Let f : R rightarrow R be defined as f(x) = 3 | vedclass

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(D) Given $f(x) = 3^{-|x|} - 3^x + \operatorname{sgn}(e^{-x}) + 2$.Since $e^{-x} > 0$ for all $x \in R$,$\operatorname{sgn}(e^{-x}) = 1$.Thus,$f(x) =

Vedclass · Accepted Answer

$f$ is neither injective nor surjective

Vedclass · Answer

(D) Given $f(x) = 3^{-|x|} - 3^x + \operatorname{sgn}(e^{-x}) + 2$.Since $e^{-x} > 0$ for all $x \in R$,$\operatorname{sgn}(e^{-x}) = 1$.Thus,$f(x) = 3^{-|x|} - 3^x + 1 + 2 = 3^{-|x|} - 3^x + 3$.Case $1$: If $x Since $f(x) = 3$ for all $x Case $2$: If $x \ge 0$,then $|x| = x$,so $f(x) = 3^{-x} - 3^x + 3$.As $x 	o \infty$,$f(x) 	o -\infty$,and at $x = 0$,$f(0) = 3^0 - 3^0 + 3 = 3$.The range of $f(x)$ for $x \ge 0$ is $(-\infty, 3]$.Combining both cases,the range of $f$ is $(-\infty, 3]$.Since the range $(-\infty, 3] 
eq R$,the function is not surjective.Therefore,$f$ is neither injective nor surjective.

List-$I$	List-$II$
$(A) f(-5)+f(0)+f(-1)$	$(I) 16$
$(B) f(f(5)+10f(-3))$	$(II) 40$
$(C) f(f(-4))$	$(III) -31$
$(D) f(f(f(1)))$	$(IV) -12$
	$(V) 19$

Let $f : R \rightarrow R$ be defined as $f(x) = 3^{-|x|} - 3^x + \operatorname{sgn}(e^{-x}) + 2$ (where $\operatorname{sgn}(x)$ denotes the signum function of $x$). Then which one of the following is correct?

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