Question:
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 :
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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Find where \(u(x)\) is undefined and where \(u^2+u-2=0\).
Updated On: Oct 31, 2026
- 4
- 3
- 2
- 1
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The Correct Option is B
Solution and Explanation
\(u^2+u-2=(u+2)(u-1)\), so \(f(u)\) is undefined at \(u=1\) and \(u=-2\).
The composite is discontinuous where \(u(x)\) itself is undefined: \(x=1\).
\(u(x)=1 \Rightarrow \dfrac{1}{x-1}=1 \Rightarrow x=2\).
\(u(x)=-2 \Rightarrow \dfrac{1}{x-1}=-2 \Rightarrow x=\dfrac12\).
Points of discontinuity: \(x=1,\ 2,\ \tfrac12\), i.e. 3 points.
Answer: (2) 3
The composite is discontinuous where \(u(x)\) itself is undefined: \(x=1\).
\(u(x)=1 \Rightarrow \dfrac{1}{x-1}=1 \Rightarrow x=2\).
\(u(x)=-2 \Rightarrow \dfrac{1}{x-1}=-2 \Rightarrow x=\dfrac12\).
Points of discontinuity: \(x=1,\ 2,\ \tfrac12\), i.e. 3 points.
Answer: (2) 3
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