Let's solve the problem such that the given expression assumes all real values.
We're looking at the expression:
y=‌To ensure that
y can take all real values, the denominator must not be zero:
2x2−3x+1≠0This can be factored as:
(2x−1)(x−1)≠0Thus,
x≠1 and
x≠‌.
For
y to take all real values for real
x, the quadratic expression
(3y+1)2−4y(y−2)≥0 must hold:
(3y+1)2−4[y(y−2)−a]≥0Simplifying the terms:
9y2+6y+1−(4y2−8y−4a)≥0This simplifies to:
5y2+14y+1+4a≥0As a quadratic in terms of
y, for it to be non-positive (invertible for any
x ), the discriminant must be negative:
The discriminant:
‌62−4×5×(1+4a)<0‌36−20(1+4a)<0‌36−20−80a<0‌16<80aThus:
a>‌Therefore, the condition for
a is:
−1<a<−‌The correct understanding implies the feasible range for
a.