Limit Continuity and Differentiability 5 Question 5

5. Let f:[1,3]R be defined as

f(x)=|x|+[x],1x<1x+|x|,1x<2x+[x],2x3,

(2019 Main, 8 April II) where, [t] denotes the greatest integer less than or equal to t. Then, f is discontinuous at

(a) four or more points

(b) only two points

(c) only three points

(d) only one point

Show Answer

Answer:

Correct Answer: 5. (b)

Solution:

  1. Since, f(x) is continuous at x=0.

limx0f(x)=f(0)f(0+)=f(0)=f(0)=0

To show, continuous at x=k

 RHL =limh0f(k+h)=limh0[f(k)+f(h)]=f(k)+f(0+)=f(k)+f(0) LHL =limh0f(kh)=limh0[f(k)+f(h)]=f(k)+f(0)=f(k)+f(0)limxkf(x)=f(k)

f(x) is continuous for all xR.



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