Functions 3 Question 5
5. If the function $f:[1, \infty) \rightarrow[1, \infty)$ is defined by $f(x)=2^{x(x-1)}$, then $f^{-1}(x)$ is
(1999, 2M)
(a) $\frac{1}{2}^{x(x-1)}$
(b) $\frac{1}{2}\left(1+\sqrt{1+4 \log _2 x}\right)$
(c) $\frac{1}{2}\left(1-\sqrt{1+4 \log _2 x}\right)$
(d) not defined
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Answer:
Correct Answer: 5. (b)
Solution:
- Given, $f(n)=\frac{n+1}{\frac{n}{2}}$, if $n$ is even,
and $\quad g(n)=n-(-1)^{n}={\begin{array}{l}n+1, \text { if } n \text { is odd } \ n-1, \text { if } n \text { is even }\end{array}\right.$
Now, $f(g(n))={\begin{array}{l}f(n+1), \text { if } n \text { is odd } \ f(n-1), \text { if } n \text { is even }\end{array}\right.$
$=\frac{\frac{n+1}{2} \text {, if } n \text { is odd }}{\frac{n-1}{2}=\frac{n}{2} \text {, if } n \text { is even }}$
$=f(x)$
$[\because$ if $n$ is odd, then $(n+1)$ is even and if $n$ is even, then $(n-1)$ is odd]
Clearly, function is not one-one as $f(2)=f(1)=1$
But it is onto function.
$[\because$ If $m \in N$ (codomain) is odd, then $2 m \in N$ (domain) such that $f(2 m)=m$ and
if $m \in N$ codomain is even, then
$$ 2 m-1 \in N \text { (domain) such that } f(2 m-1)=m \text { ] } $$
$\therefore$ Function is onto but not one-one
