JEE PYQ: Application Of Derivatives Question 67
Question 67 - 2019 (12 Apr Shift 1)
If $m$ is the minimum value of $k$ for which the function $f(x) = x\sqrt{kx - x^2}$ is increasing in the interval $[0, 3]$ and $M$ is the maximum value of $f$ in $[0, 3]$ when $k = m$, then the ordered pair $(m, M)$ is equal to:
(1) $(4, 3\sqrt{2})$
(2) $(4, 3\sqrt{3})$
(3) $(3, 3\sqrt{3})$
(4) $(5, 3\sqrt{6})$
Show Answer
Answer: (2)
Solution
$f’(x) = \frac{3kx - 4x^2}{2\sqrt{kx-x^2}} \ge 0$ for $x \in [0,3]$ requires $3k \ge 4x$ for $x \in [0,3]$, i.e., $k \ge 4$. So $m = 4$. $M = f(3) = 3\sqrt{12-9} = 3\sqrt{3}$.
BETA
