Functions Ques 55
- Let $f$ be a one-one function with domain ${x, y, z}$ and range ${1,2,3}$. It is given that exactly one of the following statements is true and the remaining two are false $f(x)=1, f(y) \neq 1, f(z) \neq 2$ determine $f^{-1}(1)$.
(1982, 2M)
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Answer:
Correct Answer: 55.(b)
Solution:
- Given, $f:[0, \infty) \rightarrow[0, \infty)$
Here, domain is $[0, \infty)$ and codomain is $[0, \infty)$. Thus, to check one-to-one
Since, $f(x)=\frac{x}{1+x} \Rightarrow f^{\prime}(x)=\frac{1}{(1+x)^{2}}>0, \forall x \in[0, \infty)$
$\therefore f(x)$ is increasing in its domain. Thus, $f(x)$ is one-one in its domain. To check onto (we find range)
$$ \begin{alignedat} & \text { Again, } \quad y=f(x)=\frac{x}{1+x} \\ & \Rightarrow \quad y+y x=x \\ & \Rightarrow \quad x=\frac{y}{1-y} \Rightarrow \frac{y}{1-y} \geq 0 \end{aligned} $$
Since $x \geq 0$, therefore $0 \leq y < 1$
i.e. Range $\neq$ Codomain
$\therefore f(x)$ is one-one but not onto.
BETA
