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  • State which of the following statements is True or False (i) If x < y and b < 0, then x/b < x/b (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

State which of the following statements is True or False

(i) If x < y and b < 0, then \frac{x}{y}< \frac{y}{b}
(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

 

 

 

 


 

Answers (1)

(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

 

 

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