Let f(x)= x^2-2+1 x , where is a real constant. The smallest for … - Vidyadip

MCQ

Let $f(x)=\alpha x^2-2+\frac{1}{x}$, where $\alpha$ is a real constant. The smallest $\alpha$ for which $f(x) \geq 0$ for all $x > 0$ is

A
$\frac{2^2}{3^3}$
B
$\frac{2^3}{3^3}$
C
$\frac{2^4}{3^3}$
✓
$\frac{2^{\sqrt{3}}}{3^3}$

✓

Answer

Correct option: D.

$\frac{2^{\sqrt{3}}}{3^3}$

d
(d)

We have, $f(x)=\alpha x^2-2+\frac{1}{x}$

$f(x)-\frac{\alpha x^3-2 x+1}{x}, x>0$

$f(x) \geq 0$ if $\alpha x^3-2 x+1>0$

Let $g(x)=\alpha x^3-2 x+1$

$\therefore \quad g^{\prime}(x)=3 \alpha x^2-2=0$

$\Rightarrow x=\pm \sqrt{\frac{2}{3 \alpha}}$

Clearly, $x=\sqrt{\frac{2}{3 \alpha}}$ is point of minima.

$\therefore g\left(\sqrt{\frac{2}{3 \alpha}}\right) \geq 0 \Rightarrow g\left(\sqrt{\frac{2}{3 \alpha}}\right)=\alpha\left(\frac{2}{3 \alpha}\right)^{3 / 2}$

$-2\left(\frac{2}{3 \alpha}\right)^{1 / 2}+1 \geq 0$

$\Rightarrow \quad\left(\frac{2}{3 \alpha}\right)^{1 / 2}\left(\frac{2}{3}-2\right)+1 \geq 0$

$\Rightarrow\left(\frac{2}{3 \alpha}\right)^{1 / 2} \leq \frac{3}{4} \Rightarrow \frac{2}{3 \alpha} \leq \frac{9}{16} \Rightarrow \alpha \geq \frac{32}{27}$

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