Question

The value of the function f(x) = `(x^2 - 3x + 2)/(x^2 + x - 6)` lies in the interval

पर्याय

• `(-oo, oo) - {1/5, 1}`

• `(-∞, ∞) - {1/5, 1}`

• `(-∞, ∞)`

• `(-oo, oo)`

• `(-oo, oo) - {1}`

• `(-∞, ∞) - {1}`

• None of these

• None of these

Answer 1

`(-oo, oo)`

Explanation:

f(x) is defined if x² + x – 6 = 0

i.e. (x + 3)(x – 2) ≠ 0

i.e. x ≠ – 3, 2

∴ Domain (f) = `(- oo, oo) - {- 3, 2}`

Let y = `(x^2 - 3x + 2)/(x^2 + x - 6)`

⇒ x²y + xy – 6y = x² – 3x + 2

⇒ x²(y – 1) + x(y + 3) – (6y + 2) = 0

For x to be real, (y + 3)² + 4(y – 1)(6y + 2) ≥ 0

⇒ 25y² – 10y + 1 ≥ 0

i.e. (5y – 1)² ≥ 0

Which is true for all real y.

Range of f = `(-oo, oo)`

Answer 2

`bb((-∞, ∞)`

Explanation:

f(x) is defined if x² + x – 6 ≠ 0

i.e. (x + 3)(x – 2) ≠ 0 i.e x ≠ –3, 2

∴ Domain (f) = (–∞, ∞) – {–3, 2}

Let y = `(x^2 - 3x + 2)/(x^2 + x - 6)`

`\implies` x²y + xy – 6y = x² – 3x + 2

`\implies` x²(y – 1) + x(y + 3) – (6y + 2) = 0

For x to be real, (y + 3)² + 4(y – 1)(6y + 2) ≥ 0

`\implies` 25y² – 10y + 1 ≥ 0 i.e (5y – 1)² ≥ 0

Which is true for all real y.

Range of f = (–∞, ∞)