MHT CET · Maths · Continuity and Differentiability
Let \(\mathrm{f}(x)= \begin{cases}\frac{x^4-5 x^2+4}{|(x-1)(x-2)|} & , x \neq 1,2 \\ 6 & , x=1 \\ 12 & , x=2\end{cases}\) then \(\mathrm{f}(x)\) is continuous on the set
- A \(\mathbb{R}-\{1\}\)
- B \(\mathbb{R}-\{2\}\)
- C \(\mathbb{R}\)
- D \(\mathbb{R}-\{1,2\}\)
Answer & Solution
Correct Answer
(D) \(\mathbb{R}-\{1,2\}\)
Step-by-step Solution
Detailed explanation
\(x^4-5x^2+4 = (x^2-1)(x^2-4) = (x-1)(x+1)(x-2)(x+2)\) For \(x \neq 1,2\), \(f(x) = \frac{(x-1)(x+1)(x-2)(x+2)}{|(x-1)(x-2)|}\)
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