Lời giải:
a) Khi $m=\sqrt{2}$ thì: \(y=f(x)=2x\)
\(f(1007)=2.1007=2014\)
b) Ta có:
\(f(-1)=m^2(-1)=-m^2\Rightarrow f(f(-1))=f(-m^2)=m^2(-m^2)=-m^4\)
\(f(2)=m^2.2=2m^2\) \(\Rightarrow f(f(2))=f(2m^2)=m^2.2m^2=2m^4\)
\(f(4)=m^2.4=4m^2\)
Để \(f(f(-1))+f(f(2))-f(4)=0\)
\(\Leftrightarrow -m^4+2m^4-4m^2=0\)
\(\Leftrightarrow m^4-4m^2=0\)
\(\Leftrightarrow m^2(m^2-4)=0\Rightarrow m^2-4=0\) (do $m\neq 0$)
\(\Rightarrow m^2=4\Rightarrow m=\pm 2\)