Question:

Let \(f:\R→\R\) be a function such that f(x + y) = f(x) + f(y) for all x, y ∈ \(\R\), and \(g:\R→(0,\infin)\) be a function such that g(x + y) = g(x)g(y) for all x, y ∈ \(\R\). If \(f(\frac{-3}{5})=12\) and \(g(\frac{-1}{3})=2\), then the value of \((f(\frac{1}{4})+g(-2)-8)g(0)\) is ______.

Updated On: May 16, 2025
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Correct Answer: 51

Approach Solution - 1

Functional Equations 

The functional equations imply:

Given:

  • f(x + y) = f(x) + f(y) ⇒ f(x) = kx
  • g(x + y) = g(x)g(y) ⇒ g(x) = ax

Step 1: Solve for k and a

From the equation \( f(-3) = 12 \), we get:

\[ k \cdot \left( -\frac{3}{5} \right) = 12 \quad \Rightarrow \quad k = -20 \]

Similarly:

Substituting \( g\left( -\frac{1}{3} \right) = 2 \), we get:

\[ a \cdot \left( -\frac{1}{3} \right) = 2 \quad \Rightarrow \quad a = \frac{1}{8} \]

Step 2: Calculate function values

  • \(f\left(\frac{1}{4}\right) = -20 \frac{1}{4} = -5\)
  • \(g(-2) = \left( \frac{1}{8} \right)^{-2} = 64\)
  • g(0) = 1

Step 3: Final Calculation

Now, calculate:

\[ f\left( \frac{1}{4} \right) + g(-2) - 8 \cdot g(0) = -5 + 64 - 8 \cdot 1 = 51 \]

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Approach Solution -2

To solve the problem, analyze the functional equations and given values.

Given:
- \(f: \mathbb{R} \to \mathbb{R}\) with \(f(x + y) = f(x) + f(y)\) (Cauchy functional equation, additive function).
- \(g: \mathbb{R} \to (0, \infty)\) with \(g(x + y) = g(x) g(y)\) (exponential-type function).
- \(f\left(-\frac{3}{5}\right) = 12\)
- \(g\left(-\frac{1}{3}\right) = 2\)

Step 1: Determine \(f(x)\)
For additive functions, \(f(x) = kx\) for some constant \(k\).
Given: \[ f\left(-\frac{3}{5}\right) = k \times \left(-\frac{3}{5}\right) = 12 \implies k = \frac{12}{-\frac{3}{5}} = 12 \times \left(-\frac{5}{3}\right) = -20 \] So, \[ f(x) = -20 x \]

Step 2: Determine \(g(x)\)
For multiplicative function with positive values, \(g(x) = a^x\) for some \(a > 0\).
Given: \[ g\left(-\frac{1}{3}\right) = a^{-\frac{1}{3}} = 2 \implies a^{\frac{1}{3}} = \frac{1}{2} \implies a = \left(\frac{1}{2}\right)^3 = \frac{1}{8} \] Thus, \[ g(x) = \left(\frac{1}{8}\right)^x = 8^{-x} \]

Step 3: Compute required expression:
\[ (f(\frac{1}{4}) + g(-2) - 8) \times g(0) \] Calculate each term: \[ f\left(\frac{1}{4}\right) = -20 \times \frac{1}{4} = -5 \] \[ g(-2) = 8^{-(-2)} = 8^2 = 64 \] \[ g(0) = 8^{0} = 1 \] So, \[ (f(\frac{1}{4}) + g(-2) - 8) g(0) = (-5 + 64 - 8) \times 1 = (51) \times 1 = 51 \]

Final Answer:
\[ \boxed{51} \]

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