JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let \(\mathrm{f}: R \rightarrow R\) be function defined as \(f ( x )=\left\{\begin{array}{cc}3\left(1-\frac{| x |}{2}\right) & \text { if }| x | \leq 2 \text { } \\ 0 & \text { if }| x |>2 \text { }\end{array}\right.\) Let \(g: R \rightarrow R\) be given by \(g(x)=f(x+2)-f(x-2)\). If \(n\) and \(m\) denote the number of points in \(R\) where \(\mathrm{g}\) is not continuous and not differentiable, respectively, then \(\mathrm{n}+\mathrm{m}\) is equal to \(....\)
- A \(4\)
- B \(3\)
- C \(2\)
- D \(1\)
Answer & Solution
Correct Answer
(A) \(4\)
Step-by-step Solution
Detailed explanation
\(f(x-2)\rightarrow =\frac{3 x}{2} \quad\quad -4 \leq x \leq-2\) \(\quad \quad \quad \quad \quad \quad -\frac{3 x}{2} \quad -2\,\) \(\quad \quad \quad \quad \quad \quad 0 \quad \quad \quad x \in(-\infty,-4) \cup(0,+\infty)\)…
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