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

Question

Aicharya · Accepted Answer

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

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

Solution