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If $f(x)=a x^2+b x+c$ satisfies $f(1)+2 f(2)=0$ and $2 \mathrm{f}(1)+\mathrm{f}(2)=0$, then $3 \mathrm{a}+\mathrm{b}=$
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$f(x)=a x^2+b x+c$ (Given)
$$
\Rightarrow \mathrm{f}(1)=\mathrm{a}+\mathrm{b}+\mathrm{c}
$$
$$
2 \mathrm{f}(2)=2(4 \mathrm{a}+2 \mathrm{~b}+\mathrm{c})
$$
since $f(1)+f(2)=0$
$$
\Rightarrow 9 \mathrm{a}+5 \mathrm{~b}+3 \mathrm{c}=0...(i)
$$
similarly
$$
6 a+4 b+3 c=0...(ii)
$$
now subtracting (ii) from (i) we get $3 a+b=0$
$$
\Rightarrow \mathrm{f}(1)=\mathrm{a}+\mathrm{b}+\mathrm{c}
$$
$$
2 \mathrm{f}(2)=2(4 \mathrm{a}+2 \mathrm{~b}+\mathrm{c})
$$
since $f(1)+f(2)=0$
$$
\Rightarrow 9 \mathrm{a}+5 \mathrm{~b}+3 \mathrm{c}=0...(i)
$$
similarly
$$
6 a+4 b+3 c=0...(ii)
$$
now subtracting (ii) from (i) we get $3 a+b=0$
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