QJEEAdv16Ma40af: JEE Advanced 2016 Mathematics - Class 12 JEEAdv16Ma40af Solution: The least value of a ∈ R for which 4 a x 2 + 1 x ≥ 1 , , for…

Question

Verified Solution • 2025–26 Session moderate

The least value of a

\in R

for which

4 a x^{2} + \frac{1}{x} \geq 1,

, for all

x > 0

. is

Accepted Answer

It is given that

4 α x^{2} + \frac{1}{x} \geq 1 \forall x > 0

That is,

4 \alpha x^2 \geq 1-\frac{1}{x}

4 α \geq (\frac{1}{x^{2}} - \frac{1}{x^{3}}) \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{matrix} (4 α) \geq (\frac{1}{x^{2}} - \frac{1}{x^{3}}) \\ 4 α \geq (\frac{4}{9} - \frac{8}{27}) \end{matrix}

That is,

α \geq \frac{1}{27}

and hence the least value of

\alpha

is

α_{least} = \frac{1}{27}

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