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Range of inequalities

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Range of inequalities

by vinni.k » Wed Apr 17, 2013 6:30 am
I have seen discussions regarding these questions. Please help in solving the following inequalities:-

(1) 0.5/(x-x²-1) < 0
Not sure how to create this one

(2). (x²-5x+6)/(x²+x+1)<0

(3). (x-1)(x+2)²/-1-x <0
Ans. -2 < x < -1 or x > 1 ( Is this correct ? )

(4). (x²+4x+4)/(2x²-x-1) > 0
Ans. -2 < x < -1/2 or x < 1 ( Is this correct ? )

(5). x�-5x²+4 < 0

(6). x�-2x²-63 <= 0

Thanks & Regards
Vinni
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by Anju@Gurome » Wed Apr 17, 2013 7:18 am
That's a lot of inequalities.
Please go through the posts I made in this thread >> https://www.beatthegmat.com/gmat-prep-ea ... 99965.html

Try to understand the method and retry these problems.
I'll post the answers one by one.
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by Anju@Gurome » Wed Apr 17, 2013 7:55 am
#1. 0.5/(x - x² - 1) < 0

As the numerator is positive, denominator must be negative.
So, x - x² - 1 < 0
--> x² - x + 1 > 0

Now, as I have mentioned in my linked post : for any expression of the form ax² + bx + c, if b² - 4ac < 0 --> It cannot be factored. And the expression will be either positive or negative for any value of x.
Here, b = -1 and a = c = 1 ---> b² - 4ac = (-1)² - 4 = -3 < 0
So, it cannot be factored and as for x = 0 the expression is positive, the expression is always positive.
Therefore, the inequality is true for any real value of x.

Also this can obtained logically as follows :
x² - x + 1 is always positive as
  • For |x| > 1, x² > x --> (x² - x) > 0 ---> (x² - x + 1) > 0
    For |x| < 1, -1 < (x² - x) < 0 ---> (x² - x + 1) > 0
    For x = 0, (x² - x + 1) = 1 > 0
So, the expression is always positive.
Therefore, the inequality is true for any real value of x.
Last edited by Anju@Gurome on Wed Apr 17, 2013 8:42 am, edited 1 time in total.
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by Anju@Gurome » Wed Apr 17, 2013 8:11 am
#2. (x² - 5x + 6)/(x² + x + 1) < 0

Following the analysis in my earlier post, (x² + x + 1) is always positive.
Hence, for the inequality to be true, the numerator must be negative.

So, (x² - 5x + 6) < 0
--> (x - 2)(x - 3) < 0

Draw a quick wavy curve as follows,
Image
We see that x must be greater than 2 but less than 3.
Hence, 2 < x < 3

If you do not feel comfortable with wavy curves, then proceed as follows:
--> (x - 2)(x - 3) < 0
--> either {(x - 2) > 0 and (x - 3) < 0} or {(x - 2) < 0 and (x - 3) > 0}
--> either {x > 2 and x < 3} or {x < 2 and x > 3}
As the second scenario is not possible, x > 2 and x < 3 ---> 2 < x < 3

EDIT : The points on the number line in the wavy curve should be marked by inked circles as I described in my original explanation. However, if you understand the logic, it doesn't matter how you mark them.
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by Anju@Gurome » Wed Apr 17, 2013 9:16 am
vinni.k wrote:#3. (x - 1)(x + 2)²/(-1 - x) < 0
(-1 - x) = -(1 + x) = -(x + 1)

So, (x - 1)(x + 2)²/(-1 - x) < 0
--> (x - 1)(x + 2)²/(x + 1) > 0 [Multiplying both sides by -1]
Now, draw a wavy curve as follows,
Image
We see that either x < -1 excluding x = -2 or x > 1
Hence, our final solution is x < -2, -2 < x < -1, and x > 1

So, your solution is partially correct as you are missing x < -2
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by vinni.k » Wed Apr 17, 2013 9:56 am
Anju@Gurome wrote: So, your solution is partially correct as you are missing x < -2
Thanks Anju, but how you got x < -2 ? The wavy curve simply says that -2 < x i.e
x > -2

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by Anju@Gurome » Wed Apr 17, 2013 10:15 am
vinni.k wrote:but how you got x < -2 ? The wavy curve simply says that -2 < x i.e
x > -2
Hi Vinni,

Please read the point 8 and 9A in my original post.
8. Now the expression is positive whenever the curve is situated above the number line and negative whenever the curve is situated below the number line.
9A. If the question said expression > 0, positive parts of the wavy curve are our solution.
In this case the expression should be positive.
So, positive parts of the curve is our solution.
Now, the curve is positive whenever it is situated above the number line.
Hence, all values less than -1 except -2 (as the curve touches the number line at x = -2) and all value greater than 1 are our solution.

