Question

Let $f\left(x\right)=\frac{\sqrt{x^2-\left|x\right|+1}}{\sqrt{9-x^2}}$. Its domain does not contain $D$, where $D=$?

• A. $\left(-3,-1\right)$
• B. $\left(1,3\right)$
• C. $\left(-3,3\right)$
• D. $\left[0\right]\cup [1,3)$

Answer 1

The correct option is C

$\left(-3,3\right)$

Explanation for the correct answer

Gievn that: $f\left(x\right)=\frac{\sqrt{x^2-\left|x\right|+1}}{\sqrt{9-x^2}}$

$f\left(x\right)$ is defined in the interval:

$\Rightarrow 9>x^2$,

$\Rightarrow x^2\ge \left|x\right|-1$,

Thus we have the range as,$\Rightarrow -3<x<3\mathrm{and}x\in R$

So, we can say that the domain of $f\left(x\right)$ is $\left(-3,3\right)$

We see that $f\left(x\right)$ does not include $-3\mathrm{and}3$.

Hence, $D$ is a set, corresponding to $\left(-3,3\right)$, for which $f\left(x\right)$ is zero.

Therefore, the correct answer is option (C),