Question

The least natural number $a$ for which $x+a{x}^{-2}>2,\forall x\in (0,\infty )$ is ?

• A. $1$
• B. $5$
• C. $2$
• D. None of the above.

Answer 1

The correct option is C $2$

$a>\frac{(2-x)}{x^{-2}}\Rightarrow a>x^2(2-x)$

We'll draw the graph of $x^2(2-x)=2x^2-x$ for $x\in (0,\infty )$

Max value of $f\left(x\right)=2x^2-x^3$ occurs at $f^{\prime }\left(x\right)=0\Rightarrow 4x-3x^2=0\Rightarrow x(4-3x)=0\Rightarrow x=\frac{4}{3}$

$\therefore f\left(\frac{4}{3}\right)=\frac{32}{27}\Rightarrow a$ has to be greater than $\frac{32}{27}$

Hence, the least natural number $a$ can take will be $2$