Question

Suppose f(x) and g(x) are two continuous functions defined for $0\le x\le 1$.
Given $f\left(x\right)={\int }_0^1{e}^{x+t}.f\left(t\right)dt$ and $g\left(x\right)={\int }_0^1{e}^{x+t}.g\left(t\right)dt+x$.

The value of g(0)-f(0) equals

• A. $\frac{2}{3-e^2}$
  
• B. $\frac{3}{e^2-2}$
  
• C. $\frac{1}{e^2-1}$
  
• D. 0

Answer 1

The correct option is A

$\frac{2}{3-e^2}$

$g\left(x\right)=e^x\underset{}{{\int }_0^1{e}^t.g\left(t\right)dt}+x$

$g\left(x\right)=B{e}^x+x$ ...(2) $\Rightarrow g\left(t\right)=B{e}^t+t$

where $B={\int }_0^1{e}^{t}g\left(t\right)dt;B={\int }_0^1{e}^t(B{e}^t+t)dt;B=B{\int }_0^1{e}^{2t}dt+{\int }_0^1{e}^t.tdt$

but $\int e^{2t}dt=\frac{1}{2}(e^2-1)$ and ${\int }_0^{1}t.e^t.dt=1$

$\therefore B=\frac{B}{2}(e^2-1)+1\Rightarrow 2B=B(e^2-1)+2\Rightarrow 3B=B{e}^2+2\Rightarrow B=\frac{2}{3-e^2}$

$\therefore$ from (2)

$g\left(x\right)=\left(\frac{2}{3-e^2}\right)e^x+x;$

$g\left(0\right)=\frac{2}{3-e^2}$