IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

rooney

Let log_[x-1]sin(x) = y ( base is always positive. So, [x-1] >0 . So, x>2 ]

Sin x = [x - 1]^y
-1 <= Sin x <= 1

-1 <= [x - 1]^y <= 1

Now, [x - 1]^y is always positive and greater than 0.
0 < [x - 1]^y<=1
[ x-1] always will be an integer. Thus for [x - 1]^y to be less than equal to 1, y will always have to be less than equal to 0. Since for y = 0, the [x - 1]^y = 1. So, for [x - 1]^y to become less in value, y will have to become MORE NEGATIVE. So, more negative from 0 is - infinity.

Thus , y

( -

, 0 ]

Take care

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