anonymous
  • anonymous
Prove that the range of \[1 + \sin ^{2} x\] is between 1 and 2, inclusive. How do I show this?
Mathematics
jamiebookeater
  • jamiebookeater
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AravindG
  • AravindG
sin x maximum vallue is 1
AravindG
  • AravindG
so max value of exp will be 2
AravindG
  • AravindG
now minimum value when x=0 sin x=0

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AravindG
  • AravindG
so minimum value =1
AravindG
  • AravindG
so 1+x^2 has values [1,2]
AravindG
  • AravindG
gt tht?
AravindG
  • AravindG
becaus sin x max value is 1
anonymous
  • anonymous
sinx , minimum value =-1, sin^2x , minimum value =0 ( sin^2x) cant be negative and maximum value of sin^2x=1....put sin^2x=1 for maximum value and put sin^2x=0 for minimum value
anonymous
  • anonymous
yes maximum value is 2
AravindG
  • AravindG
so to maximise sin^x+1 ,,,,sin x maximum should be there
AravindG
  • AravindG
ie 1+1=2

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