Question:

Step 1: Let $f(x) = 2 + Ax^2 + Bx$ where $A = \int_{-1}^{1} \frac{t}{2} f(t)\,dt$ and $B = \int_{-1}^{1} \frac{9}{14} f(t)\,dt$. Step 2: Compute $B = \frac{9}{14}\int_{-1}^{1}(2 + At^2 + Bt)\,dt$. Since $Bt$ is odd, $\int_{-1}^1 Bt\,dt = 0$. $$B = \frac{9}{14}\left[2t + \frac{At^3}{3}\right]_{-1}^{1} = \frac{9}{14}\left(4 + \frac{2A}{3}\right)$$ Step 3: Compute $A = \int_{-1}^{1}\frac{t}{2}(2 + At^2 + Bt)\,dt$. Odd terms vanish: $\int_{-1}^1 t\,dt = 0$, $\int_{-1}^1 t^2 \cdot B\,dt$ is even but $Bt^2$ integrated... Actually $\frac{t}{2}\cdot Bt = \frac{B}{2}t^2$ is even. $$A = \int_{-1}^1 \frac{t}{2}(2+At^2+Bt)\,dt = \int_{-1}^1 \left(t + \frac{At^3}{2} + \frac{Bt^2}{2}\right)dt$$ $$= 0 + 0 + \frac{B}{2}\cdot\frac{2}{3} = \frac{B}{3}$$ Step 4: Substitute $A = B/3$ into the equation for $B$: $$B = \frac{9}{14}\left(4 + \frac{2(B/3)}{3}\right) = \frac{9}{14}\left(4 + \frac{2B}{9}\right) = \frac{36}{14} + \frac{2B}{14} = \frac{18}{7} + \frac{B}{7}$$ $$B - \frac{B}{7} = \frac{18}{7} \Rightarrow \frac{6B}{7} = \frac{18}{7} \Rightarrow B = 3, \quad A = 1$$ Step 5: So $f(x) = x^2 + 3x + 2 = (x+1)(x+2)$. Step 6: Check options: - (a) Rolle's Theorem on $[-2,-1]$: $f(-2) = 0$, $f(-1) = 0$, $f$ is continuous and differentiable. ✓ - (b) $\lim_{x\to 0} f(x) = f(0) = 2 \neq 0$. ✗ - (c) $f$ is a polynomial, so continuous and derivable on $R$. ✓ - (d) $f(x) = (x+1)(x+2)$ is a parabola opening upward with minimum at $x = -3/2$; it has a minimum but no maximum (goes to $+\infty$). So maximum does not exist. ✓ Correct options: (a), (c), (d), i.e., options 1, 3, 4.