Question

Prove that the range of \[1 + \sin ^{2} x\] is between 1 and 2, inclusive.

How do I show this?

Answer 1

sin x maximum vallue is 1

Answer 2

so max value of exp will be 2

Answer 3

now minimum value when x=0 sin x=0

Answer 4

so minimum value =1

Answer 5

so 1+x^2 has values [1,2]

Answer 6

gt tht?

Answer 7

becaus sin x max value is 1

Answer 8

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

Answer 9

yes maximum value is 2

Answer 10

so to maximise sin^x+1 ,,,,sin x maximum should be there

Answer 11

ie 1+1=2