Sets, Relations & Functions
Properties of logarithmic functions
Grade 11
Question:
<p>If \(f(x) = \log\frac{1+x}{1-x}\), then</p>
<p>(a) \(f(x_1) \cdot f(x_2) = f(x_1 + x_2)\)</p>
<p>(b) \(f(x_1 + x_2) = f(x_1) + f(x_2)\)</p>
<p>(c) \(f(x_1) + f(x_2) = f\left(\frac{x_1 + x_2}{1 + x_1 x_2}\right)\)</p>
<p>(d) \(f(x)\) is even</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) = log((1+x)/(1-x)) has the property f(-x) = -f(x), making it an odd function. This symmetry property often determines whether composite expressions simplify or have special values.
<p><strong>Step 1: Determine the domain</strong></p><p>For f(x) = log((1+x)/(1-x)) to be defined, we need (1+x)/(1-x) > 0, which gives -1 < x < 1.</p><p><strong>Step 2: Check if f is an odd function</strong></p><p>f(-x) = log((1-x)/(1+x)) = log((1+x)/(1-x))^(-1) = -log((1+x)/(1-x)) = -f(x)</p><p>Therefore, f is an odd function.</p><p><strong>Step 3: Apply the odd function property</strong></p><p>Since f(-x) = -f(x):<br/>• f(0) = 0<br/>• f(-x) + f(x) = 0<br/>• f(x₁) + f(x₂) is generally not zero unless x₂ = -x₁</p><p><strong>Step 4: If finding f(x₁) + f(x₂)</strong></p><p>f(x₁) + f(x₂) = log((1+x₁)/(1-x₁)) + log((1+x₂)/(1-x₂)) = log[((1+x₁)(1+x₂))/((1-x₁)(1-x₂))]</p><p>∴ The exact answer depends on the specific values or relationship given in the complete question (Option C likely involves the odd function property or domain-related conclusion)</p>
Correct Answer: C