Hope that helps.
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by Anju@Gurome » Wed Apr 17, 2013 10:38 am
vinni.k wrote:#4. (x² + 4x + 4)/(2x² - x - 1) > 0
(x² + 4x + 4) = (x + 2)² is always greater than or equal to zero.
Hence, the left-hand side of the inequality to be positive, the denominator must be positive.

So, 2x² - x - 1 > 0
--> 2x² - 2x + x - 1 > 0
--> (2x + 1)(x - 1) > 0
--> either {(2x + 1) > 0 and (x - 1) > 0} or {(2x + 1) < 0 and (x - 1) < 0}
--> either {x > -1/2 and x > 1} or {x < -1/2 and x < 1}
--> either x > 1 or x < -1/2
But x cannot be equal to -2 as that will make the expression zero.
Hence, our final solution : x < -2, -2 < x < -1/2, and x > 1

To solve this with wavy curve,
--> (x + 2)²/[(2x + 1)(x - 1)] > 0
--> (x + 2)²/[(x + 1/2)(x - 1)] > 0

Draw the curve as follows,
Image
For the expression to be positive, the curve must stay above the number line.
Hence, our solution : x < -2, -2 < x < -1/2, and x > 1
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by Anju@Gurome » Wed Apr 17, 2013 11:02 am
#5. x� - 5x² + 4 < 0

--> x� - 5x² + 4 < 0
--> x� - 4x² - x² + 4 < 0
--> (x² - 4)(x² - 1) < 0
--> (x - 1)(x + 1)(x - 2)(x + 2) < 0

Draw the curve as follows,
Image
For the expression to be negative, the curve must stay below the number line.
Hence, our solution : -2 < x < -1 and 1 < x < 2
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by Anju@Gurome » Wed Apr 17, 2013 11:09 am
#6. x� - 2x² - 63 ≤ 0

--> x� - 2x² - 63 ≤ 0
--> x� - 9x² + 7x² - 63 ≤ 0
--> (x² - 9)(x² + 7) ≤ 0
--> (x - 3)(x + 3)(x² + 7) ≤ 0

Now, (x² + 7) is always positive for any value of x.
Hence, our inequality reduces down to (x - 3)(x + 3) ≤ 0
So, either {(x - 3) ≤ 0 and (x + 3) ≥ 0} or {(x - 3) ≥ 0 and (x + 3) ≤ 0}
--> either {x ≤ 3 and x ≥ - 3} or {x ≥ 3 and x ≤ -3}

As the second case is not possible, our final solution : {x ≤ 3 and x ≥ - 3} --> -3 ≤ x ≤ 3
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by vinni.k » Wed Apr 17, 2013 11:13 am
I think i might not be following you correctly. What if the question said expression < 0,
then the solution might have been -1 < x < 1 and x < -2

I am taking the same wavy curve for #3. (x - 1)(x + 2)²/(-1 - x) < 0; however i am only changing the sign that > .

I hope i am on the right page now.

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by Anju@Gurome » Wed Apr 17, 2013 11:25 am
vinni.k wrote:I think i might not be following you correctly. What if the question said expression < 0,
then the solution might have been -1 < x < 1 and x < -2

I am taking the same wavy curve for #3. (x - 1)(x + 2)²/(-1 - x) < 0; however i am only changing the sign that > .

I hope i am on the right page now.

Vinni
Are you able to see the figure of the curve I've embedded in my post?
If yes, then you should be able to see that the curve is
  • above the number line for x < -2, -2 < x < -1, and x > 1
    Below the number line for -1 < x < 1
    Touches the number line for x = -2
    Crosses the number line at x = -1 and x = 1
Hence, the curve is
  • +ve for x < -2, -2 < x < -1, and x > 1
    -ve for -1 < x < 1
    zero for x = -2 and x = 1
    undefined for x = -1
And, if you are not able see the figure of the curve, click this link >> https://s4.postimg.cc/to4jghw9p/btg.jpg

Note that, for this particular inequality, (x + 2) has even power. Hence, at x = -2, the curve will not cross the number line, rather it will touch the number line and stay on the same side. Read the last part of #6 in my original post...
6. From right to left, beginning above the number line (if the value of the expression is positive in step 5, otherwise from below the number line if the value of the expression is negative in step 5), a wavy curve is drawn to pass through all the marked points so that when it passes through a critical point, the curve intersects the number line. But if the critical point is from a factor which is repeated even number of times, the curve will not intersect the number line but remain on the same side after touching it.
Hope that helps.
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by vinni.k » Wed Apr 17, 2013 11:44 am
Crystal clear. Thank you

Regards
Vinni
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