MHT CET · Maths · Definite Integration
\(\int_{-5}^{5}\left[\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}\right] d x=\)
- A 0
- B 1
- C \(3 e^{5}\)
- D \(2 e^{5}\)
Answer & Solution
Correct Answer
(A) 0
Step-by-step Solution
Detailed explanation
\(\text {Let } f(x)=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}} \)
\( =\frac{e^{x}+\frac{1}{e^{x}}}{e^{x}-\frac{1}{e^{x}}}=\frac{e^{2 x}+1}{e^{2 x}-1} \text { and } f(-x)=\frac{e^{-x}+e^{x}}{e^{-x}-e^{x}}=\) \(\frac{\frac{1}{e^{x}}+e^{x}}{\frac{1}{e^{x}}-e^{x}}=\frac{1+e^{2 x}}{1-e^{2 x}} \)
\( \therefore f(x)=-f(x)\)
Thus \(f(x)\) is an odd function.
\(\therefore \int_{-5}^{5} \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}} d x=0\)
\( =\frac{e^{x}+\frac{1}{e^{x}}}{e^{x}-\frac{1}{e^{x}}}=\frac{e^{2 x}+1}{e^{2 x}-1} \text { and } f(-x)=\frac{e^{-x}+e^{x}}{e^{-x}-e^{x}}=\) \(\frac{\frac{1}{e^{x}}+e^{x}}{\frac{1}{e^{x}}-e^{x}}=\frac{1+e^{2 x}}{1-e^{2 x}} \)
\( \therefore f(x)=-f(x)\)
Thus \(f(x)\) is an odd function.
\(\therefore \int_{-5}^{5} \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}} d x=0\)
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