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Question

# Consider f(x)=sgn(sinx)+{x};2≤x≤4 and g(x)=−2+∣x−3∣; where {.} denotes fractional part function and sgn denotes signum function. Then limx→3gof(x) is equal to

A
-1
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B
0
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C
1
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D
does not exist
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Solution

## The correct option is D does not existFirst write f(x) and g(x) in different intervals as per the definition.f(x)=⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩x−1,2≤x<3x−2,3≤x<πx−3,x=πx−4,π<x≤4g(x)={1−x,x<3x−5,x≥3At x=3, RHL=gof(3+)=limh→0g(f(3+h))=limh→0g(3+h−2)=limh→0g(1+h)=limh→01−(1+h)=limh→0−h=0LHL=gof(3−)=limh→0g(f(3−h))=limh→0g(3−h−1)=limh→0g(2−h)=limh→01−(2−h)=limh→0−1+h=−1Since LHL≠RHL, limit does not exist.

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