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Question: If \[[ \text{ }]\] denotes the greatest integer function then \[f(x)=[x]+[x+\dfrac{1}{2}]\] A. is...

If [ ][ \text{ }] denotes the greatest integer function then f(x)=[x]+[x+12]f(x)=[x]+[x+\dfrac{1}{2}]
A. is continuous at x=12x=\dfrac{1}{2}
B. is discontinuous at x=12x=\dfrac{1}{2}
C. lim⁡x→12+f(x)=2\displaystyle \lim_{x \to \dfrac{1}{2}+}f(x)=2
D. lim⁡x→12−f(x)=1\displaystyle \lim_{x \to \dfrac{1}{2}-}f(x)=1

Explanation

Solution

Possibility of discontinuity of greatest integer function [x+a][x+a] is only when x+ax+a becomes integer , we can’t directly say that it is discontinuous at that point for that we have to check using continuity equation ,so in given question f(x)=[x]+[x+12]f(x)=[x]+[x+\dfrac{1}{2}], we first check continuity on x=12x=\dfrac{1}{2}
By using equation
For continuity f(x)=f(x−)=f(x+)f(x)=f({{x}^{-}})=f({{x}^{+}})
So will put x=12x=\dfrac{1}{2} and check accordingly which option satisfies

Complete step-by-step answer:
We are given a greatest integer function then f(x)=[x]+[x+12]f(x)=[x]+[x+\dfrac{1}{2}] , we have check for its continuity and for checking a functions continuity we have one formula that is f(x)=f(x−)=f(x+)f(x)=f({{x}^{-}})=f({{x}^{+}}) and we know that possibility of discontinuity of greatest integer function
[x+a][x+a] is only when x+ax+a becomes an integer, given in the question all option lies around x=12x=\dfrac{1}{2}so we will check for x=12x=\dfrac{1}{2} there can be discontinuity
Using greatest integer function property [x]=0[x]=0, when x is between [0,1)[0,1) and [x]=1[x]=1 when x is between [1,2)[1,2)
f(12)=[12]+[12+12]=0+1=1f(\dfrac{1}{2})=[\dfrac{1}{2}]+[\dfrac{1}{2}+\dfrac{1}{2}]=0+1=1
Similarly, on putting x=12+x={{\dfrac{1}{2}}^{+}}
f(12+)=[12+]+[12++12]=0+1=1f({{\dfrac{1}{2}}^{+}})=[{{\dfrac{1}{2}}^{+}}]+[{{\dfrac{1}{2}}^{+}}+\dfrac{1}{2}]=0+1=1
Similarly, on putting x=12−x={{\dfrac{1}{2}}^{-}}
f(12−)=[12−]+[12−+12]=0+0=0f({{\dfrac{1}{2}}^{-}})=[{{\dfrac{1}{2}}^{-}}]+[{{\dfrac{1}{2}}^{-}}+\dfrac{1}{2}]=0+0=0, this time in greatest integer function value becomes less then 1 result into 0
We can see very clearly that f(12)=f(12−)≠f(12+)f(\dfrac{1}{2})=f({{\dfrac{1}{2}}^{-}})\ne f({{\dfrac{1}{2}}^{+}}) , we confirms that f(x)=[x]+[x+12]f(x)=[x]+[x+\dfrac{1}{2}] is discontinuous at x=12x=\dfrac{1}{2}
Now checking other option like (c) and (d)
In which we are asked to find the limit at x→12x\to \dfrac{1}{2}
lim⁡x→12+f(x)=2\displaystyle \lim_{x \to \dfrac{1}{2}+}f(x)=2 , for this put x=12+x={{\dfrac{1}{2}}^{+}} in given equation that we have already putted an got
f(12+)=[12+]+[12++12]=0+1=1f({{\dfrac{1}{2}}^{+}})=[{{\dfrac{1}{2}}^{+}}]+[{{\dfrac{1}{2}}^{+}}+\dfrac{1}{2}]=0+1=1
Similarly
lim⁡x→12−f(x)=1\displaystyle \lim_{x \to \dfrac{1}{2}-}f(x)=1, for this put x=12−x={{\dfrac{1}{2}}^{-}} in given equation that we have already putted an got
f(12−)=[12−]+[12−+12]=0+0=0f({{\dfrac{1}{2}}^{-}})=[{{\dfrac{1}{2}}^{-}}]+[{{\dfrac{1}{2}}^{-}}+\dfrac{1}{2}]=0+0=0
So, both options are wrong and

So, the correct answer is “Option (B)".

Note: Some of the student might think that for checking continuity of this f(x)=[x]+[x+12]f(x)=[x]+[x+\dfrac{1}{2}], they just check by equating greatest integer to 0 and getting value like here we get x=0x=0 and x=−12x=-\dfrac{1}{2} but we have to check it by equating greatest integer with all integer possible.