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JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let \(f\) be a composite function of \(x\) defined by \(f\left( u \right) = \frac{1}{{{u^2} + u - 2}}\,,\,u\left( x \right) = \frac{1}{{x - 1}}\) . Then the number of points \(x\) where \(f\) is discontinuous is
- A \(4\)
- B \(3\)
- C \(2\)
- D \(1\)
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
\(\mu \left( x \right) = \frac{1}{{x - 1}}\), which is dicontious at \(x=1\) \(f\left( u \right) = \frac{1}{{{u^2} + u - 2}} = \frac{1}{{\left( {u + 2} \right)\left( {u - 1} \right)}}\), which is dicontious at \(u=-2,1\) when \(u=-2\), then…
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