AP EAMCET · Maths · Functions
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 \(f\left(\frac{2 a}{1+a^2}\right)>0\) is
- A \((0, \infty)-\{1\}\)
- B \((-\infty, 0)-\{-1\}\)
- C \((-\infty, \infty)-\{-1,1\}\)
- D \((-1,1)\)
Answer & Solution
Correct Answer
(B) \((-\infty, 0)-\{-1\}\)
Step-by-step Solution
Detailed explanation
Given \(f(a)=\log \left|\frac{1-a}{1+a}\right|, a \neq\{-1,1\}\) Now \(f\left(\frac{2 a}{1+a^2}\right)>0 \Rightarrow \log \left|\frac{1-\frac{2 a}{1+a^2}}{1+\frac{2 a}{1+a^2}}\right|>0\)…
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