JEE Advanced 2016 Paper 1Question 1
The least value of $\alpha\in\mathbb{R}$ for which $4\alpha x^2+\frac{1}{x}\ge1$, for all $x>0$, is
Differentiation and its Application
Solution
Concept. Put $\alpha$ alone on one side; the least $\alpha$ is the maximum of the other side. Formulas used. (1) $\alpha\ge h(x)$ for all $x>0$ $\iff$ $\alpha\ge\max h$Step 1 - Isolate $\alpha$. For $x>0$: $$4\alpha x^2\ge1-\frac1x\ \Longrightarrow\ \alpha\ge\frac{x-1}{4x^3}=h(x).$$ Step 2 - Maximise $h$. $$\frac{d}{dx}\frac{x-1}{x^3}=\frac{x^3-3x^2(x-1)}{x^6}=\frac{3-2x}{x^4}.$$ This is $0$ at $x=\frac32$, positive before and negative after: a maximum. Step 3 - Value. $h\left(\frac32\right)=\dfrac{1/2}{4\cdot\frac{27}{8}}=\dfrac{1}{27}$. By (1), the least $\alpha$ is $\frac1{27}$. Answer: (C)