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Question

# A function is defined as follows f(x)=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩1,when−∞<x<01+sinx,when0≤x<π22+(x−π2)2whenπ2≤x<∞ continuity of f(x) is

A
f(x) is continuous at x=π2
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B
f(x) is continuous at x=0
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C
f(x) is discontinuous at x=0
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D
f(x) is continuous over the whole real number
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Solution

## The correct option is D f(x) is continuous over the whole real numberContinuity at x=0LHL =limx→0−f(x)=limx→0(1)=1RHL =limx→0+f(x)=limx→0(1+sinx)=1And f(0)=1+sin0=1∴ LHL = RHL =f(0)So, f(x) is continuous at x=0.Continuity at x=π2LHL =limx→π2−f(x)=limx→π2(1+sinx)=1+1=2RHL =limx→π2+f(x)=limx→π22+(x−π2)2=2+(π2−π2)2=2And f(π2)=2+(π2−π2)2=2∴LHL=RHL=f(π2)So, f(x) is continuous at x=π2.Hence, f(x) is continuous over the whole real number.

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