IPMAT Indore 2022 · Quantitative Ability · Question 37

Step 1 — substitute x=0. The functional equation becomes

f\left( f\left( y \right) \right)=0\cdot f\left( 0 \right)+y=y

So f is its own inverse — in particular f is a bijection on the non-negative integers.

Step 2 — substitute y=0.

f\left({x}^{2}+f\left( 0 \right) \right)=xf\left( x \right)

Step 3 — put x=0 into (2).

f\left( f\left( 0 \right) \right)=0

But from (1) with y=0 we also have f\left( f\left( 0 \right) \right)=0 — consistent, no new information yet.

Step 4 — show f\left( 0 \right)=0. Let c=f\left( 0 \right). Applying f to both sides of f\left( c \right)=0 and using (1):

f\left( f\left( c \right) \right)=c \Rightarrow f\left( 0 \right)=c

again consistent. Now use (2) with x=1:

f\left( 1+c \right)=f\left( 1 \right)

Since f is injective (from (1)), 1+c=1, hence

c=f\left( 0 \right)=0

Step 5 — evaluate the requested expression.

{\left[ f\left( 0 \right) \right]}^{2}+f\left( 0 \right)={0}^{2}+0=0

(For completeness, the function satisfying the equation is f\left( x \right)=x, which indeed gives f\left({x}^{2}+y \right)=x\cdot x+y ✔.)

Answer: Option 2 — 0

Shortcut. Substitute x=0 first — it is almost always the most productive move in a functional equation. It yields f\left( f\left( y \right) \right)=y immediately, which tells you f is a bijection and its own inverse.

Injectivity is then the tool that pins f\left( 0 \right): from f\left( 1+f\left( 0 \right) \right)=f\left( 1 \right) you may cancel f, giving f\left( 0 \right)=0.

A legitimate shortcut under exam pressure: guess the obvious candidate f\left( x \right)=x, verify it satisfies f\left({x}^{2}+y \right)=xf\left( x \right)+y, and read off f\left( 0 \right)=0. Then {\left[ f\left( 0 \right) \right]}^{2}+f\left( 0 \right)=0. Note that the expression {t}^{2}+t vanishes only at t=0 or t=-1, and f maps into non-negative integers — so 0 is the only possible answer among the options.