Answer :

(i) False

Explanation:

Given,

x < y

Now, b < 0, and multiplying/dividing by a negative inequality inverts the inequality sign

(ii) False

Explanation:

We know, If xy > 0

Then, either x > 0 and y > 0

Or

x < 0 and y < 0

(iii) True

Explanation:

We know, If xy > 0

Then, either x > 0 and y > 0

Or

x < 0 and y < 0

(iv) False

Explanation:

We know, If xy < 0

Then, either x < 0 and y > 0

Or

x > 0 and y < 0

(v) True

Explanation:

x < -5, i.e.

x ∈ (-∞, -5) [1]

and

x < -2, i.e.

x ∈ (-∞, -2) [2]

Taking intersection from [1] and [2], we get

x ∈ (-∞, -5)

(vi) False

Explanation:

x < -5, i.e.

x ∈ (-∞, -5) [1]

and

x > 2, i.e.

x ∈ (2, ∞) [2]

From [1] and [2], we get

x has no common solution

(vii) True

Explanation:

x > -2, i.e.

x ∈ (-2, ∞) [1]

and

x < 9, i.e.

x ∈ (-∞, 9) [2]

Taking intersection from [1] and [2], we get

x ∈ (-2, 9)

(viii) True

Explanation:

|x| > 5

Hence, there are two cases,

x > 5

⇒ x ∈ (5, ∞) [1]

and

-x > 5

⇒ x < -5

⇒ x ∈ (-∞, -5) [2]

From [1] and [2], we get

⇒ x ∈ (-∞, -5) ∪ (5, ∞)

(ix) True

Explanation:

|x| ≤ 4

Hence, there are two cases,

x ≤ 4

⇒ x ∈ (-∞, 4] [1]

and

-x ≤ 4

⇒ x ≥ -4

⇒ x ∈ [-4, ∞) [2]

From [1] and [2], we get

⇒ x ∈ [-4, 4]

(x) True

Explanation:

Line is x = 3, and

since origin i.e.

(0, 0) satisfies the inequality x < 3, the above graph is correct

(xi) True

Explanation:

Since, x ≥ 0 represents the positive value of x, the region must be on positive side of line x = 0 i.e. y-axis

(xii) false

Explanation:

Since, y ≤ 0 represents the negative value of y, the region must be on negative side of line y = 0 i.e. x-axis

(xiii) False

Explanation:

x ≥ 0 and y ≤ 0 represents the 4^{th} quadrant, while the region shaded is first quadrant

(xiv) False

Explanation:

x ≥ 0 implies that, region is on the left side of y axis, and

y ≤ 1 implies that, region is below the line y = 1, therefore graph must be

(xv) True

Explanation:

If we take any point above the line x + y = 0, say (3, 2) it satisfy the inequality

x + y ≥ 0 [as, 3 + 2 = 5 > 0]

Hence, region should be above the line x + y = 0

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