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Bunuel
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The entire range of values of x is marked by the dark region on the nu 03 Aug 2021, 01:30
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66% (01:42) correct 34% (01:40) wrong based on 293 sessions

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The entire range of values of x is marked by the dark region on the number line, as shown above. Which of the following expressions describes the range of values of x?


A. \((3+x)(2−x) ≤ 0\)

B. \((3−x)(2+x) ≥ 0\)

C. \((x−3)(x+2) ≤ 0\)

D. \((x+3)(2−x) ≥ 0\)

E. \((3−x)(2−x) ≥ 0\)



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D
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The entire range of values of x is marked by the dark region on the nu 03 Aug 2021, 02:02
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Explanation:

D. (x+3)(2−x) ≥ 0
x ≤ 2
x ≥ -3

x ∈ (-3,2)

IMO-D
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The entire range of values of x is marked by the dark region on the nu 03 Aug 2021, 03:08
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D. (x+3)(2−x) ≥0

Expanding D further

= - (x+3) (x-2) ≥ 0
= (x+3) (x-2) <= 0
x= -3 and x = 2 (two end points)

IMO D
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The entire range of values of x is marked by the dark region on the nu 03 Aug 2021, 03:13
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Expression for the given range is (x+3)(x-2) < = 0

multiplying by -1
(X+3)(2-x) >=0
Option (D)

Bunuel

The entire range of values of x is marked by the dark region on the number line, as shown above. Which of the following expressions describes the range of values of x?

A. (3+x)(2−x) ≤ 0
B. (3−x)(2+x) ≥ 0
C. (x−3)(x+2) ≤ 0
D. (x+3)(2−x) ≥ 0
E. (3−x)(2−x) ≥ 0


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The entire range of values of x is marked by the dark region on the nu 03 Aug 2021, 03:32
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Bunuel

The entire range of values of x is marked by the dark region on the number line, as shown above. Which of the following expressions describes the range of values of x?

A. (3+x)(2−x) ≤ 0
B. (3−x)(2+x) ≥ 0
C. (x−3)(x+2) ≤ 0
D. (x+3)(2−x) ≥ 0
E. (3−x)(2−x) ≥ 0


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Seeing the image above, it's clear that the range is: -3<=x<=2
Using wavy curve method:
D)(x + 3)(2-x)>=0
(-x-3)(x-2)<=0
x = -3, 2
the range is: -3<=x<=2
D IMO
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The entire range of values of x is marked by the dark region on the nu 03 Aug 2021, 05:39
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To do this question efficiently (under 2 mins)
By the pic, -3 <= x <= 2 => x-2 <= 0 and x+3 >= 0
Hence, (x-2)(x+3) <= 0
Therefore, eliminate B, C and E.
A: (x+3)(2-x) <= 0
D: (x+3)(2-x) >= 0
The correct answer is D.
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The entire range of values of x is marked by the dark region on the nu 03 Aug 2021, 11:56
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Bunuel

The entire range of values of x is marked by the dark region on the number line, as shown above. Which of the following expressions describes the range of values of x?

A. (3+x)(2−x) ≤ 0
B. (3−x)(2+x) ≥ 0
C. (x−3)(x+2) ≤ 0
D. (x+3)(2−x) ≥ 0
E. (3−x)(2−x) ≥ 0


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Easy question but lengthy one. Two ways of solving - (1) substitute values (2) Actually solve inequality

A. (3+x)(2−x) ≤ 0 --> -2 does not satisfy the number line
B. (3−x)(2+x) ≥ 0 --> -3 does not satisfy the number line
C. (x−3)(x+2) ≤ 0 --> -3 does not satisfy the number line
D. (x+3)(2−x) ≥ 0 --> All values satisfy the inequality. Keep it
E. (3−x)(2−x) ≥ 0 --> All values satisfy the inequality. Keep it

Now between D and E. In both equations both factors either have to be positive or negative
D. (x+3)(2−x) ≥ 0
This implies x <= -3 AND x <= 2 Not possible
OR
x >= -3 AND x <=2 Bang on

E. (3−x)(2−x) ≥ 0
This implies x >= 3 AND x >= 2 which essentially is x>=3 Does not satisfy the number line
OR x < =3 and x < =2 which essentially is x <=2 Does not satisfy the number line

So answer is D­
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The entire range of values of x is marked by the dark region on the nu 03 Aug 2021, 22:10
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With the line, we can say the equation (x+3)(x-2)<= 0
or, (x+3)(2-x) > =0

So, I think D. :)
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The entire range of values of x is marked by the dark region on the nu 18 Mar 2025, 21:47
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