The number of points at which the function f( | vedclass

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(C) Let $u = \sqrt{2+|x|}$. Since $|x| \ge 0$,we have $u \ge \sqrt{2}$.Then $|x| = u^2 - 2$.The numerator becomes $\sqrt{11 + (u^2 - 2) - 6u} = \sqrt{

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$2$

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(C) Let $u = \sqrt{2+|x|}$. Since $|x| \ge 0$,we have $u \ge \sqrt{2}$.Then $|x| = u^2 - 2$.The numerator becomes $\sqrt{11 + (u^2 - 2) - 6u} = \sqrt{u^2 - 6u + 9} = \sqrt{(u-3)^2} = |u-3|$.The denominator is $6 - 2u = 2(3-u)$.Thus,$f(x) = \frac{|u-3|}{2(3-u)}$.For $u For $u > 3$,$f(x) = \frac{u-3}{2(3-u)} = -\frac{1}{2}$.The function is discontinuous where the denominator is zero,i.e.,$6 - 2u = 0 \Rightarrow u = 3$.Substituting back,$\sqrt{2+|x|} = 3 \Rightarrow 2+|x| = 9 \Rightarrow |x| = 7$.This gives $x = 7$ and $x = -7$.At these two points,the left-hand limit is $\frac{1}{2}$ and the right-hand limit is $-\frac{1}{2}$ (or vice versa),so the function is discontinuous.Thus,there are $2$ points of discontinuity.

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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