Q31 / 50 · Manipal (MET) 2024 · Mathematics
If expression $x+\frac{1}{x^2},|x\gt0|$ attains is minimum value at $x=\alpha$, then $\alpha^3$ is
Correct answer: 2
Solution
$\because x+\frac{1}{x^2}=\frac{x}{2}+\frac{x}{2}+\frac{1}{x^2}$
$\text {and } \mathrm{AM} \geq \mathrm{GM}$
$\Rightarrow \frac{x}{2}+\frac{x}{2}+\frac{1}{x^2} \geq\left(\frac{x}{2} \times \frac{x}{2} \times \frac{1}{x^2}\right)^{\frac{1}{3}}$
$\Rightarrow \frac{x}{2}+\frac{x}{2}+\frac{1}{x^2} \geq\left(\frac{1}{4}\right)^{\frac{1}{3}}$
Minimum value of $x+\frac{1}{x^2}=\sqrt[3]{\frac{1}{4}}$
and for minimum value
$\frac{x}{2}=\frac{1}{x^2} \Rightarrow x^3=2$
Hence, $\alpha^3=2$
Hence, the correct answer is 2.