AP EAPCET 18 May 2024 Shift 1 Solved Paper

Let's solve the problem such that the given expression assumes all real values.
We're looking at the expression:
y=‌

x+a

2x2−3x+1

To ensure that y can take all real values, the denominator must not be zero:
2x2−3x+1≠0
This can be factored as:
(2x−1)(x−1)≠0
Thus, x≠1 and x≠‌

1

2

.
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]≥0
Simplifying the terms:
9y2+6y+1−(4y2−8y−4a)≥0
This simplifies to:
5y2+14y+1+4a≥0
As 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)<0
‌36−20(1+4a)<0
‌36−20−80a<0
‌16<80a
Thus:
a>‌

1

2

Therefore, the condition for a is:
−1<a<−‌

1

2

The correct understanding implies the feasible range for a.