# From the given definition of function $f(n)$, find $f^{-1}(-100)$

$$f: \{1,2,3..\}\rightarrow \{\pm 1,\pm 2, \pm 3..\}$$ is defined by $$f(n)=\begin {matrix} \frac n2~~\text{if n is even} \\ -\frac{n-1}{2}~~\text{if n is odd}\end{matrix}$$

$$-100$$ is even, so situation one would apply in which case

$$y=\frac n2$$ $$n= 2y$$ $$f^{-1}(n)=2n$$

$$f^{-1}(-100)=-200$$

But the given answer is $$201$$ which would be the case if we solved using situation 2. Why should we use situation 2 is the number given is even?

• Your answer is $-200$ which is not even in the domain of the function. – SarGe Jun 29 at 11:16
• Note that for even $n$, $f(n)$ is positive, and for odd $n$, $f(n)\leq 0$ – Matti P. Jun 29 at 11:19
• @Doubtnut I know that. Still doesn’t get the answer – Aditya Jun 29 at 11:20
• Simply, you don't choose which branch of $f$ based on its output, so you wouldn't choose which branch of $f^{-1}$ based on its input. – AlexanderJ93 Jun 29 at 11:32

As the domain of the function $$f$$ is positive integers, you will get positive value of $$f$$ only by first case i.e. $$n$$ should be even.

Now, it is given that for a particular $$n'$$, $$f(n') =-100$$. Here, $$f$$ has yielded a negative value which is possible only for second case. So, you've $$-\left(\frac{n'-1}{2}\right)=-100\\\therefore \quad n'=201$$

because you are looking at the inverse function, so you think of it as: $$y=f^{-1}(-100)\Rightarrow -100=f(y)$$ now since $$f(y)<0$$ it has to be situation two since our input y is defined as $$y>0$$

• Then what is $n$ being odd or even have to do with it? – Aditya Jun 29 at 11:21
• when finding the inverse it isnt – Henry Lee Jun 29 at 11:47

Both -200 and 201 will give -100 when the function is applied. But notice that domain is restricted to positive numbers so that inverse exists. So 201 is the correct solution.

• How would you arrive at 201 organically – Aditya Jun 29 at 11:22
• Use the fact that $f(f^{-1}(-100))=-100$. Then assume $f^{-1}(-100)$ to be even and see what value you reach and assume it to be odd and see what value you reach. Pick the one in your domain of $f$ ,and you are sure to get only one in the domain,as the function is invertible.. – Bhaswat Jun 29 at 11:26