IPMAT Indore 2022 · Quantitative Ability · Question 6

Step 1 — recognise the structure. This is a weighted sum of distances from x to the points 0,1,2,4,6,10, with weights

Total weight =1+2+1+1+1+2=8.

Step 2 — use the weighted-median rule. A sum ∑{w}_{i}\left| x-{a}_{i}\right| is minimised at the weighted median — the point where the running total of weights first reaches half the total, here 4.

The cumulative weight reaches 4 at x=2.

Step 3 — evaluate there.

f\left( 2 \right)=\left| 2 \right|+2\left| 1 \right|+\left| 0 \right|+\left| -2 \right|+\left| -4 \right|+2\left| -8 \right|

=2+2+0+2+4+16=26

Step 4 — confirm it is the minimum. Just left of x=2 the slope is 1+2-1-1-1-2=-2<0 (decreasing); just right of it the slope is 1+2+1-1-1-2=0. So the function stops decreasing at x=2 — and a numerical sweep confirms 26 as the least value. The graph shows the minimum is attained along the whole flat stretch 2\le x\le 4.

Answer: 26

Shortcut. Never differentiate a sum of moduli — find the weighted median. Count weights from the left until you pass half the total (4 of 8 here); you land on x=2.

A useful cross-check is the slope test: at any point, slope = (weight to the left) − (weight to the right). The minimum is wherever that changes sign. Here left-weight becomes 4 and right-weight 4 at x=2, so the slope hits zero — and a flat stretch means every x\in \left[ 2,4 \right] gives the same value 26.

Treat a repeated term like 2\left| x-1 \right| as a point of weight 2, not as a separate point. That is the only subtlety in the question.