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:

  1. 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 $\frac{1}{27}$.



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