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Question

# Let f(x)=⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩(1+|cosx|)ab|cosx|,nπ<x<(2n+1)π2ea.eb,x=(2n+1)π2ecot2xcot8x,(2n+1)π2<x<(n+1)πIf f(x) is continuous in ((nπ),(n+1)π,then)

A
a = 1, b = 2
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B
a = 2, b = 2
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C
a = 2, b = 3
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D
a = 3, b = 4
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Solution

## The correct option is B a = 2, b = 2LHL=limx→(nπ+π2)f(x)=limh→0f(nπ+π2−h)=limh→0(1+∣∣cos(nπ+π2)∣∣)ab∣∣cos[nπ+π2−h]∣∣=limh→0(1+|sin(nπ−h)|)ab|sin(nπ−h)|=limh→0(1+|(−1)nsinh|)ab|1(−1)nsinh|=limh→0(1+sinh)absinh=eab V.F.=f(nπ+π2)=ea.eb [∵limh→0(1+h)hn]=eλRHL=limx→(nπ+π2)+f(x)=limh→0f(nπ+π2+h)=limh→0ecot(2nπ+π+2h)cot(8nπ+4π+8h) =elimh→0(cot2hcot8h)elimh→0tan8htan2h=e4limh→0tan8htan2h.limh→02htan2h=e4.1.1=e4∵f(x) continuous in (nπ,(n+1)π)∴LHL=RHL=V.Feab=e4=ea+b⇒ab=4=a+b⇒a=b=2

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