Question

Assertion (A) The domain of the real function $f$ defined by $f(x)=\sqrt{x-1}$ is $R-\{1\}$
Reason (R) The range of the function defined by $f(x)=\sqrt{x-1}$ is $[0, \infty)$.

• A. Both A and R are correct; $R$ is the correct explanation of $A$
• B. Both $A$ and $R$ are correct; $R$ is not the correct explanation of $A$.
• C. $A$ is correct; $R$ is incorrect
• D. $R$ is correct; $A$ is incorrect

Answer 1

Answer: (D)

We have, $f(x)=\sqrt{x-1}$
$f(x)$ is defined, if $x-1 \geq 0$
i.e., $ x \geq 1$
$\therefore $ Domain of $f=[1, \infty)$
Hence, $A$ is incorrect.
Let $f(x) = y$
Then, $y = \sqrt{x - 1}$
$\Rightarrow y^2=x-1 $
$ \Rightarrow x =y^2+1$
Since, $y \geq 0$ and $x \in[1, \infty) $
$\Rightarrow$ Range of $f=[0, \infty)$.