Question

For what value of $n$ is$\left| {n - 1} \right| + 1$ equal to $0$?
(A) $0$ (B) $1$ (C) $2$ (D) There is no such value of $n$

Answer 1

Hint- For each non-negative value of$x$ , $f(x)$ is equal to the positive value of $x$. But for negative values of $x$, $f(x)$ is equal to negative value of $x$.
There is a concept of critical values that we need to understand when using mod function. Critical values are the value where our expression inside the mod becomes equal to $0$. It is the point where our modulus function changes its nature. As for $f(x) = \left| x \right|$, the function changes its nature from $- x$ to $x$ as we reach at point $x = 0$.

Complete step by step solution-
Let $f(n) = \left| {n - 1} \right| + 1$
We have to find the value of $n$, for which $f(n) = 0$.
First we have to find Critical value and for that expression inside the mod should be equal to zero.
$\therefore n - 1 = 0$
$\Rightarrow n = 1$ (Critical value)
Now, we will have two cases,
1.$n < 1$
$\therefore \left| {n - 1} \right| = - (n - 1)$
$f(n) = 0$
$\Rightarrow - (n - 1) + 1 = 0$
$\Rightarrow - n + 2 = 0$
$\Rightarrow n = 2$
Here $n = 2$ which is greater than one and it is not possible in this case.
2.$n \ge 1$
$\therefore \left| {n - 1} \right| = (n - 1)$
$f(n) = 0$
$\Rightarrow (n - 1) + 1 = 0$
$\Rightarrow n = 0$
Here $n = 0$ which is less than one and it is not possible in this case.
Hence there is no such value of n for which $f(n) = \left| {n - 1} \right| + 1$will be equal to$0$.
Answer-(D)

Note: Here we have two cases as there is only one critical value i.e. $1$. If we will have more than one critical point, our cases may increase and then we need to open mod carefully with positive or negative sign depending upon the value inside the mod.