Functions
Functional equation and intersection area
MJAT_TS2_P2
Grade 12
Question:
Let $f:\mathbb{R}\to\mathbb{R}$ be a non-constant function satisfying $f(2a)+2f(b)=f(f(a+b))$ for all $a,b\in\mathbb{R}$, with $f(2)=-8$. Define:
$S_1=\{(x,y)\in\mathbb{R}^2: y=f(x)\}$ and $S_2=\{(x,y)\in\mathbb{R}^2: y^2=32x\}$.
Let $S_1\cap S_2$ consist of two points $A$ and $B$. The area of triangle $\triangle OAB$ (where $O=(0,0)$) is:
Step-by-Step Solution
Key Concept: From the functional equation with $a=b=0$: $3f(0)=f(f(0))$. Try $f(x)=cx+d$ (linear). $f(2a)+2f(b)=2ca+d+2cb+2d=2c(a+b)+3d$. $f(f(a+b))=c(c(a+b)+d)+d=c^2(a+b)+cd+d$. Equating: $2c=c^2\Rightarrow c=2$ and $3d=2d+d$ ✓. So $f(x)=2x+d$. $f(2)=4+d=-8\Rightarrow d=-12$. $f(x)=2x-12$.
Area of $\triangle OAB$ with $O=(0,0)$, $A=(2,-8)$, $B=(18,24)$: $=\frac{1}{2}|2\cdot 24-18\cdot(-8)|=\frac{1}{2}|48+144|=\frac{192}{2}=\mathbf{96}$.
Correct Answer: 96