Question

The interval on which $f(x) = \sqrt {1 - {x^2}} $ is continuous is:
$A.(0,\infty ) $
$B.(1,\infty ) $
$C.[ - 1,1] $
$D.( - \infty , - 1) $

Answer 1

Hint:
The square root function is an increasing function. It is a one function, and it has an inverse function present inside the square root is always positive. So, looking into its domain can help us get the result. And refer to the options given for the question. As to know whether the values at the end are included or excluded.

Complete answer:
To be continuous on an interval we should check its domain as, the function inside the square root function $1 - {x^2}$ is always greater than equal to zero. A negative quantity cannot go inside the square root function.
Now,
 $1 - {x^2} \geqslant 0 $
 $\Rightarrow - {x^2} \geqslant - 1 $
 $\Rightarrow {x^2} \leqslant 1 $
multiply with minus sign on both sides.
Also, have a look on the brackets that will be used as we know that the sign we have is greater than equal to so we will be using square bracket which is close interval , and $x$ is in the range of $ - 1$ to $1$
So we would have a closed interval between $ - 1$ to $1$ .
Hence, the function is continuous for all $x$ between $[ - 1,1]$ .

Option C is the correct answer.

Note:
Be careful with positive or negative signs while exchanging the quantity from the Left-Hand side of the Equation to the Right Hand Side of the Equation. As we know that when we multiply negative with negative it becomes positive and the sign of inequality changes. A curve bracket means the values at the end are excluded and a square bracket means the values at end are included.