Q. 315.0( 2 Votes )

State which of th

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 4th 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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