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If $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}, g: \mathrm{R} \rightarrow \mathrm{R}$ defined by $f(x)=x^{2}-3 x+4$ and $g(x)=2 x+1$, then the
value of $x$ for which $f(x)=f \circ g)(x)$ is
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value of $x$ for which $f(x)=f \circ g)(x)$ is
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Verified Answer
The correct answer is:
$-1, \frac{2}{3}$
$($ fog $)(x)=f[g(x)]=f(2 x+1)=(2 x+1)^{2}-3(2 x+1)+4$
$\quad=4 x^{2}+4 x+1-6 x-3+4=4 x^{2}-2 x+2$
Given $f(x)=(f \circ g)(x)$
$\therefore x^{2}-3 x+4=4 x^{2}-2 x+2$
$3 x^{2}+x-2=0 \Rightarrow 3 x^{2}+3 x-2 x-2=0$
$3 x(x+1)-2(x+1)=0 \Rightarrow(x+1)(3 x-2)=0$
$\therefore x=-1, \frac{2}{3}$
$\quad=4 x^{2}+4 x+1-6 x-3+4=4 x^{2}-2 x+2$
Given $f(x)=(f \circ g)(x)$
$\therefore x^{2}-3 x+4=4 x^{2}-2 x+2$
$3 x^{2}+x-2=0 \Rightarrow 3 x^{2}+3 x-2 x-2=0$
$3 x(x+1)-2(x+1)=0 \Rightarrow(x+1)(3 x-2)=0$
$\therefore x=-1, \frac{2}{3}$
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