FIND THE RANGE

Asked by vk1994 | 28th Apr, 2010, 12:09: PM

Expert Answer:

Dear student,

the solution of your quesry is as follows -

Since (y-1) is there in the denominator, therefore x will tend to infinity when y=1. So, x will not be defined at y=1. => y ≠ 1.

Also, for x to be real, (9y2-11-6y) must be greater than or equal to 0.

therefore,

We note that (1+121/2)/3 > 1,

therefore the range of y is .

Answered by  | 28th Apr, 2010, 04:03: PM

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