Question

The interval on which $f\left(x\right)=\sqrt{1-x^2}$ is continuous is:

• A. $(0,\infty )$
• B. $(1,\infty )$
• C. $[-1,1]$
• D. $(-\infty ,-1)$

Answer 1

The correct option is C $[-1,1]$

The function will be continuous on an interval where it is completely defined.
Since, we know, a negative quantity cannot go inside the square root sign,
Hence, $1-x^2>=0$
Or,
$x^2<=1$
Hence, the function is continuous for all $x$ lying between $[-1,1]$