Q131 / 240 · KCET 2026 · Mathematics
$f(x) = (x-1)^2$ 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$\sqrt{x} - 1$
- Correct answerC$\sqrt{x} + 1$
- D$\sqrt{x - 1}$
Solution
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: $\sqrt{x} + 1$
Hence, the correct answer is option: 'C'.