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Let $f(x)=\mathrm{A} x^2+\mathrm{B} x, g(x)=\mathrm{L} x^2+\mathrm{M} x+\mathrm{N}$. Given that $f(2)-g(2)=1, f(3)-g(3)=4, f(4)-g(4)=9$. Then a root of $f(x)-g(x)=0$ is
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We have, $\mathrm{f}(\mathrm{x})=\mathrm{Ax}^2+\mathrm{Bx}$
$\mathrm{g}(\mathrm{x})=\mathrm{Lx}^2+\mathrm{Mx}+\mathrm{N}$
now $\mathrm{f}(2)-\mathrm{g}(2)=1$ (Given)
$\mathrm{f}(3)-\mathrm{g}(3)=4$ (Given)
$f(4)-g(4)=9$ (Given)
On solving (i), (ii) and (iii) we get $\mathrm{A}-\mathrm{L}=1, \mathrm{~B}-\mathrm{M}=-2$ and $\mathrm{N}=-1$
Now $\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x})=0$ $\Rightarrow x^2-2 x+1=0 \Rightarrow(x-1)^2=0 \Rightarrow x=1$
$\mathrm{g}(\mathrm{x})=\mathrm{Lx}^2+\mathrm{Mx}+\mathrm{N}$
now $\mathrm{f}(2)-\mathrm{g}(2)=1$ (Given)
$\mathrm{f}(3)-\mathrm{g}(3)=4$ (Given)
$f(4)-g(4)=9$ (Given)
On solving (i), (ii) and (iii) we get $\mathrm{A}-\mathrm{L}=1, \mathrm{~B}-\mathrm{M}=-2$ and $\mathrm{N}=-1$
Now $\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x})=0$ $\Rightarrow x^2-2 x+1=0 \Rightarrow(x-1)^2=0 \Rightarrow x=1$
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