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$\frac{2^4}{3^3}$
- ✓$\frac{2^{\sqrt{3}}}{3^3}$
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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$\overline{O P} \cdot \overline{O Q}+\overline{O R} \cdot \overline{O S}=\overline{O R} \cdot \overline{O P}+\overline{O Q} \cdot \overline{O S}=\overline{O Q} \cdot \overline{O R}+\overline{O P} \cdot \overline{O S}$
Then the triangle $P Q R$ has $S$ as its