JEE Main 2022 (Online) 28th July Morning ShiftQuestion 1
If the minimum value of $f(x)=\frac{5x^2}{2}+\frac{\alpha}{x^5}$, $x>0$, is $14$, then the value of $\alpha$ is equal to :
Differentiation and its Application
Solution
Concept. Find the critical point in terms of $\alpha$ and set the minimum value equal to $14$. Formulas used. (1) $f'(x)=5x-\dfrac{5\alpha}{x^6}$Step 1 - Critical point. By (1), $f'=0$ gives $x^7=\alpha$. $f'<0$ before and $f'>0$ after, so it is the minimum. Step 2 - Minimum value. With $x^7=\alpha$, $\frac{\alpha}{x^5}=x^2$, so $$f_{\min}=\frac52x^2+x^2=\frac72x^2=14\ \Longrightarrow\ x^2=4,\ x=2.$$ Step 3 - Value. $\alpha=x^7=128$. Answer: (C)