The function f(x)=1x2−3|x|+2 is discontinuous at the points

f(x) discontinuous when x2−3|x|+2=0|x|2−3|x|+2=0(|x|−1)(|x|−2)=0|x|=1, |x|=2=0⇒x=±1, x=±2 ∴x=±1, ±2

The function f(x)=1x2−3|x|+2 is discontinuous at the points

f(x) discontinuous when x2−3|x|+2=0|x|2−3|x|+2=0(|x|−1)(|x|−2)=0|x|=1, |x|=2=0⇒x=±1, x=±2 ∴x=±1, ±2

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The function $f (x) = \frac{1}{x^{2} - 3 | x | + 2}$ is discontinuous at the points

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$x = \pm 1, \pm 2$

x = 1, 2

$R - {1, 2}$

answer is B.

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Detailed Solution

f(x) discontinuous when $x^{2} - 3 | x | + 2 = 0 | x |^{2} - 3 | x | + 2 = 0 (| x | - 1) (| x | - 2) = 0$

$| x | = 1, | x | = 2 = 0 \Rightarrow x = \pm 1, x = \pm 2 ∴ x = \pm 1, \pm 2$

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The function $f (x) = \frac{1}{x^{2} - 3 | x | + 2}$ is discontinuous at the points