1
Question
If f(x) is continuous on [−π,π], where
f(x)=⎧⎪
⎪⎨⎪
⎪⎩−2sinx,for−π≤−π2αsinx+β,for−π2<x<π2cosxforπ2≤x≤π
then α and β are
If f(x) is continuous on [−π,π], where
f(x)=⎧⎪
⎪⎨⎪
⎪⎩−2sinx,for−π≤−π2αsinx+β,for−π2<x<π2cosxforπ2≤x≤π
then α and β are
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Solution
The correct option is D 1,−1
Given, f(x)=⎧⎪
⎪⎨⎪
⎪⎩−2sinx,for−π≤x≤−π2αsinx+β,for−pi2<x<π2cosxforπ2≤x≤πAt x=π2,LHL =f(π2−0)=limx→π−2(αsinx+β) =limh→0{αsin(π2−h)+β}=α+βand f(π2)=cosπ2=0∵ f(x) is continuous at x=π2.∴α+β=0 ...(i)At x=−π2,RHL =f(−π2+0)=limx→−π2(αsinx+β)=limh→0[αsin(−π2+h)+β]=−α+βand f(−π2)=−2sinπ2=−2∵ f(x) is continuous at x=−π2∴−α+β=−2 ...(ii)On solving Eqs. (i) and (ii), we getα=1 and β=−1
Given,
f(x)=⎧⎪
⎪⎨⎪
⎪⎩−2sinx,for−π≤x≤−π2αsinx+β,for−pi2<x<π2cosxforπ2≤x≤π
At x=π2,
LHL =f(π2−0)=limx→π−2(αsinx+β)
=limh→0{αsin(π2−h)+β}
=α+β
and f(π2)=cosπ2=0
∵ f(x) is continuous at x=π2.
∴α+β=0 ...(i)
At x=−π2,
RHL =f(−π2+0)=limx→−π2(αsinx+β)
=limh→0[αsin(−π2+h)+β]
=−α+β
and f(−π2)=−2sinπ2=−2
∵ f(x) is continuous at x=−π2
∴−α+β=−2 ...(ii)
On solving Eqs. (i) and (ii), we get
α=1 and β=−1
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