Question:medium

If the function \( f(x) \) satisfies \[ f'(x) = f(x) - \pi x + \pi, \] then the possible value of \( f(1) \) is

Show Hint

For linear differential equations of the form \( y' + Py = Q \), use the integrating factor method to reduce it into exact derivative form.
Updated On: Feb 15, 2026
  • \( \pi + \frac{1}{6} \)
  • \( \pi - \frac{1}{6} \)
  • \( \frac{\pi}{2} + 1 \)
  • \( 1 - \frac{1}{2} \)
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Topic
This question requires solving a first-order linear ordinary differential equation. The equation is of the form $y' + P(x)y = Q(x)$, which can be solved systematically using the method of integrating factors.
Step 2: Key Approach - Integrating Factor Method
First, we rearrange the equation into the standard linear form: \[ f'(x) - f(x) = \pi - \pi x \] Here, $P(x) = -1$ and $Q(x) = \pi - \pi x$. The integrating factor is given by $I(x) = e^{\int P(x) dx}$. After finding $I(x)$, we multiply the entire equation by it, which makes the left side an exact derivative of $(I(x) \cdot f(x))$. We then integrate both sides to find the general solution for $f(x)$.
Step 3: Detailed Calculation
A. Find the integrating factor:
\[ I(x) = e^{\int -1 \,dx} = e^{-x} \] B. Multiply the DE by the integrating factor:
\[ e^{-x} f'(x) - e^{-x} f(x) = (\pi - \pi x) e^{-x} \] The left side is the derivative of $(e^{-x} f(x))$: \[ \frac{d}{dx} \left( e^{-x} f(x) \right) = (\pi - \pi x) e^{-x} \] C. Integrate both sides with respect to x:
\[ e^{-x} f(x) = \int (\pi - \pi x) e^{-x} dx \] We solve the integral on the right using integration by parts ($\int u \, dv = uv - \int v \, du$). Let $u = \pi - \pi x$ and $dv = e^{-x} dx$. Then $du = -\pi dx$ and $v = -e^{-x}$. \[ \int (\pi - \pi x) e^{-x} dx = (\pi - \pi x)(-e^{-x}) - \int (-e^{-x})(-\pi dx) \] \[ = -(\pi - \pi x)e^{-x} - \pi \int e^{-x} dx = -(\pi - \pi x)e^{-x} - \pi(-e^{-x}) + C \] \[ = -\pi e^{-x} + \pi x e^{-x} + \pi e^{-x} + C = \pi x e^{-x} + C \] So, we have: \[ e^{-x} f(x) = \pi x e^{-x} + C \] D. Solve for f(x) and evaluate f(1):
Multiply by $e^x$ to isolate $f(x)$: \[ f(x) = \pi x + C e^x \] Now, find the value at $x=1$: \[ f(1) = \pi(1) + C e^1 = \pi + Ce \] The solution must be of the form $\pi$ plus some constant. Looking at the options, option (A) is $\pi + 1/6$. This is a possible value if the constant of integration $C$ is such that $Ce = 1/6$. Without initial conditions, $C$ is arbitrary, so any value of the form $\pi + \text{const.}$ is possible. Option (A) is the only one that fits this structure.
Step 4: Final Answer
The general solution for $f(1)$ is $\pi + Ce$. Option (A) is a possible value. \[ \boxed{f(1) = \pi + \frac{1}{6}} \]
Was this answer helpful?
0