MHT CET · Maths · Functions
If \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\), such that \(f(x)=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}\), then \(\mathrm{f}\) is
- A a periodic function
- B an even function
- C an odd function
- D a neither even nor odd function
Answer & Solution
Correct Answer
(C) an odd function
Step-by-step Solution
Detailed explanation
\(f(x)=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}\)
\(=\frac{e^{x}+\frac{1}{e^{x}}}{e^{x}-\frac{1}{e^{x}}}=\frac{e^{2 x}+1}{e^{2 x}-1}\)
\(f(-x)=\frac{e^{-x}+e^{x}}{e^{-x}-e^{x}}\)
\(=\frac{\frac{1}{e^{x}}+e^{x}}{\frac{1}{e^{x}}-e^{x}}=\frac{1+e^{2 x}}{1-e^{2 x}}=\frac{1+e^{2 x}}{-\left(e^{2 x}-1\right)}\)
\(\therefore f(-x)=-f(x)\)
\(\therefore \mathrm{f}(\mathrm{x})\) is an odd function.
\(=\frac{e^{x}+\frac{1}{e^{x}}}{e^{x}-\frac{1}{e^{x}}}=\frac{e^{2 x}+1}{e^{2 x}-1}\)
\(f(-x)=\frac{e^{-x}+e^{x}}{e^{-x}-e^{x}}\)
\(=\frac{\frac{1}{e^{x}}+e^{x}}{\frac{1}{e^{x}}-e^{x}}=\frac{1+e^{2 x}}{1-e^{2 x}}=\frac{1+e^{2 x}}{-\left(e^{2 x}-1\right)}\)
\(\therefore f(-x)=-f(x)\)
\(\therefore \mathrm{f}(\mathrm{x})\) is an odd function.
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