CAT 2022 Slot 1 · Quant · Question 22

The condition a\le x fixes the sign of the first modulus, so \left| x-a \right|=x-a.

The condition x\le 100 fixes the second, so \left| x-100 \right|=100-x.

Adding those two, \left( x-a \right)+\left( 100-x \right)=100-a, a constant.

So f\left( x \right)=100-a+\left| x-a-50 \right|, and only the last modulus varies with x.

To maximise f, maximise \left| x-a-50 \right| over the allowed range a\le x\le 100.

A modulus over an interval is largest at one of the endpoints.

At x=a the value is \left| a-a-50 \right|=50.

At x=100 the value is \left| 100-a-50 \right|=\left| 50-a \right|, which is at most 50 for every a between 0 and 100.

So the largest value of the last modulus is 50, whatever a is.

Therefore the maximum of f is 100-a+50=150-a.

Setting 150-a=100 gives a=50 — the fourth option.

Checking at a=50: f\left( x \right)=\left| x-50 \right|+\left| x-100 \right|+\left| x-100 \right|, whose largest value on 50\le x\le 100 is 100, reached at x=50.

Shortcut. The first two moduli cancel into the constant 100-a the moment the given range is used, so the whole question is one modulus on an interval. Its peak is 50 regardless of a, leaving the single equation 150-a=100.