IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

kislay

find the range of
log_[x-1]sin(x) ,where[.] denotes greatest integer function.
ans is (-infinity,0]
i want the working

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

spideyunlimited

base is greater than 0 and not equal to 1.
So
[x-1] > 0 and [x-1] not equal to 1
so x >= 2 and x not equal to 2
so the intersection is:
x > 2

log_[x-1]sin(x) = y

[x-1] ^y = sin x

now as x>2

so [x-1] will vary from (1, infinity) .. 1 is not included.

so for the least value of [x-1] .. which is. tending to 1 from right hand side.

y will be 0, as sinx can have values between -1 and 1.

similarly for the highest value of [x-1] which is infinity, the power y will tend to -infinity to make the net value 1. ( as[x-1]'s value increases, the power increases in the negative)

so the range (y) = [0, -infinity)

iitkgp_bipin

Well done Rooney and Gaurav.

kishan12

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

Now take log
-infas [x-1] is +ve,divide it throughout and u have your ans.
i have copy paste half of rooney (17)'s solution.......
But this is how i got and looks a bit simpler,isn't it??

rooney

@kishan12

Yes man. The language part of the question is thought in mind only . Converting into written text is a bit combersome. lol

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