If f(x)=2x^{2}+bx+c,f(0)=3 and f(2)=1,then (f | vedclass

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(D) Given $f(x) = 2x^{2} + bx + c$.Since $f(0) = 3$,we have $2(0)^{2} + b(0) + c = 3$,which gives $c = 3$.Now,$f(x) = 2x^{2} + bx + 3$.Given $f(2) = 1

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$3$

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(D) Given $f(x) = 2x^{2} + bx + c$.Since $f(0) = 3$,we have $2(0)^{2} + b(0) + c = 3$,which gives $c = 3$.Now,$f(x) = 2x^{2} + bx + 3$.Given $f(2) = 1$,we have $2(2)^{2} + b(2) + 3 = 1$.$8 + 2b + 3 = 1 \Rightarrow 2b + 11 = 1 \Rightarrow 2b = -10 \Rightarrow b = -5$.Thus,$f(x) = 2x^{2} - 5x + 3$.First,calculate $f(1) = 2(1)^{2} - 5(1) + 3 = 2 - 5 + 3 = 0$.Now,$(f \circ f)(1) = f(f(1)) = f(0)$.Since $f(0) = 3$,we get $(f \circ f)(1) = 3$.

If $f(x)=2x^{2}+bx+c$,$f(0)=3$ and $f(2)=1$,then $(f \circ f)(1)=$

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