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if (x2 -5x+3)(y2 +y+1)<2y for all y belongs to the set of real numbers,then set of all values of x is

6 years ago


Answers : (1)


Since y2+y+1 is always positive, we can divide by this quantity on both sides to get the inequality


x2 - 5x + 3 < 2y/(y2+y+1) or equivalently x2 - 5x + 3 < 2/(y+1/y+1)


When y is non-negative, min val of RHS is zero.


If y<0 then y + 1/y <=-2 so, y+1/y+1 <=-1 and hence RHS >=-2.


Hence we need x2 - 5x + 3<-2 or x2 - 5x + 5<0 which is easy to solve

6 years ago

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