IPMAT Indore 2023 · Quantitative Ability · Question 24

Step 1 — identify the form of the remainder. The divisor \left( x-1 \right)\left( x-2 \right) has degree 2, so the remainder has degree at most 1:

p\left( x \right)=\left( x-1 \right)\left( x-2 \right)\,q\left( x \right)+\left( \alpha x+\beta \right)

Step 2 — use the two given facts. By the Remainder Theorem, p\left( 1 \right)=2 and p\left( 2 \right)=1. Substituting x=1 and x=2 makes the product term vanish:

x=1: \alpha +\beta =2

x=2: 2\alpha +\beta =1

Step 3 — solve. Subtracting the first from the second:

\alpha =-1 \Rightarrow \beta =3

\text{Remainder}=-x+3=3-x

Step 4 — check. At x=1: 3-1=2 ✔ At x=2: 3-2=1 ✔

Answer: Option 2 — 3-x

Shortcut. Check the options against the two conditions — it is faster than solving. The remainder must give 2 at x=1 and 1 at x=2:

Done in two substitutions.

If you prefer to derive it, note that the remainder is the unique linear interpolation through the points \left( 1,2 \right) and \left( 2,1 \right) — a line of slope -1 through \left( 1,2 \right), i.e. y=3-x. Seeing it as interpolation makes the general case (k divisors, degree k-1 remainder) obvious.