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The number of points at which the function $f(x)=\frac{\sqrt{11+|x|-6 \sqrt{2+|x|}}}{6-2 \sqrt{2+|x|}}$ is discontinuous in $(-\infty, \infty)$ is
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The correct answer is:
2
$f(x)=\sqrt{\frac{11+|x|-6 \sqrt{2+|x|}}{6-2 \sqrt{2+|x|}}}$
This function will be discontinuous where function is undefined.
So, it is undefined, if $\frac{11+|x|-6 \sqrt{2+|x|}}{6-2 \sqrt{2+|x|}} < 0$ and $6-2 \sqrt{2+|x|}=0$ and $2+|x| < 0$
Now, $2+|x| < 0 \Rightarrow|x| < -2$
Not possible.
$$
\begin{aligned}
& \text { Again } 6-2 \sqrt{2+|x|}=0 \\
& \Rightarrow \quad-2 \sqrt{2+|x|}=-6 \\
& \Rightarrow \quad \sqrt{2+|x|}=3 \Rightarrow 2+|x|=9 \\
& \Rightarrow \quad|x|=7 \quad \Rightarrow \quad x= \pm 7 \\
&
\end{aligned}
$$
Hence, numbers of points is 2 .
This function will be discontinuous where function is undefined.
So, it is undefined, if $\frac{11+|x|-6 \sqrt{2+|x|}}{6-2 \sqrt{2+|x|}} < 0$ and $6-2 \sqrt{2+|x|}=0$ and $2+|x| < 0$
Now, $2+|x| < 0 \Rightarrow|x| < -2$
Not possible.
$$
\begin{aligned}
& \text { Again } 6-2 \sqrt{2+|x|}=0 \\
& \Rightarrow \quad-2 \sqrt{2+|x|}=-6 \\
& \Rightarrow \quad \sqrt{2+|x|}=3 \Rightarrow 2+|x|=9 \\
& \Rightarrow \quad|x|=7 \quad \Rightarrow \quad x= \pm 7 \\
&
\end{aligned}
$$
Hence, numbers of points is 2 .
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