tmua.me Extreme Set A Paper 2 · Question 5

A student wishes to prove that there is no real-valued function $f$ , defined for every real number, such that

f(x+f(y))=f(x)+y^2\qquad\text{for all real }x,y. \tag{$*$}

The student assumes that such a function exists and writes the following proof, with one line missing.

Suppose $f(u)=f(v)$ . Substituting $y=u$ and $y=v$ into $(*)$ gives $u^2=v^2$ .

[Missing line]

Hence, if $u\ne v$ , then $f(t)=0$ for every real $t$ , which is incompatible with $(*)$ . It follows that $f$ is one-to-one. But taking $x=0$ , first with $y=1$ and then with $y=-1$ , gives

f(f(1))=f(f(-1)).

Applying the one-to-one property twice gives $1=-1$ , a contradiction.

Which of the following fills the gap by valid reasoning, without assuming in advance that $f$ is one-to-one?

AFor every real

t

, translating the inputs preserves equality, so

f(u+t)=f(v+t)

. Hence

(u+t)^2=(v+t)^2

, giving

2(u-v)t=0

BFor every real

t

(*)

gives

f(u+f(t))=f(v+f(t))

. Cancelling

f

gives

u+f(t)=v+f(t)

, so

u=v

CSince

u^2=v^2

, either

u=v

u=-v

. In the latter case,

(*)

with

x=u

and

y=v

gives

f(u+f(v))=f(u)+u^2

, so

f(u)=0

DFor every real

t

(*)

gives

f(u+f(t))=f(v+f(t))

. Applying the implication already established for equal function values, then subtracting

u^2=v^2

, gives

2(u-v)f(t)=0

EThe equality

f(u)=f(v)

gives

f(u-v)=0

. Taking

y=u-v

(*)

therefore gives

f(x)=f(x)+(u-v)^2

, so

u=v

FFor every real

t

(*)

gives

f(u+f(t))=f(v+f(t))

, and hence

(u+f(t))^2=(v+f(t))^2

. Taking square roots gives

u+f(t)=v+f(t)

, so

u=v

examday

tmua.me Extreme Set A Paper 2 · Q5