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Let $f$ be a composite function of $x$ defined by $f(u)=\frac{1}{u^2+u-2}, u(x)=\frac{1}{x-1}$.
Then the number of points $x$ where $f$ is discontinuous is :
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Then the number of points $x$ where $f$ is discontinuous is :
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Verified Answer
The correct answer is:
3
3
$\mu(x)=\frac{1}{x-1}$, which is discontinous at $x=1$
$$
f(u)=\frac{1}{u^2+u-2}=\frac{1}{(u+2)(u-1)},
$$
which is discontinous at $u=-2,1$
when $u=-2$, then $\frac{1}{x-1}=-2 \Rightarrow x=\frac{1}{2}$
when $u=1$, then $\frac{1}{x-1}=1 \Rightarrow x=2$
Hence given composite function is discontinous at three points, $x=1, \frac{1}{2}$ and 2 .
$$
f(u)=\frac{1}{u^2+u-2}=\frac{1}{(u+2)(u-1)},
$$
which is discontinous at $u=-2,1$
when $u=-2$, then $\frac{1}{x-1}=-2 \Rightarrow x=\frac{1}{2}$
when $u=1$, then $\frac{1}{x-1}=1 \Rightarrow x=2$
Hence given composite function is discontinous at three points, $x=1, \frac{1}{2}$ and 2 .
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