Question

For $x>0$, which of the following is true :

• A. $1+2\ln x<x^2$
• B. $1+2\ln x>x^2$
• C. $1+2\ln x\le x^2$
• D. $1+2\ln x\ge x^2$

Answer 1

The correct option is C $1+2\ln x\le x^2$

Let $f\left(x\right)=x^2-2\ln x-1$
$\Rightarrow f^{\prime }\left(x\right)=2x-\frac{2}{x}$
$\Rightarrow x-\frac{1}{x}=0$
$\Rightarrow x^2=1$
$\Rightarrow x=1\{\because x=-1\text{not in domain}\}$
The derivative is negative to the left of this point and positive to the right of this point.
Hence, the function has a minimum at this point i.e.
$f\left(1\right)=1-2\ln 1-1=0$
Thus, $f\left(x\right)\ge 0$ for $x>0$
$\therefore x^2-2\ln x-1\ge 0$
$\Rightarrow 1+2\ln x\le x^2$