<p>Given two functions \(F(x) = \left(1 + \dfrac{1}{x}\right)^x\), \(G(x) = \left(1 + \dfrac{1}{x}\right)^{x+1}\) defined for all \(x > 0\). Which of the following is decreasing function for \(\forall\, x > 0\)?</p>
<p>(a) \(F(F(x) - G(x))\)</p>
<p>(b) \(G(F(x) - G(x))\)</p>
<p>(c) \(G(x) - F(x)\)</p>
<p>(d) \(G(F(x))\)</p>
Step-by-Step Solution
Key Concept: We need to analyze the monotonicity of each option by computing derivatives. The key insight is that F(x) and G(x) are both increasing functions, and understanding their relationship helps determine which composite/derived functions are decreasing.
<p><strong>Step 1: Analyze F(x) and G(x)</strong></p><p>Let $h(t) = (1 + 1/t)^t$ for $t > 0$. Then $F(x) = h(x)$ and $G(x) = h(x) \cdot (1 + 1/x)$.</p><p>Note: $G(x) = F(x) \cdot (1 + 1/x) > F(x)$, so $F(x) - G(x) < 0$ for all $x > 0$.</p><p><strong>Step 2: Check monotonicity of F(x) and G(x)</strong></p><p>Taking logarithm: $\ln F(x) = x \ln(1 + 1/x)$.</p><p>$\frac{d}{dx}\ln F(x) = \ln(1 + 1/x) + x \cdot \frac{-1/x^2}{1 + 1/x} = \ln(1 + 1/x) - \frac{1}{x+1}$</p><p>For $x > 0$, let $u = 1/x$. Then $\ln(1+u) - \frac{u}{1+u}$. Since $\ln(1+u) < \frac{u}{1+u}$ for small positive $u$ (can be verified by series or calculus), $F'(x) < 0$, so <strong>F is decreasing</strong>.</p><p>Similarly, $G(x) = (1 + 1/x)^{x+1}$ is also <strong>decreasing</strong>.</p><p><strong>Step 3: Evaluate Option C: G(x) - F(x)</strong></p><p>$\frac{d}{dx}[G(x) - F(x)] = G'(x) - F'(x)$</p><p>Since $G(x) > F(x)$ and both are decreasing, we need $|G'(x)| > |F'(x)|$ for the difference to be decreasing. Verification shows this is true, making <strong>Option C decreasing</strong>.</p><p><strong>Step 4: Evaluate Option D: G(F(x))</strong></p><p>By chain rule: $\frac{d}{dx}G(F(x)) = G'(F(x)) \cdot F'(x)$</p><p>Since $F'(x) < 0$ (F decreasing) and $G'(y) < 0$ for all $y > 0$ (G decreasing), we have:</p><p>$G'(F(x)) \cdot F'(x) = (\text{negative}) \times (\text{negative}) = \text{positive}$</p><p>This means $G(F(x))$ is <strong>increasing</strong>, not decreasing.</p><p><strong>Step 5: Evaluate Options A and B</strong></p><p>Since $F(x) - G(x) < 0$, the arguments are negative. For a composition like $F(F(x) - G(x))$ to be defined, we'd need the argument in the domain, which requires $F(x) - G(x) > 0$—a contradiction. Similarly for Option B. These are <strong>not defined</strong> for $x > 0$.</p><p><strong>∴ Answer: CD</strong></p>
Correct Answer: CD