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Question

If the function f(x) defined asf(x)=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩(sinx+cosx)cscx,−π2<x<0a,x=0e1/x+e2/x+e3/xae−2+1/x+be−1+3/x,0<x<π2 is continuous at x=0, then

A
a=e,b=1
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B
a=1,b=e
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C
a=1e,b=1
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D
None of these
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Solution

The correct option is A a=e,b=1We have, limx→0−f(x)=limh→0[sin(−h)+cos(−h)]csch=limh→0[(cosh−sinh)]−csch=limh→0(1+(cosh−sinh−1))1cosh−sinh−1.cosh.sinh−1−sinh=[limy→0(1+y)1y]limh→0cosh−sinh−1−sinhNow, limh→0cosh−sinh−1−sinh(00)=limh→0−sinh−cosh−cosh=0−1−1=1Thus, limx→0−f(x)=eNow, we have,limx→0+f(x)=limh→0e1h+e2h+e3hae−2+1−h+be−1+3h=limh→0e−2h+e−1h+1(ae−2)e−2h+(be−1)=0+0+1(ae−2)0+(be−1)=ebIf f is continuous at x=0, then e=a=eb given a=e and b=1

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