Question

If \[ f(x)=1-2x+\int_{0}^{x} e^{x-t} f(t)\,dt \] and \[ g(x)=\int_{0}^{x} (f(t)+2)^{11}(t+12)^{17}(t-4)^4\,dt, \] If local minima and local maxima of \(g(x)\) occur at \(x=p\) and \(x=q\) respectively, find \(|p|+q\).

• A. \(12\)
• B. \(15\)
• C. \(9\)
• D. \(20\)

Hint:

For extrema of an integral-defined function, differentiate using FTC and analyze sign changes using multiplicity of roots.

Answer 1

The Correct Option is B

Step 1: Solve for \(f(x)\)
Differentiate the given equation: \[ f(x)=1-2x+\int_{0}^{x} e^{x-t} f(t)\,dt \] \[ \Rightarrow f'(x)=-2+\int_{0}^{x} e^{x-t} f(t)\,dt + f(x) \] Using the original equation: \[ \int_{0}^{x} e^{x-t} f(t)\,dt = f(x)-1+2x \] \[ \Rightarrow f'(x)=-2+f(x)-1+2x+f(x) \] \[ f'(x)-2f(x)=2x-3 \] This is a linear differential equation. Solving, we obtain: \[ f(x)=x-1 \]
Step 2: Differentiate \(g(x)\)
By Fundamental Theorem of Calculus: \[ g'(x)=(f(x)+2)^{11}(x+12)^{17}(x-4)^4 \] Substitute \(f(x)=x-1\): \[ g'(x)=(x+1)^{11}(x+12)^{17}(x-4)^4 \]
Step 3: Critical points
\[ g'(x)=0 \Rightarrow x=-1,\,-12,\,4 \] Multiplicity:

\(x=-1\): odd power (11) \(\Rightarrow\) sign change
\(x=-12\): odd power (17) \(\Rightarrow\) sign change
\(x=4\): even power (4) \(\Rightarrow\) no sign change

Step 4: Identify maxima and minima


At \(x=-12\): local maximum

At \(x=-1\): local minimum

At \(x=4\): point of inflection
Thus, \[ p=-1,\quad q=12 \]
Step 5: Compute required value
\[ |p|+q = 1+12 = \boxed{15} \]