TS EAMCET · Maths · Functions
If \(\quad e^{f(x)}=\frac{10+x}{10-x}, x \in(-10,10) \quad\) and \(f(x)=k f\left(\frac{200 x}{100+x^2}\right)\), then \(k\) is equal to
- A 0.5
- B 0.6
- C 0.7
- D 0.8
Answer & Solution
Correct Answer
(A) 0.5
Step-by-step Solution
Detailed explanation
We have, \(e^{f(x)}=\frac{10+x}{10-x}\) \(f(x)=\log \frac{10+x}{10-x}\) Given that \(f(x)=k f\left(\frac{200 x}{100+x^2}\right)\) \(\Rightarrow \quad \log \frac{10+x}{10-x}=k \cdot \log \left\{\frac{10+\frac{200 x}{100+x^2}}{10-\frac{200 x}{100+x^2}}\right\}\)…
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