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
    $A = ln(e^2 -3) + 2$

Answer

Correct option: A.
$A = ln(e^2 -3) -2$
a
$\phi(x)=\mathrm{A} \mathrm{e}^{\mathrm{x}}+\mathrm{x}$

$\mathrm{A}=\int_{0}^{1} \mathrm{e}^{\mathrm{t}}\left\{\mathrm{A} \mathrm{e}^{\mathrm{t}}+\mathrm{t}\right\} \mathrm{dt}$

$\left.=\mathrm{A} \frac{\mathrm{e}^{2 \mathrm{t}}}{2}+\mathrm{te}^{\mathrm{t}}-\mathrm{e}^{\mathrm{t}}\right)_{0}^{1}$

$=\frac{\mathrm{A}}{2}\left(\mathrm{e}^{2}-1\right)+(\mathrm{e}-0)-(\mathrm{e}-1)$

$A=\frac{A}{2}\left(e^{2}-1\right)+1$

$A\left(1-\frac{e^{2}-1}{2}\right)=1 \Rightarrow A=\frac{2}{3-e^{2}}$

$\phi(x)=\frac{2}{3-e^{2}} e^{x}+x$

$\phi \left( {\ln \left( {{e^2} - 3} \right)} \right) = 2 + \ln \left( {{e^2} - 3} \right)$

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