Functions 3 Question 2
2. If $f(x)=\sin x+\cos x, g(x)=x^{2}-1$, then $g{f(x)}$ is invertible in the domain
(2004, 1M)
(a) $0, \frac{\pi}{2}$
(b) $-\frac{\pi}{4}, \frac{\pi}{4}$
(c) $-\frac{\pi}{2}, \frac{\pi}{2}$
(d) $[0, \pi]$
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Answer:
Correct Answer: 2. (d)
Solution:
- We have, $f(x)=\frac{|x-1|}{x}=\begin{array}{cc}-\frac{(x-1)}{x}, & \text { if } 0<x \leq 1 \ \frac{x-1}{x}, & \text { if } x>1\end{array}$
$$ \begin{array}{ll} \frac{1}{x}-1, & \text { if } 0<x \leq 1 \\ 1-\frac{1}{x}, & \text { if } x>1 \end{array} $$
Now, let us draw the graph of $y=f(x)$
Note that when $x \rightarrow 0$, then $f(x) \rightarrow \infty$, when $x=1$, then $f(x)=0$, and when $x \rightarrow \infty$, then $f(x) \rightarrow 1$
Clearly, $f(x)$ is not injective because if $f(x)<1$, then $f$ is many one, as shown in figure.
Also, $f(x)$ is not surjective because range of $f(x)$ is $[0, \infty[$ and but in problem co-domain is $(0, \infty)$, which is wrong.
$\therefore f(x)$ is neither injective nor surjective
