If $f(x)=x^{2}-3 x+4$ and $f(x)=f(2 x+1)$, then $x=$

If $f(x)=x^{2}-3 x+4$ and $f(x)=f(2 x+1)$, then $x=$
  1. $-1, \frac{2}{3}$
  2. $-1, \frac{3}{2}$
  3. $1, \frac{3}{2}$
  4. $1, \frac{2}{3}$

Solution

$\begin{array}{ll}\text { Here } \mathrm{f}(2 \mathrm{x}+1) & =(2 \mathrm{x}+1)^{2}-3(2 \mathrm{x}+1)+4 \\ & =4 \mathrm{x}^{2}+4 \mathrm{x}+1-6 \mathrm{x}-3+4 \\ \therefore \quad \mathrm{f}(2 \mathrm{x}+1) & =4 \mathrm{x}^{2}-2 \mathrm{x}+2 \\ \text { Given } \mathrm{f}(\mathrm{x})=\mathrm{f}(2 \mathrm{x}+1) \\ \therefore \mathrm{x}^{2}-3 \mathrm{x}+4=4 \mathrm{x}^{2}-2 \mathrm{x}+2 \\ \therefore \quad 3 \mathrm{x}^{2}+\mathrm{x}-2 \quad=0 \Rightarrow 3 \mathrm{x}^{2}+3 \mathrm{x}-2 \mathrm{x}-2=0 \\ \quad 3 \mathrm{x}(\mathrm{x}+1)-2(\mathrm{x}+1)=0 \Rightarrow(\mathrm{x}+1)(3 \mathrm{x}-2)=0 \\ \therefore \mathrm{x}=-1, \frac{2}{3}\end{array}$

Asked in: MHT CET 2020 (15 Oct Shift 1)

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