JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let \(f(x)=\begin{cases} e^{x-1}, & x<0 \\ x^2-5x+6, & x \geq 0 \end{cases}\) and \(g(x)=f(|x|)+|f(x)|\). If the number of points where \(g\) is not continuous and is not differentiable are \(\alpha\) and \(\beta\) respectively, then \(\alpha+\beta\) is equal to ______
- A 2
- B 4
- C 6
- D 8
Answer & Solution
Correct Answer
(B) 4
Step-by-step Solution
Detailed explanation
We are given the function: \(f(x) = \begin{cases} e^{x-1}, & x 0\). Thus, \(f(|x|) = (-x)^2 - 5(-x) + 6 = x^2 + 5x + 6\). Also, for \(x 0\), so \(|f(x)| = e^{x-1}\). Therefore, for \(x 0\), we can rewrite \(g(x)\) by analyzing the sign of \(x^2 - 5x + 6 = (x-2)(x-3)\):…
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