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Question

Find the value of p & q for which f(x)⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪1sin3x3cos2x,ifx<π2p,ifx=π2iscontinousatx=π2q(1sinx)(π2x)2,ifx>π2

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Solution

f(x)=⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪1sin3x3cos2xifx<π2pifx=π2I(1sinx)(π2x)2ifx>π2
f(x2)=f(π2)=f(π+2)
f(π2)=3sin2xcosx3×2cosxsinx
sinx2=12 [L hospital rule]
f(π2)=p=12P=12
(π+2)=q(1sinx)(π2x)2
Applying 1 Hospital rule.
f(π+2)=q(cosx)2(π2x)x2
Apply are more
f(π+2)=qsinx8x=q8
q8=12
q=4


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