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 :

Updated On: Apr 12, 2026
  • \( \frac{13}{8} \)
  • \( \frac{11}{16} \)
  • \( \frac{15}{32} \)
  • \( \frac{17}{64} \)
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The Correct Option is B

Solution and Explanation

Concept: If a function satisfies \[ f(x+y)=f(x)f(y) \] then it must be exponential in nature: \[ f(x)=e^{kx} \] Also use differentiation of integral equations to convert them into differential equations. Step 1: {Differentiate the equation.} Given \[ x^2g(x)=\int_1^x(t^2f(t)-tg(t))dt \] Differentiate both sides: \[ 2xg(x)+x^2g'(x)=x^2f(x)-xg(x) \] Step 2: {Simplify.} \[ x^2g'(x)+3xg(x)=x^2f(x) \] Divide by \(x^2\): \[ g'(x)+\frac{3}{x}g(x)=f(x) \] Step 3: {Use the form of \(f(x)\).} Since \(f(x+y)=f(x)f(y)\) and \(f(0)\neq0\), \[ f(x)=e^{kx} \] Substituting into differential equation and solving gives \[ g(x)=\frac{1}{x^3}\int x^3f(x)\,dx \] Step 4: {Evaluate using the given limits.} Solving and substituting \(x=2\): \[ g(2)=\frac{11}{16} \]
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