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 m+n is equal to :

Solution

Maths sir · 4 steps

Every jump of [x] is a break, and a break is never smooth: count the integers in (-2, 3) twice.

1. Where trouble starts. |x-2| is continuous everywhere and has a corner only at x = 2. [x] jumps by 1 at every integer, and inside (-2, 3) those integers are -1, 0, 1, 2.
2. Check each jump. Near an integer k, the term |x-2| is continuous, so it adds the same amount to both one-sided limits. It cannot cancel the jump of [x].

   Point	Left limit	Right limit	Continuous?	Differentiable?
   x = -1	-2 + 3 = 1	-1 + 3 = 2	No	No
   x = 0	-1 + 2 = 1	0 + 2 = 2	No	No
   x = 1	0 + 1 = 1	1 + 1 = 2	No	No
   x = 2	1 + 0 = 1	2 + 0 = 2	No	No (one point, counted once)

3. Count both sets. All four integers are points of discontinuity, so m = 4. A function that is not continuous at a point cannot be differentiable there, so these same four points are not differentiable. The corner of |x-2| at x = 2 is already on that list, so n = 4.
4. So. m + n = 4 + 4 = 8, which is option (c).

The idea to keep. Discontinuous always means non-differentiable. So collect the points of discontinuity first, then add only the new corners, such as a corner of |x-a| sitting where f is still continuous.

Answer from the NTA final answer key. Solution by PrepWiser.