Question

Prove that the expression 8x-4/x^2+2x-1 cannot have values between 2 to 4 in its domain

Answer 1

given expression, f(x) = (8x - 4)/(x² + 2x + 1)

for f(x) to be defined, (x² + 2x - 1) ≠ 0

x ≠ {-2 ± √(2² + 4)}/2 = -1 ± √2

so, domain of f(x) $\in$ R - {-1-√2, -1 + √2}

now, f(x) = y = (8x - 4)/(x² + 2x - 1)

or, y(x² + 2x -1) = 8x - 4

or, yx² + x(2y - 8) + (4 - y) = 0

for real value of x,discriminant, D = (2y - 8)² - 4(4 - y)y ≥ 0

or, (2y - 8)² - 4y(4 - y) ≥ 0

or, 4y² + 64 -32y - 16y + 4y² ≥ 0

or, 8y² - 48y + 64 ≥ 0

or, y² - 6y + 8 ≥ 0

or, y² - 4y - 2y + 8 ≥ 0

or, y(y - 4) -2(y - 4) ≥ 0

or, (y - 2)(y - 4) ≥ 0

or, y ≥ 4 or, y ≤ 2

or, y $\in$ R - (2, 4)

hence it is clear that given expression cannot have values between 2 to 4 in its domain.

Answer 2

Answer:

1

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