Question

If $f\left(x\right)=x+{\int }_0^1(x{y}^2+x^{2}y)f\left(y\right)dy$, and $f\left(x\right)=\frac{A{x}^2+Bx}{119}$ then $\left[\frac{A+B}{100}\right]$ = (where [.] is G.I.F) ___

Answer 1

$f\left(x\right)=x+x{\int }_0^1{y}^{2}f\left(y\right)dy+x^2{\int }_0^{1}yf\left(y\right)dy$

$=x[1+{\int }_0^1{y}^{2}f\left(y\right)dy]+x^2\left[{\int }_0^{1}yf\left(y\right)dy\right]$

f(x) is quadratic expression $f\left(x\right)=ax+b{x}^2$ or $f\left(y\right)=ay+b{y}^2$ . . . (1)

$a=1+{\int }_0^1{y}^{2}f\left(y\right)dy$

$=1+{\int }_0^1{y}^2(ay+b{y}^2)dy$

$=1+{[\frac{a{y}^4}{4}+\frac{b{y}^5}{5}]}_0^1$ $a=1+(\frac{a}{4}+\frac{b}{5})$ . . . (2)

20a = 20 + 5a + 4b 15a - 4b = 20

$b={\int }_0^{1}yf\left(y\right)dy={\int }_0^{1}y(ay+b{y}^2)dy$

$={(\frac{a{y}^3}{3}+\frac{b{y}^4}{4})}_0^1\Rightarrow b=\frac{a}{3}+\frac{b}{4}$

12b = 4a + 3b 9b - 4a = 0 . . . .(3) from (2) and (3)

$a=\frac{180}{119}$ and $b=\frac{80}{119}$

$f\left(x\right)=\frac{180}{119}x+\frac{80}{119}{x}^2=\frac{A{x}^2+Bx}{119}$

A = 80 B = 180

A + B = 260

$\left[\frac{A+B}{100}\right]=2$