CUET · MATHS · PYQ PAPER 2023
If \(f(x)=\frac{1}{2 x+1}, x \neq-\frac{1}{2}\), then \(f[f(x)]\) is :
- A \(\frac{2 x+1}{2 x+3}\), provided \(x \neq-\frac{1}{2}\) and \(x \neq-\frac{3}{2}\)
- B \(\frac{2 x+1}{2 x+3}\)
- C \(\frac{3 x+3}{2 x+1}\)
- D \(\frac{2 x+3}{2 x+1}\)
Answer & Solution
Correct Answer
(A) \(\frac{2 x+1}{2 x+3}\), provided \(x \neq-\frac{1}{2}\) and \(x \neq-\frac{3}{2}\)
Step-by-step Solution
Detailed explanation
\(f[f(x)] = \frac{1}{2\left(\frac{1}{2x+1}\right)+1}\) \(f[f(x)] = \frac{1}{\frac{2}{2x+1}+1}\) \(f[f(x)] = \frac{1}{\frac{2+(2x+1)}{2x+1}}\) \(f[f(x)] = \frac{1}{\frac{2x+3}{2x+1}}\) \(f[f(x)] = \frac{2x+1}{2x+3}\), provided \(x \neq-\frac{1}{2}\) and \(x \neq-\frac{3}{2}\)
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