Applications of Derivatives
Rolle's Theorem / Mean Value Theorem Application
nta_pyq_2024_jan
Grade 12
Question:
Let $g:\mathbb{R}\to\mathbb{R}$ be a non-constant twice differentiable function such that $g'\left(\dfrac{1}{2}\right)=g'\left(\dfrac{3}{2}\right)$. If a real valued function $f$ is defined as $f(x)=\dfrac{1}{2}[g(x)+g(2-x)]$, then
$f''(x)=0$ for at least two $x$ in $(0,2)$
$f''(x)=0$ for exactly one $x$ in $(0,1)$
$f''(x)=0$ for no $x$ in $(0,1)$
$f'\left(\frac{3}{2}\right)+f'\left(\frac{1}{2}\right)=1$
Step-by-Step Solution
Key Concept: $f'(x)=\frac{g'(x)-g'(2-x)}{2}$. $f'(1/2)=\frac{g'(1/2)-g'(3/2)}{2}=0$ and $f'(3/2)=\frac{g'(3/2)-g'(1/2)}{2}=0$. By Rolle's theorem applied to $f'$ on $[1/2,3/2]$, $f''=0$ for some $c\in(1/2,3/2)$. Since $f'$ has roots in both $(1/2,1)$ and $(1,3/2)$, $f''=0$ at least twice in $(0,2)$.
$f'(x)=\frac{g'(x)-g'(2-x)}{2}$. $f'(1/2)=0$, $f'(3/2)=0$, $f'(1)=0$. By Rolle's theorem: $f''=0$ for some $c_1\in(1/2,1)$ and $c_2\in(1,3/2)$. So $f''=0$ at least twice in $(0,2)$.
Correct Answer: 1