let the integral equation is
$$\phi (x)-\frac{e}{2}\int_{-1}^{1}xe^{t}\phi(t)dt=f(x)$$
then
1).there exist a continous function $f:[-1,1]\to (0,\infty)$ for which solution exist.
2).there exist a continous function $f:[-1,1]\to (-\infty,0)$ for which solution exist.
3).for $f(x)=e^{-x}(1-3x^2)$ ,a solution exist
4).for $f(x)=e^{-x}(x+x^3+x^5)$ ,a solution exist
how to solve this question ,i can only solve simple questions in integral equations but this very tough ,
$\text{EDIT}$
After so much of reading of integral equation,i try to solve the question
$\textbf{the solution i tried}$-$$\phi (x)=\frac{e}{2}\int_{-1}^{1}xe^{t}\phi(t)dt+f(x)$$
$$\phi (x)=\frac{e}{2}x\int_{-1}^{1}e^{t}\phi(t)dt +f(x)$$
$$\phi (x)=\frac{e}{2}xC+f(x) \;\;\;\;\;\;\;....1$$ where $C=\int_{-1}^{1}e^{t}\phi(t)dt$
$\textbf{from (1)}$
$$C=\int_{-1}^{1}e^{t}(\frac{e}{2}tC+f(t))$$
$$C=\int_{-1}^{1}e^{t}\frac{e}{2}tCdt+\int_{-1}^1 e^tf(t)dt$$
$$C=\frac{e}{2}C\int_{-1}^{1}e^{t}tdt+\int_{-1}^1 e^tf(t)dt$$
further doing calculation i get $$C=C+\int_{-1}^1 e^tf(t)dt$$
$$\implies \int_{-1}^1 e^tf(t)dt=0$$ after that i pick the option 'c' and 'd' and they satisfied the integral so they are sutiable for $f(x)$ and my option are also right but still don't know how to work for option 'a' and 'b'
please help.
Thankyou