FIND THE RANGE
Asked by vk1994 | 28th Apr, 2010, 12:09: PM
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.
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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