CAT 2022 Slot 1 — QA Question 22

Simplify on $a\le x\le 100$:

$$\begin{aligned} &f(x)=(x-a)+(100-x)+|x-a-50|\\ &\Rightarrow\ f(x)=(100-a)+|x-a-50| \end{aligned}$$

$a=0$: $f(x)=100+|x-50|$. At $x=100$, $f=150\gt 100$. Rejected.

$a=25$: $f(x)=75+|x-75|$. At $x=100$, $f=100$ but at e.g. $x=100$ it is $100$; however the simplified form only holds for $x\ge a$, and checking the full range gives values exceeding $100$ near the other end. Rejected.

$a=50$:

$$\begin{aligned} &f(x)=50+|x-100|\\ &\text{for }50\le x\le 100:\ |x-100|=100-x\le 50\\ &\Rightarrow\ f(x)\le 50+50=100,\ \text{equality at }x=50 \end{aligned}$$

Maximum is exactly $100$. Correct.

$a=100$: the range forces $x=100$ only, giving $f(x)=0+|100-150|=50$. Maximum is $50$, not $100$. Rejected.