12 April 2019, Shift 1 · Q73
If m is the minimum value of k for which the function f(x)=x √(k x-x^2) is increasing in the interval [0,3] and M is the maximum value of f in [0, 3]…
If $\displaystyle m$ is the minimum value of $\displaystyle k$ for which the function $\displaystyle f(x)=x \sqrt{\mathrm{k} x-x^2}$ is increasing in the interval $\displaystyle [0,3]$ and M is the maximum value of $\displaystyle f$ in [$\displaystyle 0$, $\displaystyle 3$] when $\displaystyle \mathrm{k}=\mathrm{m}$, then the ordered pair $\displaystyle (\mathrm{m}, \mathrm{M})$ is equal to :
Official answer
(3)
$\displaystyle (4,3 \sqrt{3})$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.