Functions
Functional equation — quadratic form
MJAT_TS6_P1
Grade 12
Question:
Let $f:\mathbb{R}\to\mathbb{R}$ be differentiable such that $f(x+y)-f(x-y)=4xy+10y$ for all $x,y\in\mathbb{R}$ and $f(1)=2$. Then:
A) Minimum value of $f(x)$ is $-\dfrac{41}{4}$
B) Minimum value of $f(f(x))$ is $-\dfrac{41}{4}$
C) The equation $f(x)=K$ has exactly two solutions when $K\in(0,4)$
D) The equation $f(x)=4$ has exactly 3 solutions
Step-by-Step Solution
Key Concept: Setting $x=y$: $f(2x)-f(0)=4x^2+10x$. Setting $x=0$: $f(y)-f(-y)=10y$. From the functional equation, $f'(x)=2x+5$ (by differentiating w.r.t. $y$ and setting $y=0$). So $f(x)=x^2+5x+C$. Using $f(1)=1+5+C=2\Rightarrow C=-4$. $f(x)=x^2+5x-4$.
A ✓, B ✓, D ✓. Answer: A, B, D.
Correct Answer: ABD