Question:
Let \(f(x) = \displaystyle\int_0^x e^t f(t)\,dt + e^x\) be a differentiable function for all \(x \in R\). Then \(f(x)\) equals :
Let \(f(x) = \displaystyle\int_0^x e^t f(t)\,dt + e^x\) be a differentiable function for all \(x \in R\). Then \(f(x)\) equals :
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Differentiate to get \(f' = e^x(f+1)\) with \(f(0)=1\), then separate variables.
Updated On: Nov 4, 2026
- \(e^{(e^x - 1)}\)
- \(2e^{e^x} - 1\)
- \(2e^{e^x - 1} - 1\)
- \(e^{e^x} - 1\)
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The Correct Option is C
Solution and Explanation
At \(x=0\): \(f(0) = 0 + e^0 = 1\).
Differentiate: \(f'(x) = e^x f(x) + e^x = e^x\,(f(x) + 1)\).
Separate: \(\frac{f'(x)}{f(x)+1} = e^x\), so \(\ln(f(x)+1) = e^x + C\).
Using \(f(0) = 1\): \(\ln 2 = 1 + C\), \(C = \ln 2 - 1\).
\(f(x) + 1 = 2e^{e^x - 1}\), so \(f(x) = 2e^{e^x - 1} - 1\).
Answer: (3) \(2e^{e^x - 1} - 1\)
Differentiate: \(f'(x) = e^x f(x) + e^x = e^x\,(f(x) + 1)\).
Separate: \(\frac{f'(x)}{f(x)+1} = e^x\), so \(\ln(f(x)+1) = e^x + C\).
Using \(f(0) = 1\): \(\ln 2 = 1 + C\), \(C = \ln 2 - 1\).
\(f(x) + 1 = 2e^{e^x - 1}\), so \(f(x) = 2e^{e^x - 1} - 1\).
Answer: (3) \(2e^{e^x - 1} - 1\)
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