MHT CET · Maths · Functions
If \(\mathrm{f}(x)=\frac{x}{2-x}, \mathrm{~g}(x)=\frac{x+1}{x+2}\), then (gogof) \((x)=\)
- A \(\frac{6+x}{10-2 x}\)
- B \(\frac{6-x}{10+2 x}\)
- C \(\frac{6+x}{10+2 x}\)
- D \(\frac{6-x}{10-2 x}\)
Answer & Solution
Correct Answer
(D) \(\frac{6-x}{10-2 x}\)
Step-by-step Solution
Detailed explanation
\(f(x)=\frac{x}{2-x}, g(x)=\frac{x+1}{x+2}\)
\((\) gof \()(x)=\frac{\frac{x}{2-x}+1}{\frac{x}{2-x}+2}\)
\(\begin{aligned}
& \text { (gogof) }(x)=\frac{\frac{\frac{x}{2-x}+1}{x}+1}{\frac{\frac{x}{2-x}+2}{\frac{x}{2-x}+2}+2} \\
& =\frac{\frac{x+2-x}{x+4-2 x}+1}{\frac{x+2-x}{x+4-2 x}+2}
\end{aligned}\)
\(\begin{aligned} & =\frac{\frac{2}{4-x}+1}{\frac{2}{4-x}+2} \\ & =\frac{2+4-x}{2+8-2 x} \\ & =\frac{6-x}{10-2 x}\end{aligned}\)
\((\) gof \()(x)=\frac{\frac{x}{2-x}+1}{\frac{x}{2-x}+2}\)
\(\begin{aligned}
& \text { (gogof) }(x)=\frac{\frac{\frac{x}{2-x}+1}{x}+1}{\frac{\frac{x}{2-x}+2}{\frac{x}{2-x}+2}+2} \\
& =\frac{\frac{x+2-x}{x+4-2 x}+1}{\frac{x+2-x}{x+4-2 x}+2}
\end{aligned}\)
\(\begin{aligned} & =\frac{\frac{2}{4-x}+1}{\frac{2}{4-x}+2} \\ & =\frac{2+4-x}{2+8-2 x} \\ & =\frac{6-x}{10-2 x}\end{aligned}\)
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