Question

Let \(f:\mathbb{R}\to\mathbb{R}\) be such that \(f(x+y)=f(x)f(y)\) for all \(x,y\in\mathbb{R}\) and \(f(0)\neq0\). Let \(g:[1,\infty)\to\mathbb{R}\) be a differentiable function such that \[ x^2g(x)=\int_1^x\big(t^2f(t)-tg(t)\big)\,dt \] Then \(g(2)\) is equal to :

• A. \( \frac{13}{8} \)
• B. \( \frac{11}{16} \)
• C. \( \frac{15}{32} \)
• D. \( \frac{17}{64} \)

Answer 1

The Correct Option is B

Step 1: Understanding the Question
We are given a functional equation for \(f(x)\) and an integral equation involving \(f(x)\) and another function \(g(x)\). We need to first determine the function \(f(x)\). Then, we solve the integral equation to find \(g(x)\) and finally evaluate \(g(2)\).
Step 2: Key Formula or Approach
1. Solve the functional equation for \(f(x)\).
2. Use the Leibniz rule (a part of the Fundamental Theorem of Calculus) to differentiate the integral equation. If \( H(x) = \int_{a(x)}^{b(x)} K(t) dt \), then \( H'(x) = K(b(x))b'(x) - K(a(x))a'(x) \).
3. Solve the resulting first-order linear differential equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \) using an integrating factor (I.F.), which is \( e^{\int P(x) dx} \).
Step 3: Detailed Explanation
Part 1: Solving for f(x)
Given \( f(xy) = f(x)f(y) \) and \( f(0) \neq 0 \).
Let \(x=0\) in the equation: \( f(0) = f(0)f(y) \).
Since \( f(0) \neq 0 \), we can divide both sides by \( f(0) \), which gives \( f(y) = 1 \) for all \( y \in \mathbb{R} \).
So, the function is \( f(x) = 1 \).
Part 2: Solving the integral equation for g(x)
Substitute \( f(t) = 1 \) into the given integral equation: \[ x^2 g(x) = \int_{1}^{x} (t^2(1) - tg(t))dt \] To solve for \(g(x)\), we differentiate both sides with respect to \(x\).
Using the product rule on the left side and the Leibniz rule on the right side: \[ \frac{d}{dx}(x^2 g(x)) = \frac{d}{dx} \int_{1}^{x} (t^2 - tg(t))dt \] \[ 2x g(x) + x^2 g'(x) = (x^2 - xg(x)) \cdot 1 - 0 \] \[ 2x g(x) + x^2 g'(x) = x^2 - xg(x) \] Rearrange the terms to form a differential equation: \[ x^2 g'(x) + 3x g(x) = x^2 \] Since the domain of \(g\) is \( [1, \infty) \), \(x \neq 0\). We can divide by \(x\): \[ x g'(x) + 3 g(x) = x \implies g'(x) + \frac{3}{x} g(x) = 1 \] This is a linear first-order differential equation with \(P(x) = 3/x\) and \(Q(x) = 1\).
The integrating factor (I.F.) is \( e^{\int (3/x) dx} = e^{3\ln x} = e^{\ln(x^3)} = x^3 \).
Multiplying the DE by the I.F.: \[ x^3 g'(x) + 3x^2 g(x) = x^3 \] The left side is the derivative of \( (I.F.) \cdot g(x) \): \[ \frac{d}{dx} (x^3 g(x)) = x^3 \] Integrating both sides: \[ x^3 g(x) = \int x^3 dx = \frac{x^4}{4} + C \] \[ g(x) = \frac{x}{4} + \frac{C}{x^3} \] To find the constant \(C\), we use the initial condition from the integral equation. At \(x=1\), the integral is zero.
\[ (1)^2 g(1) = \int_1^1 (t^2 - tg(t)) dt = 0 \implies g(1) = 0 \] Substitute \(x=1, g(1)=0\) into the solution for \(g(x)\): \[ 0 = \frac{1}{4} + \frac{C}{1^3} \implies C = -\frac{1}{4} \] So, the function \(g(x)\) is: \[ g(x) = \frac{x}{4} - \frac{1}{4x^3} \] Part 3: Final Calculation
We need to find \( g(2) \). \[ g(2) = \frac{2}{4} - \frac{1}{4(2^3)} = \frac{1}{2} - \frac{1}{4 \cdot 8} = \frac{1}{2} - \frac{1}{32} = \frac{16}{32} - \frac{1}{32} = \frac{15}{32} \] Step 4: Final Answer
The value of \( g(2) \) is \( \frac{15}{32} \).