The number of points of discontinuity of f(x)={[cosπx],0≤x≤1|2x−3|[x−2],1<x≤2 is/are where [.] denotes the greatest integral function
A
Two
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B
Three
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C
Four
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D
Zero
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Solution
The correct option is C Three Consider x∈[0,1] From the graph given, it is clear that [cosπx] is discontinuous at x=0,1/2 (1) Now, consider x∈(1,2]. f(x)=[x−2]|2x−3| For x∈(1,2),[x−2]=−1, and for x=2,[x−2]=0. Also, |2x−3|=0 or x=3/2. Therefore, x=3/2 and 2 may be the points at which f(x) is discontinuous. f(x)=⎧⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪⎨⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪⎩1,x=0x,0<x≤12−1,12<x≤1−(3−2x),1<x≤3/2−(2x−3),3/2<x≤20,x=2 Thus, f(x) is continuous when x∈[0,2]−{0,1/2,2}.