KCET · Maths · Functions
\(f(x) = (x-1)^2\) for \(x \geq 1\), \(g(x)\) is a function whose graph is the reflection of the graph of \(f(x)\) in the line \(y = x\), then \(g(x)\) is
- A \(-\sqrt{x} + 1\)
- B \(\sqrt{x} - 1\)
- C \(\sqrt{x} + 1\)
- D \(\sqrt{x - 1}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{x} + 1\)
Step-by-step Solution
Detailed explanation
The graph of \(g(x)\) is the reflection of the graph of \(f(x)\) in the line \(y = x\), which means \(g(x)\) is the inverse function of \(f(x)\).
Let \(y = f(x) = (x-1)^2\) for \(x \geq 1\).
Taking the square root on both sides, we get:
\(\sqrt{y} = |x-1|\)
Since \(x \geq 1\), we have \(x-1 \geq 0\), so \(|x-1| = x-1\).
\(\sqrt{y} = x - 1\)
\(x = \sqrt{y} + 1\)
Interchanging \(x\) and \(y\) to find the inverse function, we get:
\(y = \sqrt{x} + 1\)
Therefore, \(g(x) = \sqrt{x} + 1\).
Answer: \(\sqrt{x} + 1\)
Let \(y = f(x) = (x-1)^2\) for \(x \geq 1\).
Taking the square root on both sides, we get:
\(\sqrt{y} = |x-1|\)
Since \(x \geq 1\), we have \(x-1 \geq 0\), so \(|x-1| = x-1\).
\(\sqrt{y} = x - 1\)
\(x = \sqrt{y} + 1\)
Interchanging \(x\) and \(y\) to find the inverse function, we get:
\(y = \sqrt{x} + 1\)
Therefore, \(g(x) = \sqrt{x} + 1\).
Answer: \(\sqrt{x} + 1\)
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