Answer
a) $1$ b) $2x+2y-1$
Work Step by Step
Since, we have $L(x,y)=f(x,y)+f_x(x-a)+f_y(y-b)$ ...(1)
a ) Now,
$f_x=2x \implies f_x(0,0)=0$; and $f_y=2y \implies f_y(0,0)=0$
and $f(0,0)=1$
Now eq. (1) becomes $L(0,0)=1+0(x-0)+0(y-0)=1$
b) Since, we have $L(x,y)=f(x,y)+f_x(x-a)+f_y(y-b)$ ...(1)
Now,
$f_x=2x \implies f_x(1,1)=2$; and $f_y=2y \implies f_y(1,1)=2$
and $f(1,1)=3$
Now eq. (1) becomes $L(1,1)=3+2(x-1)+2(y-1)=2x+2y-1$
Hence, a) $1$ b) $2x+2y-1$
