JEE MAIN - Mathematics (2025 - 24th January Evening Shift - No. 17)
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 :
6
9
8
7
Explanation
$$f(x)=[x]+|x-2| \quad-2< x<3$$
$$f(x)=\left\{\begin{array}{cc} -x, & -2< x<-1 \\ -x+1, & -1 \leq x<0 \\ -x+2, & 0 \leq x<1 \\ -x+3, & 1 \leq x<2 \\ x, & 2 \leq x<3 \end{array}\right.$$
So $f(x)$ is not continuous at 4 points and not differentiable at 4 point
So $\mathrm{m}+\mathrm{n}=4+4=8$
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