Application of Derivatives 4 Question 14
####15. The least value of $\alpha \in R$ for which $4 \alpha x^{2}+\frac{1}{x} \geq 1$, for all $x>0$, is
(a) $\frac{1}{64}$
(b) $\frac{1}{32}$
(c) $\frac{1}{27}$
(d) $\frac{1}{25}$
(2016 Adv.)
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Answer:
Correct Answer: 15. (c)
Solution:
- Here, to find the least value of $\alpha \in R$, for which $4 \alpha x^{2}+\frac{1}{x} \geq 1$, for all $x>0$.
i.e. to find the minimum value of $\alpha$ when $y=4 \alpha x^{2}+\frac{1}{x} ; x>0$ attains minimum value of $\alpha$.
$\therefore \quad \frac{d y}{d x}=8 \alpha x-\frac{1}{x^{2}}$
Now,
$ \frac{d^{2} y}{d x^{2}}=8 \alpha+\frac{2}{x^{3}} $
When
At $x=\frac{1}{8 \alpha}^{1 / 3}, \frac{d^{2} y}{d x^{2}}=8 \alpha+16 \alpha=24 \alpha$, Thus, $y$ attains minimum when $x=\frac{1}{8 \alpha}^{1 / 3} ; \alpha>0$.
$\therefore y$ attains minimum when $x=\frac{1}{8 \alpha}^{1 / 3}$.
$ \begin{aligned} & \text { i.e. } \quad 4 \alpha \frac{1}{8 \alpha}^{2 / 3}+(8 \alpha)^{1 / 3} \geq 1 \\ & \Rightarrow \quad \alpha^{1 / 3}+2 \alpha^{1 / 3} \geq 1 \\ & \Rightarrow \quad 3 \alpha^{1 / 3} \geq 1 \Rightarrow \alpha \geq \frac{1}{27} \end{aligned} $
Hence, the least value of $\alpha$ is $\frac{1}{27}$.
