JEE Advance - Mathematics (2016 - Paper 1 Offline - No. 6)
The least value of a $$ \in R$$ for which $$4a{x^2} + {1 \over x} \ge 1,$$, for all $$x>0$$. is
$${1 \over {64}}$$
$${1 \over {32}}$$
$${1 \over {27}}$$
$${1 \over {25}}$$
Explanation
It is given that
$$ 4 \alpha x^2+\frac{1}{x} \geq 1 \forall x>0 $$
That is, $4 \alpha x^2 \geq 1-\frac{1}{x}$
$$ 4 \alpha \geq\left(\frac{1}{x^2}-\frac{1}{x^3}\right) \forall x>0 $$
Now, let us consider that
$$ f(x)=\frac{1}{x^2}-\frac{1}{x^3} $$
Therefore, $f^{\prime}(x)=\frac{-2}{x^3}+\frac{3}{x^4}=0$
When $x=3 / 2$ :
$$ \begin{gathered} (4 \alpha) \geq\left(\frac{1}{x^2}-\frac{1}{x^3}\right) \\\\ 4 \alpha \geq\left(\frac{4}{9}-\frac{8}{27}\right) \end{gathered} $$
That is, $$ \alpha \geq \frac{1}{27} $$
and hence the least value of $\alpha$ is
$$ \alpha_{\text {least }}=\frac{1}{27} $$
$$ 4 \alpha x^2+\frac{1}{x} \geq 1 \forall x>0 $$
That is, $4 \alpha x^2 \geq 1-\frac{1}{x}$
$$ 4 \alpha \geq\left(\frac{1}{x^2}-\frac{1}{x^3}\right) \forall x>0 $$
Now, let us consider that
$$ f(x)=\frac{1}{x^2}-\frac{1}{x^3} $$
Therefore, $f^{\prime}(x)=\frac{-2}{x^3}+\frac{3}{x^4}=0$
When $x=3 / 2$ :
$$ \begin{gathered} (4 \alpha) \geq\left(\frac{1}{x^2}-\frac{1}{x^3}\right) \\\\ 4 \alpha \geq\left(\frac{4}{9}-\frac{8}{27}\right) \end{gathered} $$
That is, $$ \alpha \geq \frac{1}{27} $$
and hence the least value of $\alpha$ is
$$ \alpha_{\text {least }}=\frac{1}{27} $$
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