Advertisements
Advertisements
प्रश्न
Solve `(x^2 - 4)/(x^2 - 2x - 15) ≤ 0`
Advertisements
उत्तर
The given inequality is f(x) = `(x^2 - 4)/(x^2 + 4x - 15)`
= `((x + 2)(x - 2))/(x^2 - 5x + 3x - 15)`
= `((x + 2)(x - 2))/(x(x - 5) + 3(x - 5))`
f(x) = `((x + 2)(x - 2))/((x + 3)(x - 5)) ≤ 0`
[The critical numbers of f(x) are those values of x for which f(x) = 0, and those values of x for which f(x) is not defined. When x = – 3, 5. f(x) = ∞ ⇒ f(x) is not defined.]
The critical numbers are x = – 2 , 2, – 3, 5
Divide the number line into five intervals
(- ∞, – 3), (- 3, – 2), (- 2, 2), (2, 5), (5, ∞)
(i) (– ∞, – 3)
When x < – 3 say x = – 4
The factor x + 2 = – 4 + 2 = – 2 < 0
The factor x – 2 = – 4 – 2 = 6 < 0
The factor x + 3 = – 4 + 3 = – 1 < 0
The factor x – 5 = – 4 – 5 = – 9 < 0
Thus x + 2 < 0, x + 3 < 0, x – 2 < 0, x – 5 < 0
∴ `(x^2 - 4)/(x^2 + 4x - 15) > 0`
Thus `(x^2 - 4)/(x^2 + 4x - 15) ≤ 0` is not true in the interval (– ∞, – 3).
∴ It has no solution in (– ∞, – 3)
(ii) (– 3, – 2)
When – 3 < x ≤ – 2 say x = – 2.5
The factor x + 2 = – 2.5 + 2 = – 0.5 < 0
The factor x – 2 = – 2.5 – 2 = – 4.5 < 0
The factor x + 3 = – 2.5 + 3 = 0.5 > 0
The factor x – 5 = – 2.5 – 5 = – 7.5 < 0
Thus x + 2 < 0, x + 3 > 0
and
x – 2 < 0
x – 5 < 0
∴ `(x^2 - 4)/(x^2 + 4x - 15) < 0`
Thus `(x^2 - 4)/(x^2 + 4x - 15) ≤ 0` is not true in the interval (– 3, – 2).
∴ It has no solution in (– 3, – 2)
(iii) (– 2, 2)
When – 2 ≤ x ≤ 2 say x = 0
The factor x + 2 = 0 + 2 = 2 > 0
The factor x – 2 = 0 – 2 = – 2 < 0
The factor x + 3 = 0 + 3 = 3 > 0
The factor x – 5 = 0 – 5 = – 5 < 0
| Interval | Sign of x + 2 |
Sign of x – 2 |
Sign of x + 3 |
Sign of x – 5 |
Sign of `((x + 2)(x - 2))/((x + 3)(x - 5))` |
| (– ∞, – 3) | – | – | – | – | + |
| (– 3, – 2) | – | – | + | – | – |
| (– 2, 2) | + | – | + | – | + |
| (2, 5) | + | + | + | – | – |
| (5, ∞) | + | + | + | + | + |
Thus x + 2 > 0
x + 3 > 0
and
x – 2 < 0
x – 5 < 0
∴ `(x^2 - 4)/(x^2 + 4x - 15) > 0`
Thus `(x^2 - 4)/(x^2 + 4x - 15) ≤ 0` is not true in the interval (– 2, – 2).
∴ It has no solution in (– 2, – 2)
(iv) (2, 5)
When 2 ≤ x < 5 say x = 3
The factor x + 2 = 3 = 3 + 2 = 5 > 0
The factor x – 2 = 3 – 2 = 1 > 0
The factor x + 3 = 3 + 3 = 6 > 0
The factor x – 5 = 3 – 5 = – 2 < 0
Thus x + 2 > 0,
x + 3 > 0
and
x – 2 > 0
x – 5 < 0
∴ `(x^2 - 4)/(x^2 + 4x - 15) < 0`
Thus `(x^2 - 4)/(x^2 + 4x - 15) ≤ 0` is not true in the interval (2, 5).
∴ It has no solution in (2, 5)
(v) (5, ∞)
When 5 < x < ∞ say x = 6
The factor x + 2 = 6 + 2 = 8 > 0
The factor x – 2 = 6 – 2 = 4 > 0
The factor x + 3 = 6 + 3 = 9 > 0
The factor x – 5 = 6 – 5 = 1 > 0
Thus
x + 2 > 0,
x + 3 > 0
and
x – 2 > 0,
x – 5 > 0
∴ `(x^2 - 4)/(x^2 + 4x - 15) > 0`
Thus `(x^2 - 4)/(x^2 + 4x - 15) ≤ 0` is not true in the interval (5, ∞).
∴ It has no solution in (5, ∞)
The given inequality f(x) = `(x^2 - 4)/(x^2 + 4x - 15) ≤ 0` has solution in the intervals (– 3, – 2]
∴ The solution set is (– 3, 2] ∪ [2, 5)
APPEARS IN
संबंधित प्रश्न
Find all values of x that satisfies the inequality `(2x - 3)/((x - 2)(x - 4)) < 0`
Resolve the following rational expressions into partial fractions
`x/((x^2 + 1)(x - 1)(x + 2))`
Resolve the following rational expressions into partial fractions
`x/((x - 1)^3`
Resolve the following rational expressions into partial fractions
`1/(x^4 - 1)`
Resolve the following rational expressions into partial fractions
`(x - 1)^2/(x^3 + x)`
Resolve the following rational expressions into partial fractions
`(x^2 + x + 1)/(x^2 - 5x + 6)`
Resolve the following rational expressions into partial fractions
`(x^3 + 2x + 1)/(x^2 + 5x + 6)`
Resolve the following rational expressions into partial fractions
`(x + 12)/((x + 1)^2 (x - 2))`
Resolve the following rational expressions into partial fractions
`(6x^2 - x + 1)/(x^3 + x^2 + x + 1)`
Resolve the following rational expressions into partial fractions
`(7 + x)/((1 + x)(1 + x^2))`
Determine the region in the plane determined by the inequalities:
3x + 5y ≥ 45, x ≥ 0, y ≥ 0
Determine the region in the plane determined by the inequalities:
2x + 3y ≤ 35, y ≥ 2, x ≥ 5.
Determine the region in the plane determined by the inequalities:
x − 2y ≥ 0, 2x − y ≤ −2, x ≥ 0, y ≥ 0
Determine the region in the plane determined by the inequalities:
2x + y ≥ 8, x + 2y ≥ 8, x + y ≤ 6
Choose the correct alternative:
The solution of 5x − 1 < 24 and 5x + 1 > −24 is
Choose the correct alternative:
The solution set of the following inequality |x − 1| ≥ |x − 3| is
