Assertion :Let f(x)=⎧⎨⎩1+xx<01+[x]+sinx0≤x≤π/23x≥π/2 f is continuous on R −{1} Reason: The greatest integer function is discontinuous at every integer.
A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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C
Assertion is correct but Reason is incorrect
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D
Both Assertion and Reason are incorrect
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Solution
The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion f(0)=1=limx→0−f(x)=limx→0−(1+x)=1 =limx→0+f(x)=limx→0+(1+[x]+sinx)=1 So f is continuous at x=0. Since [x] is right continuous but not left continuous at x=1 so also is f. f(π/2)=3=limx→π/2+f(x) limx→π/2−f(x)=limx→π/2−(1+[x]+sinx) =1+limx→π/2−[x]+limx→π/2−sinx =1+1+1=3 So f is continuous at x=π/2. Thus f is continuous on R∼{1}. Also [x] is not continuous at every x∈I