Mark (√) against the correct answer in the following: Let $f(x)=\sqrt{\log }\left(2 x-x^{2}\right)$. Then, $\operatorname{dom}(f)=?$ A. $(0,2)$ B. $[1,2]$ C. $(-\infty, 1]$ D. None of these

Home » Mark (√) against the correct answer in the following: Let . Then, A. B. C. D. None of these

Mark (√) against the correct answer in the following: Let $f(x)=\sqrt{\log }\left(2 x-x^{2}\right)$ . Then, $\operatorname{dom}(f)=?$
A. $(0,2)$
B. $[1,2]$
C. $(-\infty, 1]$
D. None of these

CBSE | Class 12 | Functions | Maths | RS Aggarwal

Mark (√) against the correct answer in the following: Let $f(x)=\sqrt{\log }\left(2 x-x^{2}\right)$ . Then, $\operatorname{dom}(f)=?$
A. $(0,2)$
B. $[1,2]$
C. $(-\infty, 1]$
D. None of these

Solution:

Option(C) is correct.
$f(x)=\sqrt{\log \left(2 x-x^{2}\right)}$
For $f(x)$ to be defined $2 x-x^2$ should be positive.
Solving inequality,
(Log taken to the opposite side of the equation becomes e)
$\begin{array}{l} \Rightarrow \mathrm{x}^2-2 \mathrm{x}+1 \leq 0 \\ \Rightarrow(\mathrm{x}-1)^2 \leq 0 \\ \Rightarrow \mathrm{x}-1 \leq 0 \\ \Rightarrow \mathrm{x} \leq 1 \end{array}$
Therefore, domain of $\mathrm{f}(\mathrm{x})$ is $(-\infty, 1)$