Question

If the interval in which the real-valued function \[ f(x) = \log\left(\frac{1+x}{1-x}\right) - 2x - \frac{x^{3}}{1-x^{2}} \] is decreasing in \( (a,b) \), where \( |b-a| \) is maximum, then {a}⁄{b} =

• A. -1
• B. 1
• C. \(\frac{2}{3}\)
• D. \(\frac{3}{2}\)

Hint:

Remember that a function is decreasing where its derivative is negative (or non-positive if we include the endpoints).

Answer 1

The Correct Option is A

To find the interval where the function is decreasing, we need to analyze its derivative. First, find the derivative of \(f(x)\): $$f'(x) = \frac{d}{dx} \left[ \log\left(\frac{1+x}{1-x}\right) - 2x - \frac{x^3}{1-x^2} \right]$$ Using the chain rule and quotient rule, we get: $$f'(x) = \frac{1-x}{1+x} \cdot \frac{(1-x) + (1+x)}{(1-x)^2} - 2 - \frac{3x^2(1-x^2) + 2x^4}{(1-x^2)^2}$$ Simplifying the expression: $$f'(x) = \frac{2}{(1-x^2)} - 2 - \frac{3x^2 - x^4}{(1-x^2)^2}$$ $$f'(x) = \frac{2(1-x^2) - 2(1-x^2)^2 - 3x^2 + x^4}{(1-x^2)^2}$$ $$f'(x) = \frac{2 - 2x^2 - 2 + 4x^2 - 2x^4 - 3x^2 + x^4}{(1-x^2)^2}$$ $$f'(x) = \frac{-x^4 - x^2}{(1-x^2)^2} = -\frac{x^2(x^2 + 1)}{(1-x^2)^2}$$ Since \(x^2(x^2 + 1)\) is always non-negative and \((1-x^2)^2\) is always positive (except at \(x = \pm 1\) where it's zero), \(f'(x)\) is always non-positive. This means the function is always decreasing, except at \(x = \pm 1\) where it's undefined. The largest interval where the function is defined and decreasing is \((-1, 1)\). Therefore, \(a = -1\) and \(b = 1\), and \(\frac{a}{b} = -1\).