Question:

Step 1: Understand the given condition and integrate it. We are given that $f''(x) = f''(5-x)$ for all $x \in [0,5]$. To find a relationship between $f'(x)$ and $f'(5-x)$, we integrate both sides with respect to $x$: $$\int f''(x)\,dx = \int f''(5-x)\,dx$$ This gives us: $$f'(x) = -f'(5-x) + C$$ where $C$ is a constant of integration. Step 2: Determine the constant of integration using initial conditions. We use the given conditions $f'(0) = 1$ and $f'(5) = 7$. Substituting $x = 0$ into the equation from Step 1: $$f'(0) = -f'(5) + C$$ $$1 = -7 + C$$ $$C = 8$$ Step 3: Establish the key relationship. From Step 2, we now have: $$f'(x) = -f'(5-x) + 8$$ Rearranging: $$f'(x) + f'(5-x) = 8$$ This is a crucial symmetry property that will help us evaluate the integral. Step 4: Set up the integral using substitution. Let $I = \int_1^4 f'(x)\,dx$. To use the symmetry property, we perform a substitution. Let $x = 5 - t$, so $dx = -dt$. When $x = 1$, we have $t = 4$; when $x = 4$, we have $t = 1$: $$I = \int_4^1 f'(5-t)(-dt) = \int_1^4 f'(5-t)\,dt$$ Step 5: Add the two expressions for the integral. We now have two expressions for $I$: - Original: $I = \int_1^4 f'(x)\,dx$ - From substitution: $I = \int_1^4 f'(5-x)\,dx$ Adding them: $$2I = \int_1^4 f'(x)\,dx + \int_1^4 f'(5-x)\,dx = \int_1^4 [f'(x) + f'(5-x)]\,dx$$ Step 6: Evaluate using the symmetry property. From Step 3, we know that $f'(x) + f'(5-x) = 8$. Substituting: $$2I = \int_1^4 8\,dx = 8(4-1) = 8 \times 3 = 24$$ Step 7: Find the final answer. Dividing both sides by 2: $$I = \frac{24}{2} = 12$$ Therefore, $\int_1^4 f'(x)\,dx = 12$. The answer is **(d) 12** (Option 4).