If \(f(x)=2 x^{2}+b x+c, f(0)=3\) and \(f(2)=1\), then \((f o f)(1)=\)

\(\begin{aligned} \text { Here } \mathrm{f}(\mathrm{x}) &=2 \mathrm{x}^{2}+\mathrm{bx}+\mathrm{c} \\ \mathrm{f}(0) &=0+0+\mathrm{c} \quad \Rightarrow 3=\mathrm{c} \\ \mathrm{f}(2) &=8+2 \mathrm{~b}+3 \end{aligned}\)
\(\begin{array}{ll}\therefore \quad \mathrm{f}=8 & +2 \mathrm{~b}+3 \Rightarrow 2 \mathrm{~b}=-10 \Rightarrow \mathrm{b}=-5 \\ \therefore \quad \mathrm{f}(\mathrm{x})= & 2 \mathrm{x}^{2}-5 \mathrm{x}+3 \\ & \mathrm{f}(1)=2-5+3=0 \\ & (\mathrm{f} 0 \mathrm{f})(\mathrm{x})=\mathrm{f}[\mathrm{f}(\mathrm{x})] \\ & (\mathrm{f} 0 \mathrm{f})(1)=\mathrm{f}[\mathrm{f}(1)] \\ & =\mathrm{f}(0)=0-0+3=3\end{array}\)