Question:

Step 1: Differentiate $x = y^3 + y^2 + y + 1$ implicitly with respect to $y$: $$\frac{dx}{dy} = 3y^2 + 2y + 1$$ So $\frac{dy}{dx} = \frac{1}{3y^2 + 2y + 1}$, i.e., $f'(x) = \frac{1}{3y^2 + 2y + 1}$. Step 2: Find $f'(0)$. When $x = 0$: $0 = y^3 + y^2 + y + 1 = (y+1)(y^2+1)$, so $y = -1$. $$f'(0) = \frac{1}{3(1) + 2(-1) + 1} = \frac{1}{2}$$ Thus $2f'(0) = 1$. ✓ Option (1) is correct. Step 3: Find $f''(0)$. Differentiate $f'(x) = (3y^2+2y+1)^{-1}$ with respect to $x$: $$f''(x) = -\frac{(6y+2)\frac{dy}{dx}}{(3y^2+2y+1)^2}$$ At $y=-1$: $6(-1)+2 = -4$, $\frac{dy}{dx} = \frac{1}{2}$, denominator $= 4$: $$f''(0) = -\frac{(-4)\cdot\frac{1}{2}}{4} = \frac{2}{4} = \frac{1}{2}$$ Option (2) is correct. ✓ Step 4: Evaluate $\int_0^4 f(x)\,dx$. When $x=0$, $y=-1$; when $x=4$, $4 = y^3+y^2+y+1 \Rightarrow y=1$. Use substitution $x = y^3+y^2+y+1$, $dx=(3y^2+2y+1)dy$: $$\int_0^4 f(x)\,dx = \int_{-1}^{1} y(3y^2+2y+1)\,dy = \int_{-1}^{1}(3y^3+2y^2+y)\,dy$$ Step 5: Evaluate the integral: $$= \left[\frac{3y^4}{4}+\frac{2y^3}{3}+\frac{y^2}{2}\right]_{-1}^{1} = \left(\frac{3}{4}+\frac{2}{3}+\frac{1}{2}\right)-\left(\frac{3}{4}-\frac{2}{3}+\frac{1}{2}\right) = \frac{4}{3}$$ Option (3) is correct. ✓ Option (4) is incorrect.