    Question

# lf f(x)=⎧⎪⎨⎪⎩−1,x<00,x=01,x>0 and g(x)=sinx+cosx, then points of discontinuity of (fog)(x) in (0,2π) are

A
π4,5π4
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B
π4,3π4
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C
π4,7π4
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D
3π4,7π4
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Solution

## The correct option is A 3π4,7π4From the definition of f(x), it is discontinuous at x=0. Therefore, fog(x) will be discontinuous whenever g(x)=0 in the interval 0<x<2π.The function g(x) can be expressed as below:g(x)=sinx+cosx=√2(1√2sinx+1√2cosx)=√2(cosπ4sinx+sinπ4cosx)=√2sin(x+π4)Now we need to determine the points where g(x)=0, since the function is discontinuous at those points:√2sin(x+π4)=0⇒x+π4=nπ,nϵZ⇒x=nπ−π4=…,−π4,3π4,7π4,11π4,…But xϵ(0,2π). So the points of discontinuity in this interval are x=3π4,7π4.  Suggest Corrections  0      Similar questions  Related Videos   MATHEMATICS
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