Question

Let $R$ be the set of real numbers and $f:R\to R$ be given by $f\left(x\right)=\sqrt{|x|}-\log (1+|x|).$ We now make the following assertions :

I. There exists a real number $A$ such that $f\left(x\right)\le A$ for all $x.$

II. There exists a real number $B$ such that $f\left(x\right)\ge B$ for all $x$.

• A. I is true and II is false
• B. I is false and II is true
• C. I and II both are true
• D. I and II both are false

Answer 1

The correct option is B I is false and II is true

$f\left(x\right)=\sqrt{|x|}-\log (1+|x|)$
The function will be symmetrical about $y$-axis.
$f\left(x\right)=\{\begin{array}{cc}\sqrt{x}-\log (1+x)& x\ge 0\\ \sqrt{-x}-\log (1-x)& x<0\end{array}$

For $x\ge 0$
$f\left(x\right)=\sqrt{x}-\log (1+x)\Rightarrow f^{\prime }\left(x\right)=\frac{1}{2\sqrt{x}}-\frac{1}{(1+x)\ln 10}$
$\therefore f^{\prime }\left(x\right)>0\forall x\ge 0$
So $f\left(x\right)$ is a increasing function in $[0,\infty )$
$f\left(0\right)=0\therefore f\left(x\right)>0\forall x\in [0,\infty )$
Also, $f$ is symmetric about $y$-axis.
So, $f\left(x\right)>0\forall x\in R$

As $x\to \infty ,f\left(x\right)\to \infty$
$\therefore$ Option (b) is correct.