# Get Answers to all your Questions

#### State which of the following statements is True or False (i) If x < y and b < 0, then  (ii) If xy > 0, then x > 0 and y < 0 (iii) If xy > 0, then x < 0 and y < 0 (iv) If xy < 0, then x < 0 and y < 0 (v) If x < –5 and x < –2, then x ∈ (– ∞, – 5) (vi) If x < –5 and x > 2, then x ∈ (– 5, 2) (vii) If x > –2 and x < 9, then x ∈ (– 2, 9) (viii) If |x| > 5, then x ∈ (– ∞, – 5) ∪ [5, ∞) (ix) If |x| ≤ 4, then x ∈ [– 4, 4] (x) Graph of x < 3 is (xi) Graph of x ≥ 0 is (xii) Graph of y ≤ 0 is(xiii) Solution set of x ≥ 0 and y ≤ 0 is(xiv) Solution set of x ≥ 0 and y ≤ 1 is (xv) Solution set of x + y ≥ 0 is

(i) It is False.

x < y, b<0     ……. (given)

Multiplication or division by -ve no. Inverts the inequality sign

Thus, x/b > y/b

(ii)  It is False.

If xy > 0, then,

Either x >0 & y > 0,

Or x < 0 & y < 0.

(iii) It is True.

If xy > 0, then,

Either x >0 & y > 0,

Or x < 0 & y < 0.

(iv) It is False.

If xy < 0, then,

Either x < 0 & y > 0,

Or x > 0 & y < 0

(v) It is True.

We know that,

x < -5 → x $\epsilon$ (-∞,-5)      ………. (i)

& x < -2 → x $\epsilon$ (-∞,-2)  ………. (ii)

Thus, x$\epsilon$ (-∞,-5)            ………… [By taking intersection from (i) & (ii)]

(vi) It is False.

We know that,

x < -5 → x $\epsilon$ (-∞,-5)      ………. (i)

& x < 2 → x $\epsilon$ (∞,2)  ………. (ii)

Therefore, x has no common solution         ……… [From (i) & (ii)]

(vii) It is True.

x > -2 → x $\epsilon$ (-∞,-2)  ………. (i)

& x < 9  x $\epsilon$ (-∞,9)      ………. (ii)

Therefore, x $\epsilon$ (-2,9)       ……… [From (i) & (ii)]

(viii) It is True.

|x|< 5

Thus, there will be two cases,

x > 5 → x $\epsilon$ (5,∞)  …… (i)

& -x > 5 → x < -5

→ x $\epsilon$ ( -∞,-2)    …… (ii)

x $\epsilon$ (-∞,-5) U (5,∞)      ……. [From (i) & (ii)]

(ix) It is True.

|x| ≤ 4,

Thus, there will be two cases,

x ≤ 4 → x $\epsilon$ (-∞,4)        …….. (i)

& -x ≤4, x ≥ -4 → x $\epsilon$ [-4,∞]       ……. (ii)

Therefore, x $\epsilon$ [-4,4]         …….. [From (i) & (ii)]

(x) It is True.

Line: x = 3 & origin is (0,0),

Thus, the inequality is satisfied & hence the above graph is correct.

(xi) It is True.

The positive value of x is represented by x ≥ 0,

Therefore, the region of line x = 0 must be on the positive side that is the y-axis.

(xii) It is False.

The negative value of y is represented by y ≤ 0,

Therefore, the region of line y = 0 must be on the negative side that is the x-axis.

(xiii) It is False.

The shaded region is the first quadrant & the 4th quadrant is represented by x ≥ 0 & y ≤ 0.

(xiv) It is False.

The region on the left side of the y-axis is implied by x ≥ 0 & the region below the line y = 1 is implied by y ≤ 1.

(xv) It is True

The inequality is satisfied if we take any point above the line x + y = 0, say (3,2)

Thus, x+y ≥ 0

….…..(since, 3 + 2 = 5 > 0)

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