MHT CET · Maths · Functions
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
- A \(1, \frac{-2}{3}\)
- B \(-1, \frac{2}{3}\)
- C \(1, \frac{2}{3}\)
- D \(-1, \frac{-2}{3}\)
Answer & Solution
Correct Answer
(B) \(-1, \frac{2}{3}\)
Step-by-step Solution
Detailed explanation
\((\) 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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