Question:

Step 1: Find $f(x)$ by differentiating $e^x f(x) + C$. We need $\dfrac{d}{dx}[e^x f(x)] = e^x(x^2-1)(x^2+4x+1)$. So $e^x f(x) + e^x f'(x) = e^x(x^2-1)(x^2+4x+1)$, giving $f(x) + f'(x) = (x^2-1)(x^2+4x+1)$. Step 2: Use the standard formula $\int e^x[h(x) + h'(x)]dx = e^x h(x) + C$. We need to write $(x^2-1)e^x(x^2+4x+1)$ in the form $e^x[h(x)+h'(x)]$. Let $f(x) = x^4 + 4x^3 + 2x^2 - 4x - 1$ (to be verified). Then $f'(x) = 4x^3 + 12x^2 + 4x - 4$. $f(x) + f'(x) = x^4 + 8x^3 + 14x^2 - 8x - 5$. Compare with $(x^2-1)(x^2+4x+1) = x^4 + 4x^3 + x^2 - x^2 - 4x - 1 = x^4 + 4x^3 - 4x - 1$. These don't match, so let us try $f(x) = x^4 + 4x^3 - 2x - 1$. $f'(x) = 4x^3 + 12x^2 - 2$. $f+f' = x^4 + 8x^3 + 12x^2 - 4x - 3 \neq (x^2-1)(x^2+4x+1)$. Step 3: Expand $(x^2-1)(x^2+4x+1) = x^4 + 4x^3 + x^2 - x^2 - 4x - 1 = x^4 + 4x^3 - 4x - 1$. Try $f(x) = x^4 + 4x^3 - 4x - 5$. $f'(x) = 4x^3 + 12x^2 - 4$. $f+f' = x^4 + 8x^3 + 12x^2 - 8x - 9 \neq x^4 + 4x^3 - 4x - 1$. Step 4: Use the identity: if $f(x) + f'(x) = (x^2-1)(x^2+4x+1)$, then $g(x) = f(x) + f'(x) = (x^2-1)(x^2+4x+1) = x^4 + 4x^3 - 4x - 1$. Step 5: Find roots of $g(x) = 0$: $x^4 + 4x^3 - 4x - 1 = 0$. Try $x = 1$: $1 + 4 - 4 - 1 = 0$. ✓ Try $x = -1$: $1 - 4 + 4 - 1 = 0$. ✓ So $(x-1)(x+1)$ divides $g(x)$. Factor: $x^4 + 4x^3 - 4x - 1 = (x^2-1)(x^2+4x+1)$. The quadratic $x^2 + 4x + 1 = 0$ gives $x = \dfrac{-4 \pm \sqrt{12}}{2} = -2 \pm \sqrt{3}$, which are non-integral. Step 6: Verify option (a): Integral roots are $x = 1$ and $x = -1$, so number of integral roots = 2. ✓ Step 7: Verify option (b): Sum of squares of integral roots $= 1^2 + (-1)^2 = 2$. ✓ Step 8: Verify option (c): If $\alpha$ is a non-integral root, then $\alpha^2 + 4\alpha + 1 = 0$. Compute $\alpha^4 + 4\alpha^3 + 2\alpha^2 + 4\alpha + 2 = \alpha^2(\alpha^2 + 4\alpha + 1) + (\alpha^2 + 4\alpha + 1) + 1 = \alpha^2 \cdot 0 + 0 + 1 = 1$. ✓ Step 9: Verify option (d): $g'(x) = 4x^3 + 12x^2 - 4$. $g'(0) = -4 \neq 4$. So option (d) is incorrect. The correct answers are (a), (b), (c).