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(B) Given $f(x) = \sqrt{|x|} - \log(1 + |x|)$.Since $f(x)$ is an even function,we consider $g(t) = \sqrt{t} - \log(1 + t)$ for $t \geq 0$,where $t = |

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$I$ is false and $II$ is true

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(B) Given $f(x) = \sqrt{|x|} - \log(1 + |x|)$.Since $f(x)$ is an even function,we consider $g(t) = \sqrt{t} - \log(1 + t)$ for $t \geq 0$,where $t = |x|$.To check the behavior of $g(t)$,we find its derivative:$g'(t) = \frac{1}{2\sqrt{t}} - \frac{1}{1 + t} = \frac{1 + t - 2\sqrt{t}}{2\sqrt{t}(1 + t)} = \frac{(\sqrt{t} - 1)^2}{2\sqrt{t}(1 + t)}$.Since $g'(t) > 0$ for all $t > 0$ $(t 
eq 1)$,the function $g(t)$ is strictly increasing for $t \geq 0$.As $t 	o \infty$,$g(t) = \sqrt{t}(1 - \frac{\log(1 + t)}{\sqrt{t}}) 	o \infty$. Thus,$f(x)$ is not bounded above,so assertion $I$ is false.At $t = 0$,$g(0) = 0$. Since $g(t)$ is increasing for $t \geq 0$,the minimum value of $g(t)$ is $g(0) = 0$. Thus,$f(x) \geq 0$ for all $x$. So,assertion $II$ is true.Therefore,$I$ is false and $II$ is true.

Let $R$ be the set of real numbers and $f: R \rightarrow R$ be given by $f(x) = \sqrt{|x|} - \log(1 + |x|)$. We now make the following assertions:
$I.$ There exists a real number $A$ such that $f(x) \leq A$ for all $x$.
$II.$ There exists a real number $B$ such that $f(x) \geq B$ for all $x$.

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Let $R$ be the set of real numbers and $f: R \rightarrow R$ be given by $f(x) = \sqrt{|x|} - \log(1 + |x|)$. We now make the following assertions:$I.$ There exists a real number $A$ such that $f(x) \leq A$ for all $x$.$II.$ There exists a real number $B$ such that $f(x) \geq B$ for all $x$.