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Find the range of values of x such that (x-5)^3 (2-4x) < 0

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Find the range of values of x such that (x-5)^3 (2-4x) < 0  [#permalink]

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New post Updated on: 23 Aug 2018, 21:25
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  45% (medium)

Question Stats:

63% (01:34) correct 37% (01:23) wrong based on 236 sessions

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Solving inequalities- Number Line Method - Practice Question #1

Find the range of values of x such that \((x-5)^3 (2-4x) < 0\)


A. \(\frac{1}{2} < x < 5\)
B. \(0 < x < \frac{1}{2}\)
C. \(x >5\)
D. \(\frac{1}{2} < x < 5\)
E. \((x < \frac{1}{2}\)) and (\(x > 5\))

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Originally posted by EgmatQuantExpert on 23 Aug 2018, 04:23.
Last edited by EgmatQuantExpert on 23 Aug 2018, 21:25, edited 1 time in total.
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Find the range of values of x such that (x-5)^3 (2-4x) < 0  [#permalink]

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New post Updated on: 27 Aug 2018, 02:05
1

Solution


Given:
    • An inequality, \((x-5)^3 (2-4x) < 0\)

To find:
    • The range of x, that satisfies the above inequality

Approach and Working:
    • So, if we observe carefully, the inequality given to us can be written as,
      o \((x-5)^2 * (x-5) * (2-4x) < 0\)
    • We know that, any number of the form N2 is always positive, expect for N = 0.
      o So, we can say that, \((x - 5)^2\) is always positive, expect for x = 5,
      o Thus, the inequality can be written as (x - 5) * (2 - 4x) < 0
    • Now, let’s multiply the inequality by -1 to make the coefficient of x, positive
      o And, note that the inequality sign must be changed as we are multiplying it by a negative number, -1
      o Thus, the inequality becomes, (x - 5) * (4x - 2) > 0
    • The zero points of this inequality are x = 5 and \(x = \frac{1}{2}\)
    • Plot these points on the number line.
      o Since, both (x – 5) and (4x – 2) > 0, the inequality will be positive, for all the points to the right of 5
      o And, it is negative, in the region, between \(\frac{1}{2}\) and 5, since (x – 5) is negative and (4x – 2) is positive in this region
      o It is again positive for all the values less than \(\frac{1}{2}\), since both (x – 5) and (4x – 2) are negative in this region

Image

Therefore, the range of x is \(x < \frac{1}{2}\) and x > 5

Hence, the correct answer is option E.

Answer: E

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Originally posted by EgmatQuantExpert on 23 Aug 2018, 04:33.
Last edited by EgmatQuantExpert on 27 Aug 2018, 02:05, edited 3 times in total.
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Re: Find the range of values of x such that (x-5)^3 (2-4x) < 0  [#permalink]

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New post 23 Aug 2018, 05:38
2
EgmatQuantExpert wrote:

Find the range of values of x such that \((x-5)^3 (2-4x) < 0\)


A. \(\frac{1}{2} < x < 5\)
B. \(0 < x < \frac{1}{2}\)
C. \(x >5\)
D. \(\frac{1}{2} < x < 5\)
E. \((x < \frac{1}{2}\)) and (\(x > 5\))


\((x-5)^3 (2-4x) < 0...........(x-5)^3(1-2x)<0\)

two cases-
1) \(x-5<0\) or \(x<5\), then \(1-2x>0\) or \(2x<1\) or \(x<\frac{1}{2}\).... thus \(x<\frac{1}{2}\)
2) \(x-5>0\) or \(x>5\), then \(1-2x<0\) or \(2x>1\) or \(x>\frac{1}{2}\).... thus \(x>5\)

combined \(x<\frac{1}{2}\) and \(x>5\)

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Re: Find the range of values of x such that (x-5)^3 (2-4x) < 0  [#permalink]

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New post 23 Aug 2018, 06:35
EgmatQuantExpert wrote:
Solving inequalities- Number Line Method - Practice Question #1

Find the range of values of x such that \((x-5)^3 (2-4x) < 0\)


A. \(\frac{1}{2} < x < 5\)
B. \(0 < x < \frac{1}{2}\)
C. \(x >5\)
D. \(\frac{1}{2} < x < 5\)
E. \((x < \frac{1}{2}\)) and (\(x > 5\))


\((x-5)^3 (2-4x) < 0\)
Or, \(-2\left(x-5\right)^3\left(2x-1\right)<0\)
Or, \(2\left(x-5\right)^3\left(2x-1\right)>0\)

Critical points:- 5, 1/2
Using wavy curve method:-
range of x:-
\(x<\frac{1}{2}\) or \(x>5\)

Ans. (E)

P.S:- In option (E) , OR sounds logical instead of AND.
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Re: Find the range of values of x such that (x-5)^3 (2-4x) < 0  [#permalink]

