2
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$f: \{1,2,3..\}\rightarrow \{\pm 1,\pm 2, \pm 3..\}$ 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?

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    $\begingroup$ Your answer is $-200$ which is not even in the domain of the function. $\endgroup$ – SarGe Jun 29 at 11:16
  • $\begingroup$ Note that for even $n$, $f(n)$ is positive, and for odd $n$, $f(n)\leq 0$ $\endgroup$ – Matti P. Jun 29 at 11:19
  • $\begingroup$ @Doubtnut I know that. Still doesn’t get the answer $\endgroup$ – Aditya Jun 29 at 11:20
  • $\begingroup$ 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. $\endgroup$ – AlexanderJ93 Jun 29 at 11:32
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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 $$-\left(\frac{n'-1}{2}\right)=-100\\\therefore \quad n'=201$$

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2
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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 $y>0$

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

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  • $\begingroup$ How would you arrive at 201 organically $\endgroup$ – Aditya Jun 29 at 11:22
  • $\begingroup$ 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.. $\endgroup$ – Bhaswat Jun 29 at 11:26

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