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Question

Let f(x)=⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪1sin3x3cos2x:x<π2p:x=π2q(1sin x)(x2x)2:x>π2  If f(x) is continuous at x=π2,(p,q)=
  1. (12,2)
  2. (1,4)
  3. (12,4)
  4. (12,7)


Solution

The correct option is C (12,4)
f[(π2)]=limh01sin3[(π2)h]3cos2[(π2)h]
=limh01cos3h3sin2h=12
f[(π2)+]=limh0q[1sin{(π2)+h}][π2{(π2)+h}]2
=limh0q(1cos h)4h2=q8
p=12=q8p=12,q=4

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