For real x, find the range of f(x)=(x-1)/(x^2+2x-2).
Answer and solution
Correct answer: A
Solution
Answer: A — (-∞, 1/6] ∪ [1/2, ∞)
1. The function is undefined at x=−1±√3. Set y=(x−1)/(x²+2x−2) and rearrange to yx²+(2y−1)x+1−2y=0.
2. For y≠0, real x requires Δ=(2y−1)²−4y(1−2y)=12y²−8y+1=(2y−1)(6y−1)≥0.
3. Thus y≤1/6 or y≥1/2. For y=0 the equation is linear: x=1, whose denominator is 1, so 0 is included.
4. No solution of the rearranged equation can have denominator zero: that would also require x−1=0, but the denominator at x=1 is 1.
5. Endpoints are attained: f(2)=1/6 and f(0)=1/2. Range=(−∞,1/6] ∪ [1/2,∞), option A.
Why this method works:
Solve for x at a fixed proposed outputy. A y belongs to the range exactly when the resulting equation has a real, permitted x. The discriminant gives the range boundary, but y=0 changes the equation from quadratic to linear and needs its own check.
Independent check and common trap:
At y=1/6 the equation has the repeated rootx=2; at y=1/2 it hasx=0. Both denominators are nonzero, so both endpoints are included. A denominator zero could not solve the rearranged equation because it would requirex=1 simultaneously.
Apply: Treat x as the unknown in yx²+(2y−1)x+1−2y=0. Its discriminant gives (2y−1)(6y−1)≥0. Separately check y=0 and denominator exclusions.
Check: At y=1/6 the equation has the repeated rootx=2; at y=1/2 it hasx=0. Both denominators are nonzero, so both endpoints are included. A denominator zero could not solve the rearranged equation because it would requirex=1 simultaneously.
1. The function is undefined at x=−1±√3. Set y=(x−1)/(x²+2x−2) and rearrange to yx²+(2y−1)x+1−2y=0.
2. For y≠0, real x requires Δ=(2y−1)²−4y(1−2y)=12y²−8y+1=(2y−1)(6y−1)≥0.
3. Thus y≤1/6 or y≥1/2. For y=0 the equation is linear: x=1, whose denominator is 1, so 0 is included.
4. No solution of the rearranged equation can have denominator zero: that would also require x−1=0, but the denominator at x=1 is 1.
5. Endpoints are attained: f(2)=1/6 and f(0)=1/2. Range=(−∞,1/6] ∪ [1/2,∞), option A.
Why this method works:
Solve for x at a fixed proposed outputy. A y belongs to the range exactly when the resulting equation has a real, permitted x. The discriminant gives the range boundary, but y=0 changes the equation from quadratic to linear and needs its own check.
Independent check and common trap:
At y=1/6 the equation has the repeated rootx=2; at y=1/2 it hasx=0. Both denominators are nonzero, so both endpoints are included. A denominator zero could not solve the rearranged equation because it would requirex=1 simultaneously.
G Strategy shortcut
Spot: Functions and Range.Apply: Treat x as the unknown in yx²+(2y−1)x+1−2y=0. Its discriminant gives (2y−1)(6y−1)≥0. Separately check y=0 and denominator exclusions.
Check: At y=1/6 the equation has the repeated rootx=2; at y=1/2 it hasx=0. Both denominators are nonzero, so both endpoints are included. A denominator zero could not solve the rearranged equation because it would requirex=1 simultaneously.
G Strategy: Spot: Functions and Range. Apply: Treat x as the unknown in yx²+(2y−1)x+1−2y=0. Its discriminant gives (2y−1)(6y−1)≥0. Separately check y=0 and denominator exclusions. Check: At y=1/6 the equation has the repeated rootx=2; at y=1/2 it hasx=0. Both denominators are nonzero, so both endpoints are included. A denominator zero could not solve the rearranged equation because it would requirex=1 simultaneously.