If the range of the function f(x) = (5-x)/(x² - 3x + 2) , x ≠ 1, 2, is (-∞ , α] ∪ [β, ∞) , then α² + β² is equal to :

Solution

Maths sir · 4 steps

Set y = f(x), turn it into a quadratic in x, and demand real roots: the discriminant gives the range.

1. Make a quadratic. Put y = (5-x)/(x²-3x+2) and cross-multiply. This gives yx² + (1-3y)x + (2y-5) = 0. A value y is in the range only if this equation has a real root x.
2. Real roots. For y ≠ 0 the discriminant must be non-negative. For y = 0 the equation gives x = 5, so 0 is in the range. Neither x = 1 nor x = 2 ever satisfies the equation (they give 0 = 4 and 0 = 3), so no extra value needs to be removed.

   (1-3y)² - 4y(2y-5) ≥ 0 ⇒ y² + 14y + 1 ≥ 0

3. Read off the ends. The roots of y² + 14y + 1 = 0 are y = -7 ± 4√3. So the range is y ≤ -7-4√3 or y ≥ -7+4√3, which gives α = -7-4√3 and β = -7+4√3.
4. So. α + β = -14 and αβ = 1, so α² + β² = (α+β)² - 2αβ = 196 - 2 = 194, which is option (d).

The idea to keep. For a rational function with a quadratic inside, write y = f(x) as a quadratic in x and force its discriminant to be ≥ 0. Then check y = 0 and the excluded x values separately.

Answer from the NTA final answer key. Solution by PrepWiser.