MCQ
Let $\phi \left( x \right) = \int\limits_0^1 {{e^x}{e^t}\phi (t)} dt + x$ and $\phi \left( {\ln \left( {{e^2} - 3} \right)} \right)$ is equal to $A$ , then
- ✓$A = ln(e^2 -3) -2$
- B$A \in (3,4)$
- C$A = e^2 -3$
- D$A = ln(e^2 -3) + 2$
$\mathrm{A}=\int_{0}^{1} \mathrm{e}^{\mathrm{t}}\left\{\mathrm{A} \mathrm{e}^{\mathrm{t}}+\mathrm{t}\right\} \mathrm{dt}$
$\left.=\mathrm{A} \frac{\mathrm{e}^{2 \mathrm{t}}}{2}+\mathrm{te}^{\mathrm{t}}-\mathrm{e}^{\mathrm{t}}\right)_{0}^{1}$
$=\frac{\mathrm{A}}{2}\left(\mathrm{e}^{2}-1\right)+(\mathrm{e}-0)-(\mathrm{e}-1)$
$A=\frac{A}{2}\left(e^{2}-1\right)+1$
$A\left(1-\frac{e^{2}-1}{2}\right)=1 \Rightarrow A=\frac{2}{3-e^{2}}$
$\phi(x)=\frac{2}{3-e^{2}} e^{x}+x$
$\phi \left( {\ln \left( {{e^2} - 3} \right)} \right) = 2 + \ln \left( {{e^2} - 3} \right)$
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