Advertisements
Advertisements
प्रश्न
Solve for x, |x + 1| + |x| > 3.
उत्तर
On L.H.S. of the given inequality, We have two terms both containing modulus. By equating the expression within the modulus to zero, we get x = –1, 0 as critical points. These critical points divide the real line in three parts as (– `oo`, – 1), [–1, 0), [0, `oo`).
Case I: When `–oo < x < –1`
|x + 1| + |x| > 3
⇒ –x – 1 – x > 3
⇒ x < –2
Case II: When –1 ≤ x < 0,
|x + 1| + |x| > 3
⇒ x + 1 – x > 3
⇒ 1 > 3 ....(Not possible)
Case III: When 0 ≤ x < `oo`,
|x + 1| + |x| > 3
⇒ x + 1 + x > 3
⇒ x > 1
Combining the results of cases (I), (II) and (III), We get x ∈ `(–oo , –2) ∪ (1, oo)`.
APPEARS IN
संबंधित प्रश्न
Solve: −4x > 30, when x ∈ Z
Solve: 4x − 2 < 8, when x ∈ R
3x − 7 > x + 1
x + 5 > 4x − 10
\[\frac{4 + 2x}{3} \geq \frac{x}{2} - 3\]
\[\frac{4x + 3}{2x - 5} < 6\]
\[\frac{5x + 8}{4 - x} < 2\]
Solve each of the following system of equations in R.
x − 2 > 0, 3x < 18
Solve each of the following system of equations in R.
2x + 5 ≤ 0, x − 3 ≤ 0
Solve each of the following system of equations in R.
11 − 5x > −4, 4x + 13 ≤ −11
Solve each of the following system of equations in R.
20. −5 < 2x − 3 < 5
Solve \[\frac{\left| x + 2 \right| - x}{x} < 2\]
Mark the correct alternative in each of the following:
If \[\frac{\left| x - 2 \right|}{x - 2}\]\[\geq\] then
Mark the correct alternative in each of the following:
If \[\left| x + 3 \right|\]\[\geq\]10, then
Solve the inequality, 3x – 5 < x + 7, when x is a natural number.
Solve `(x - 2)/(x + 5) > 2`.
Solve |3 – 4x| ≥ 9.
Solve for x, `(|x + 3| + x)/(x + 2) > 1`.
The length of a rectangle is three times the breadth. If the minimum perimeter of the rectangle is 160 cm, then ______.
If |3x – 7| > 2, then x ______ `5/3` or x ______ 3.
If –3x + 17 < –13, then ______.
If |x − 1| > 5, then ______.
If x < –5 and x > 2, then x ∈ (– 5, 2)
If – 4x ≥ 12, then x ______ – 3.
If x > y and z < 0, then – xz ______ – yz.
If p > 0 and q < 0, then p – q ______ p.
