Question

If \( f(a) = \log \left| \frac{1 - a}{1 + a} \right| \) for \( a \neq \{-1, 1\} \), then the set of values of all \( a \), for which \( f\left( \frac{2a}{1 + a^2} \right)>0 \) is:

• A. \( (0, \infty) - \{1\} \)
• B. \( (-\infty, 0) - \{ -1 \} \)
• C. \( (-\infty, \infty) - \{-1, 1\} \)
• D. \( (-1, 1) \)

Hint:

When dealing with composite functions inside logarithms, simplify step by step and carefully examine the domain and sign of the inner expressions.

Answer 1

The Correct Option is B

We are given: \[ f(a) = \log \left| \frac{1 - a}{1 + a} \right|, \quad a \neq -1, 1 \] We are required to find values of \( a \) such that: \[ f\left( \frac{2a}{1 + a^2} \right)>0 \] Let \( x = \frac{2a}{1 + a^2} \). Then: \[ f(x) = \log \left| \frac{1 - x}{1 + x} \right|>0 \Rightarrow \left| \frac{1 - x}{1 + x} \right|>1 \] This implies: \[ \left| \frac{1 - x}{1 + x} \right|>1 \Rightarrow \left| \frac{1 - x}{1 + x} \right|^2>1 \Rightarrow \left( \frac{1 - x}{1 + x} \right)^2>1 \Rightarrow \frac{(1 - x)^2}{(1 + x)^2}>1 \] \[ \Rightarrow (1 - x)^2>(1 + x)^2 \Rightarrow 1 - 2x + x^2>1 + 2x + x^2 \Rightarrow -2x>2x \Rightarrow -4x>0 \Rightarrow x<0 \] So, we want: \[ \frac{2a}{1 + a^2}<0 \] Now, observe the sign of \( \frac{2a}{1 + a^2} \): - The denominator \( 1 + a^2>0 \) for all real \( a \) - Hence, \( \frac{2a}{1 + a^2}<0 \) when \( 2a<0 \Rightarrow a<0 \) Also, we are given \( a \neq \pm 1 \) \[ \Rightarrow \boxed{a \in (-\infty, 0) \setminus \{-1\}} \]