JEE Main 2019 (Online) 12th April Morning ShiftQuestion 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:
Differentiation and its Application
Solution
Concept. $f'$ has numerator $3kx-4x^2$; it must be $\ge0$ for all $x\in[0,3]$. Formulas used. (1) $f'(x)=\sqrt{kx-x^2}+\dfrac{x(k-2x)}{2\sqrt{kx-x^2}}=\dfrac{3kx-4x^2}{2\sqrt{kx-x^2}}$Step 1 - Condition. By (1), $x(3k-4x)\ge0$ on $[0,3]$ needs $3k\ge4\cdot3$, i.e. $k\ge4$. (This also keeps $kx-x^2\ge0$ on $[0,3]$.) So $m=4$. Step 2 - Maximum. With $k=4$, $f$ increases on $[0,3]$, so $M=f(3)=3\sqrt{12-9}=3\sqrt3$. Answer: (A)