Question:
If
\( \int\left(e^{2x}+2e^x-e^{-x}-1\right)e^{\left(e^x+e^{-x}\right)}dx = g(x)\,e^{\left(e^x+e^{-x}\right)}+c \)
where \(c\) is a constant of integration, then \(g(0)\) is equal to :
If
\( \int\left(e^{2x}+2e^x-e^{-x}-1\right)e^{\left(e^x+e^{-x}\right)}dx = g(x)\,e^{\left(e^x+e^{-x}\right)}+c \)
where \(c\) is a constant of integration, then \(g(0)\) is equal to :
\( \int\left(e^{2x}+2e^x-e^{-x}-1\right)e^{\left(e^x+e^{-x}\right)}dx = g(x)\,e^{\left(e^x+e^{-x}\right)}+c \)
where \(c\) is a constant of integration, then \(g(0)\) is equal to :
Show Hint
Differentiate the right side: \(g'+g\,(e^x-e^{-x})\) must equal \(e^{2x}+2e^x-e^{-x}-1\); guess \(g=e^x+1\).
Updated On: Nov 5, 2026
- \(1\)
- \(2\)
- \(e\)
- \(e^2\)
Show Solution
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The Correct Option is B
Solution and Explanation
Let \(h(x)=e^x+e^{-x}\), so \(h'=e^x-e^{-x}\). Differentiating the RHS: \(\frac{d}{dx}\left[g\,e^{h}\right]=\left(g'+g\,(e^x-e^{-x})\right)e^h\).
So we need \(g'+g\,(e^x-e^{-x})=e^{2x}+2e^x-e^{-x}-1\).
Try \(g=e^x+1\): \(g'=e^x\), \(g(e^x-e^{-x})=e^{2x}-1+e^x-e^{-x}\). Sum \(=e^{2x}+2e^x-e^{-x}-1\). It matches.
Hence \(g(x)=e^x+1\) and \(g(0)=1+1=2\).
Answer: (2) \(2\)
So we need \(g'+g\,(e^x-e^{-x})=e^{2x}+2e^x-e^{-x}-1\).
Try \(g=e^x+1\): \(g'=e^x\), \(g(e^x-e^{-x})=e^{2x}-1+e^x-e^{-x}\). Sum \(=e^{2x}+2e^x-e^{-x}-1\). It matches.
Hence \(g(x)=e^x+1\) and \(g(0)=1+1=2\).
Answer: (2) \(2\)
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