Let $[x]$ denote the greatest integer function, and let m and n respectively be the numbers of the points, where the function $f(x)=[x]+|x-2|,-2< x<3$, is not continuous and not differentiable. Then $\mathrm{m}+\mathrm{n}$ is equal to :
Solution
<p>$f(x)=[x]+|x-2| \quad-2< x<3$</p>
<p>$$f(x)=\left\{\begin{array}{cc}
-x, & -2< x<-1 \\
-x+1, & -1 \leq x<0 \\
-x+2, & 0 \leq x<1 \\
-x+3, & 1 \leq x<2 \\
x, & 2 \leq x<3
\end{array}\right.$$</p>
<p>So $f(x)$ is not continuous at 4 points and not differentiable at 4 point</p>
<p>So $\mathrm{m}+\mathrm{n}=4+4=8$</p>
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
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