MHT CET · Maths · Functions
If \(f(x)=\frac{x}{2 x+1}\) and \(g(x)=\frac{x}{x+1}\), then \((f \circ g)(x)=\)
- A \(\frac{2 x-1}{x+1}\)
- B \(\frac{x}{3 x+1}\)
- C \(\frac{x+1}{x+2}\)
- D \(\frac{x-1}{2 x+1}\)
Answer & Solution
Correct Answer
(B) \(\frac{x}{3 x+1}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& (f \circ g)(x)=f\left[\frac{x}{x+1}\right] \\
& =\frac{\left(\frac{x}{x+1}\right)}{2\left(\frac{x}{x+1}\right)+1}=\frac{x}{x+1} \times \frac{x+1}{2 x+x+1}=\frac{x}{3 x+1}
\end{aligned}
\)
\(\begin{aligned} & (f \circ g)(x)=f\left[\frac{x}{x+1}\right] \\ & =\frac{\left(\frac{x}{x+1}\right)}{2\left(\frac{x}{x+1}\right)+1}=\frac{x}{x+1} \times \frac{x+1}{2 x+x+1}=\frac{x}{3 x+1}\end{aligned}\)
\begin{aligned}
& (f \circ g)(x)=f\left[\frac{x}{x+1}\right] \\
& =\frac{\left(\frac{x}{x+1}\right)}{2\left(\frac{x}{x+1}\right)+1}=\frac{x}{x+1} \times \frac{x+1}{2 x+x+1}=\frac{x}{3 x+1}
\end{aligned}
\)
\(\begin{aligned} & (f \circ g)(x)=f\left[\frac{x}{x+1}\right] \\ & =\frac{\left(\frac{x}{x+1}\right)}{2\left(\frac{x}{x+1}\right)+1}=\frac{x}{x+1} \times \frac{x+1}{2 x+x+1}=\frac{x}{3 x+1}\end{aligned}\)
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