Find the slope out of the functional

Put $x=11$ in the Given equation to get $f(11)+2f(121)=L(11)$

Now Put $x=121$ in the Given equation to get $f(121)+2f(11)=L(121)$

Substitute the Value of $f(11)$ to get. $2L(121)-L(11)=847\times 3$

Let $L(x)=y=m\times x +c$ where $m$ is the slope of line representing $L(x)$

Now since the line passes through origin so $c=0$

Now $y=m\times x$

Using this relation in the above equation we get $m(121\times2-11)=847\times3$

$m=\dfrac{847 \times 3 }{231}$

By this $\boxed{ m=11}$

1. Let $x=a$

   $f (a) + 2f(\frac{1331}{a}) = L \times a$

2. Let $x= \frac{1331}{a}$

   $f(\frac{1331}{a}) + 2f(a) = L \times \frac{1331} {a}$

   $2f(\frac{1331}{a}) + 4f(a) = L \times \frac{2662}{a}$

$(2) - (1)$ :

$3f(a) = L \times \frac{2662}{a} - L \times a$

$3f(x) = L \times \frac{2662}{x} - L \times x$

Since $f(11) = 847$ , then

$3f(11) = L \times \frac{2662}{11} - L \times 11$

$3 \times 847 = 242L - 11L$

$2541 = 231 \times L$

$L = \boxed{11}$

This is a linear system governed by $\left(\begin{array}{cc}1 & 2 \\2 & 1\end{array}\right)\left(\begin{array}{c}f_{11} \\ f_{11^2}\end{array}\right) = L \left(\begin{array}{cc}11 \\ 11^2\end{array}\right)$ which restricts the solution of $L$ to $847 = -\frac{L}{3}(11 - 2 \times 11^2) \implies L = 11$

$f(x) +2f(1331/x)=mx (1)$ (since l(x)is a straight line passing through origin hence c=0) here $m=slope$ and $c=intercept$ on the y axis. Replace x by 1331/x in (1) $f(1331/x) +2f(x)=m(1331/x) (2)$ adding (1) and (2) and dividing by 3 both sides $f(x)+f(1331/x)=1/3(mx+1331m/x)$ (3) put the value of f(1331/x) from (2) in (3) $1331m/x -f(x)=mx/3+1331m/3x$ $Put, x=11$ $77m=f(11)$ Hence $m=11$