Question

Let \(f\) be a differentiable function satisfying \[ f(x)=1-2x+\int_0^x (t-x)f(t)\,dt,\quad x\in\mathbb{R}, \] and let \[ g(x)=\int_0^x \{f(t)+2\}^5(t-4)^6(t+12)^7\,dt. \] If \(p\) and \(q\) are respectively the points of local minima and local maxima of \(g\), then the value of \(|p+q|\) is _______.

Hint:

In integrals defining functions, extrema are found by analysing the sign of the integrand.

Answer 1

Correct Answer: 8

Step 1: Differentiate the functional equation Differentiate once: \[ f'(x)=-2+\int_0^x(-f(t))\,dt \] Differentiate again: \[ f''(x)=-f(x) \]
Step 2: Solve the differential equation \[ f''+f=0 \Rightarrow f(x)=A\cos x+B\sin x \] Using \(f(0)=1\) and \(f'(0)=-2\): \[ A=1,\ B=-2 \] \[ f(x)=\cos x-2\sin x \]
Step 3: Find critical points of \(g(x)\) \[ g'(x)=(f(x)+2)^5(x-4)^6(x+12)^7 \] Critical points: \[ x=4,\ x=-12,\ f(x)+2=0 \] \[ \cos x-2\sin x+2=0 \Rightarrow x=\frac{\pi}{2} \]
Step 4: Nature of extrema \[ (x-4)^6\ \text{even power},\quad (x+12)^7\ \text{odd power} \] Hence: \[ \text{Local minimum at }x=-12,\quad \text{Local maximum at }x=4 \] \[ p=-12,\ q=4 \] Final Answer: \[ \boxed{8} \]

Answer 2

Concept:

• Solve for $f(x)$ using the characteristic-equation method for the differential equation, instead of writing down the general solution of $f''+f=0$ from memory.

Step 1: Differentiate the given equation twice to remove the integral.
$f(x)=1-2x+\displaystyle\int_0^x (t-x)f(t)\,dt$
Differentiating once: $f'(x)=-2-\displaystyle\int_0^x f(t)\,dt$
Differentiating again: $f''(x)=-f(x)$

Step 2: Solve $f''+f=0$ using the characteristic equation.
Assume $f(x)=e^{rx}$. Substituting gives $r^2+1=0 \Rightarrow r=\pm i$.
Complex roots $r=\pm i$ give the general solution $f(x)=A\cos x+B\sin x$.

Step 3: Use the given conditions to find $A$ and $B$.
From the original equation, $f(0)=1 \Rightarrow A=1$.
From Step 1, $f'(0)=-2 \Rightarrow B=-2$.
So $f(x)=\cos x - 2\sin x$.

Step 4: Find the critical points of $g(x)$.
$g'(x)=\{f(x)+2\}^5(x-4)^6(x+12)^7$
The critical points coming from the polynomial factors are $x=4$ and $x=-12$.

Step 5: Identify which is the minimum and which is the maximum.
The odd power $(x+12)^7$ makes $g'(x)$ change sign at $x=-12$, giving a local minimum there.
Combined with the behaviour of the remaining factors, $x=4$ gives the local maximum.
So $p=-12$, $q=4$.

Step 6: Compute the required value.
$|p+q| = |-12+4| = |-8| = 8$

Final Answer: $8$