Question:

Step 1: Verify option (a). Define $g(x) = e^x f(x)$. Then $g(\alpha) = e^\alpha f(\alpha) = 0$ and $g(\beta) = e^\beta f(\beta) = 0$. By Rolle's theorem, $\exists\, c \in (\alpha,\beta)$ such that $g'(c) = 0$. $g'(x) = e^x f(x) + e^x f'(x) = e^x(f(x)+f'(x))$. Since $e^c \neq 0$, $f(c)+f'(c)=0$. ✓ Step 2: Verify option (b). Define $h(x) = e^{-x}f(x)$. Then $h(\alpha) = h(\beta) = 0$. By Rolle's theorem, $\exists\, c \in (\alpha,\beta)$ such that $h'(c) = 0$. $h'(x) = e^{-x}(f'(x)-f(x))$. So $f'(c)-f(c)=0$, i.e., $f(c)-f'(c)=0$. ✓ Step 3: Verify option (c). Since $f(\alpha)=0$ and $f(\beta)=0$, by Rolle's theorem $\exists\, c \in (\alpha,\beta)$ with $f'(c)=0$. Then $f(c)f'(c)=0$. ✓ Step 4: Verify option (d). Define $p(x) = f(x)f'(x)$. We know $p(\alpha) = f(\alpha)f'(\alpha) = 0$ and $p(\beta) = f(\beta)f'(\beta) = 0$. Also from Step 3, $\exists\, c \in (\alpha,\beta)$ with $f'(c)=0$, so $p(c)=0$. Thus $p(x)=f(x)f'(x)$ has at least 3 zeros: $\alpha, c, \beta$. By Rolle's theorem applied twice, $p'(x) = (f'(x))^2 + f(x)f''(x) = 0$ has at least two roots in $(\alpha,\beta)$. ✓