The least value of $\alpha \in \mathbb{R}$ for which $4 \alpha x^{2}+\frac{1}{x} \geq 1$, for all $x>0$, is

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Sequences and Series

The least value of

α \in R

for which

4 α x^{2} + \frac{1}{x} \geq 1

, for all

x > 0

, is

Answer: (C)

We need

4 α x^{2} + \frac{1}{x} \geq 1

for all

x > 0

.

Let

f (x) = 4 α x^{2} + \frac{1}{x}

. For the inequality to hold for all

x > 0

, we need

min_{x > 0} f (x) \geq 1

.

Taking the derivative:

f^{'} (x) = 8 α x - \frac{1}{x ^{2}}

Setting

f^{'} (x) = 0

8 α x = \frac{1}{x ^{2}} \Rightarrow x^{3} = \frac{1}{8 α} \Rightarrow x = \frac{1}{2 α ^{1/3}}

At this critical point:

f (\frac{1}{2 α ^{1/3}}) = 4 α \cdot \frac{1}{4 α ^{2/3}} + 2 α^{1/3} = α^{1/3} + 2 α^{1/3} = 3 α^{1/3}

For the inequality to hold:

3 α^{1/3} \geq 1 \Rightarrow α^{1/3} \geq \frac{1}{3} \Rightarrow α \geq \frac{1}{27}

The least value is

α = \frac{1}{27}

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Single correct option