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 f(1) equals

• A. 0
• B. 1
• C. $e^{-1}$
• D. e

Answer 1

The correct option is A

0

$\left(\begin{array}{cc}f\left(x\right)=& e^x\underset{}{{\int }_0^1{e}^t.f\left(t\right)dt}\\ & \text{A say}\end{array}\right)$

$f\left(x\right)=A{e}^x$ ...(1) $\Rightarrow f\left(t\right)=A{e}^t$

where, $A={\int }_0^1{e}^t.f\left(t\right)dt\Rightarrow A={\int }_0^1.A{e}^{t}dt;A=A{\int }_0^1{e}^{2t}dt$

now $A[{\int }_0^1{e}^{2t}dt-1]=0\Rightarrow A=0$ as ${\int }_0^1{e}^{2t}dt\ne 0$

Hence $f\left(x\right)=0\Rightarrow f\left(1\right)=0$