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 \( a \neq \pm 1 \) \[ \Rightarrow \boxed{a \in (-\infty, 0) \setminus \{-1\}} \]

Answer 2

Step 1: Understand the given function and domain.
Given:
\[ f(a) = \log \left| \frac{1 - a}{1 + a} \right|, \] with \( a \neq \pm 1 \) to avoid division by zero.

Step 2: Analyze the inequality \( f\left( \frac{2a}{1 + a^2} \right) > 0 \).
Substitute \( x = \frac{2a}{1 + a^2} \) into the function:
\[ f(x) = \log \left| \frac{1 - x}{1 + x} \right| > 0. \]

Since logarithm is positive if and only if the argument is greater than 1:
\[ \left| \frac{1 - x}{1 + x} \right| > 1. \]

Step 3: Solve the inequality for \( x \).
Consider two cases due to absolute value:
1) \(\frac{1 - x}{1 + x} > 1\),
2) \(\frac{1 - x}{1 + x} < -1\).

Case 1:
\[ \frac{1 - x}{1 + x} > 1 \implies 1 - x > 1 + x \implies -x > x \implies -2x > 0 \implies x < 0. \] But also denominator \( 1 + x \neq 0 \implies x \neq -1 \).

Case 2:
\[ \frac{1 - x}{1 + x} < -1 \implies 1 - x < - (1 + x) \implies 1 - x < -1 - x \implies 1 < -1, \] which is false.

So only Case 1 holds with \( x < 0 \) and \( x \neq -1 \).

Step 4: Recall \( x = \frac{2a}{1 + a^2} \) and analyze its range.
Since \( 1 + a^2 > 0 \) for all real \( a \),
the sign of \( x \) depends on the numerator \( 2a \):
\[ x < 0 \iff \frac{2a}{1 + a^2} < 0 \iff a < 0. \]

Also, check if \( x = -1 \) is possible:
\[ \frac{2a}{1 + a^2} = -1 \implies 2a = -1 - a^2 \implies a^2 + 2a + 1 = 0 \implies (a + 1)^2 = 0 \implies a = -1. \]
Since \( a = -1 \) is excluded (denominator zero in original function), remove it.

Step 5: Final solution set.
The set of all \( a \) such that:
\[ a < 0 \quad \text{and} \quad a \neq -1, \] which is:
\[ (-\infty, 0) - \{ -1 \}. \]