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New post 05 Sep 2018, 02:29
EgmatQuantExpert wrote:

Solution


Given:
    • An inequality, \((x-5)^3 (2-4x) < 0\)

To find:
    • The range of x, that satisfies the above inequality[/list

    Approach and Working:
      • So, if we observe carefully, the inequality given to us can be written as,
        o \((x-5)^2 * (x-5) * (2-4x) < 0\)
      • We know that, any number of the form N2 is always positive, expect for N = 0.
        o So, we can say that, \((x - 5)^2\) is always positive, expect for x = 5,
        o Thus, the inequality can be written as (x - 5) * (2 - 4x) < 0
      • Now, let’s multiply the inequality by -1 to make the coefficient of x, positive
        o And, note that the inequality sign must be changed as we are multiplying it by a negative number, -1
        o Thus, the inequality becomes, (x - 5) * (4x - 2) > 0
      • The zero points of this inequality are x = 5 and \(x = \frac{1}{2}\)
      • Plot these points on the number line.
        o Since, both (x – 5) and (4x – 2) > 0, the inequality will be positive, for all the points to the right of 5
        o And, it is negative, in the region, between \(\frac{1}{2}\) and 5, since (x – 5) is negative and (4x – 2) is positive in this region
        o It is again positive for all the values less than \(\frac{1}{2}\), since both (x – 5) and (4x – 2) are negative in this region

Image

Therefore, the range of x is \(x < \frac{1}{2}\) and x > 5

Hence, the correct answer is option E.

Answer: E

Image



Hello EgmatQuantExpert, thanks for the wholesome explanation. I would like to know why after multiplying the whole inequality by -1, the sign in (x-5) did no change only that of 2-4x changed. Thanks.

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Re: Find the range of values of x such that (x-5)^3 (2-4x) < 0  [#permalink]

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New post 06 Sep 2018, 19:23
EgmatQuantExpert wrote:

Find the range of values of x such that \((x-5)^3 (2-4x) < 0\)

A. \(\frac{1}{2} < x < 5\)
B. \(0 < x < \frac{1}{2}\)
C. \(x >5\)
D. \(\frac{1}{2} < x < 5\)
E. \((x < \frac{1}{2}\)) and (\(x > 5\))

\(?\,\,\,:\,\,\,{\left( {x - 5} \right)^3}\left( {2 - 4x} \right) < 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\boxed{\left( {x - 5} \right)\left( {1 - 2x} \right) < 0}\)

\(\left( {x - 5} \right)\left( {1 - 2x} \right) = 0\,\,\,\,\, \Leftrightarrow \,\,\,\,x = 5\,\,{\text{or}}\,\,x = \frac{1}{2}\)

\(?\,\,:\,\,\,\left\{ {x < \frac{1}{2}} \right\}\,\,\, \cup \,\,\,\left\{ {x > 5} \right\}\)

The solution above follows the notations and rationale taught in the GMATH method.

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Re: Find the range of values of x such that (x-5)^3 (2-4x) < 0  [#permalink]

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New post 06 Sep 2018, 21:09
I thought the answer must be written in form like x < 1/2 or x > 5 as there is no way x can satisfy both the condition. The "And" here made me confused.
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Find the range of values of x such that (x-5)^3 (2-4x) < 0  [#permalink]

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New post 06 Sep 2018, 21:34
Hungluu92vn wrote:
I thought the answer must be written in form like x < 1/2 or x > 5 as there is no way x can satisfy both the condition. The "And" here made me confused.

Hi, Hungluu92vn!

Excellent observation. I have avoided to use the term "and", because EACH x cannot be simultaneously in ]-infinity, 1/2[ and in ]5, +infinity[.
That´s why, for the solution set of a inequation such as the one presented in the question stem, the proper word is OR , the proper symbol is U (reunion).

Other experts used the term AND, specifically in this question, because the word "range" was presented in the question stem!
Their interpretation: there are real values "x" that belong to ]-infinity, 1/2[ AND there are (OTHER) real values "x" that belong to ]5, +infinity[.
(It is as if "x" is a dummy variable: different "roles" each time it appears! This is not the classical way of looking into this "theme" but... you got the point!)

Regards,
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Find the range of values of x such that (x-5)^3 (2-4x) < 0  [#permalink]

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New post 09 Sep 2018, 05:21
Under 30sec approach

You directly see that x can be bigger than 5. Eliminate A,B and D.
Left with C or E. --> x can also be negative therefore eliminate C. Hence, E
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Re: Find the range of values of x such that (x-5)^3 (2-4x) < 0  [#permalink]

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New post 09 Sep 2018, 06:13
Hello Egmatquantexpert why are multiplying by -1? What is the need for it?

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Re: Find the range of values of x such that (x-5)^3 (2-4x) < 0   [#permalink] 09 Sep 2018, 06:13
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