Functions
Functional Equation
MMTS_Full_Test_07
Grade 12
Question:
$f(x)$ is differentiable satisfying $f^2(x)+f^2(y)+2(xy-1)=f^2(x+y)$ $\forall x,y\in\mathbb{R}$. Also $f(x)>0$, $f(\sqrt{2})=2$. Then $f(\sqrt{7})=$
Step-by-Step Solution
Key Concept: Guess $f(x)=\sqrt{x^2+1}$ or $f(x)=x$; check functional equation
Try $f(x)^2=x^2+c$. $f^2(x)+f^2(y)+2(xy-1)=x^2+c+y^2+c+2xy-2=(x+y)^2+2c-2$. For $=f^2(x+y)=(x+y)^2+c$: $c=2$. $f(\sqrt{2})=2$ ✓. $f(\sqrt{7})=\sqrt{9}=3$.
Correct Answer: 1