Let R be the set of real numbers and f: R rig | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 = |

Vedclass · Accepted Answer

$I$ is false and $II$ is true

Vedclass · Answer

(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 $x$.

Similar Questions

The domain of the real-valued function $f(x) = \frac{3}{4-x^2} + \log_{10}(x^3-x)$ is

The range of the function $f : R \rightarrow R$,defined by $f(x) = \frac{(x + 1)^4}{x^4 + 1}$,is

If the range of $f(x) = \frac{2x^4-14x^2-8x+49}{x^4-7x^2-4x+23}$ is $(a, b]$,then $(a + b)$ is

The domain of the function defined by $f(x) = \frac{-5}{4x^2+1} + \sqrt{x^2-4}$ is

$A$ real-valued function $f: A \rightarrow B$ defined by $f(x) = \frac{4-x^2}{4+x^2}$ for all $x \in A$ is a bijection. If $-4 \in A$,then $A \cap B =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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 $x$.