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Limits Continuity and Differentiability question

2025 · 24 Jan · Shift 2 · Q42
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Limits Continuity and Differentiability question

2025 · 24 Jan · Shift 2 · Q42

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let [x][x][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<3f(x)=[x]+|x-2|,-2\lt x\lt 3f(x)=[x]+∣x−2∣,−2<x<3, is not continuous and not differentiable. Then m+n\mathrm{m}+\mathrm{n}m+n is equal to :
  1. A
    6
  2. B
    9
  3. C
    8
  4. D
    7
View written solutionFree

Correct answer: C

  1. Given function

We need to study

f(x)=[x]+∣x−2∣,−2<x<3f(x)=[x]+|x-2|, \qquad -2<x<3f(x)=[x]+∣x−2∣,−2<x<3

where [x][x][x] is the greatest integer function.

We must find:

  • mmm = number of points where f(x)f(x)f(x) is not continuous
  • nnn = number of points where f(x)f(x)f(x) is not differentiable

Then compute m+nm+nm+n.


  1. Where can problems occur?

The function is a sum of two parts:

  • [x][x][x], which is discontinuous at every integer
  • ∣x−2∣|x-2|∣x−2∣, which is continuous everywhere but not differentiable at x=2x=2x=2

So possible special points in the interval (−2,3)(-2,3)(−2,3) are the integers inside it:

−1,0,1,2-1,0,1,2−1,0,1,2

and also x=2x=2x=2 because of the modulus corner.


  1. Continuity analysis

(i) Continuity of [x][x][x]

The greatest integer function has jump discontinuity at every integer. Thus in (−2,3)(-2,3)(−2,3), discontinuities occur at

x=−1,0,1,2x=-1,0,1,2x=−1,0,1,2

(ii) Continuity of ∣x−2∣|x-2|∣x−2∣

This is continuous for all real xxx, including x=2x=2x=2.

Therefore, the sum f(x)=[x]+∣x−2∣f(x)=[x]+|x-2|f(x)=[x]+∣x−2∣ is discontinuous exactly at

−1,0,1,2-1,0,1,2−1,0,1,2

So,

m=4m=4m=4
  1. Differentiability analysis A function can be differentiable only where it is continuous. Hence all discontinuity points are automatically non-differentiable.

So f(x)f(x)f(x) is not differentiable at least at

−1,0,1,2-1,0,1,2−1,0,1,2

Now check whether there are any additional non-differentiable points.

(i) For [x][x][x]

Between consecutive integers, [x][x][x] is constant, so differentiable there with derivative 000.

(ii) For ∣x−2∣|x-2|∣x−2∣

It is not differentiable at

x=2x=2x=2

but x=2x=2x=2 is already included above.

Thus there is no extra point besides the discontinuity points. Hence,

n=4n=4n=4
  1. Compute m+nm+nm+n
m+n=4+4=8m+n=4+4=8m+n=4+4=8
  1. Option check

The correct option is

8\boxed{8}8​

which is Option C.


  1. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.

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