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If x is a positive integer, how many values of x satisfy the..........

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If x is a positive integer, how many values of x satisfy the.......... [#permalink] New post   31 Oct 2018, 04:28
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If x is a positive integer, how many values of x satisfy the inequality, \(\frac{(x - 2)}{(x^2 – 13x + 36)} ≤ 0\)?

    A. 4
    B. 5
    C. 6
    D. 7
    E. Cannot be determined

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Re: If x is a positive integer, how many values of x satisfy the.......... [#permalink] New post   31 Oct 2018, 09:56
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EgmatQuantExpert wrote:
If x is a positive integer, how many values of x satisfy the inequality, \(\frac{(x - 2)}{(x^2 – 13x + 36)} ≤ 0\)?

A. 4
B. 5
C. 6
D. 7
E. Cannot be determined


\(\frac{(x - 2)}{(x^2 – 13x + 36)} ≤ 0\)

Factorizing denominator,

Or, \(\frac{x-2}{\left(x-4\right)\left(x-9\right)}\le \:0\)

Now, applying Wavy-curve method, we have the following ranges of x:-

\(x\leq{2}\) or 4<x<9

As 'x' is a positive integer, hence the posssible value of x:- 1,2,5,6,7,8

Therefore, SIX nos. of values of 'x' satisfy the given inequality.

Ans. (C)
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If x is a positive integer, how many values of x satisfy the.......... [#permalink] New post   02 Nov 2018, 05:12
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Solution


Given:
    • We are given that x is a positive integer
    • And, we are given an inequality, \(\frac{(x - 2)}{(x^2 – 13x + 36)} ≤ 0\)

To find:
    • We need to find out the number of values of x that satisfies the given inequality

Approach and Working:
    • If we observe the given inequality, we can see that the denominator is a quadratic expression.
    • So, let’s try to factorise the quadratic expression, \(x^2 – 13x + 36\)
      o \(x^2 – 13x + 36 = (x – 9) * (x – 4)\)

    • Thus, the inequality can be written as, \(\frac{(x - 2)}{(x – 9)(x - 4)} ≤ 0\)
    • Now, we need to multiply the numerator and denominator of LHS with (x – 9)*(x – 4)
      o \(\frac{(x - 2) (x – 9)(x - 4)}{[(x – 9)(x - 4)]^2} ≤ 0\)
      o The denominator of the above inequality is always greater than 0 and x ≠ 4 or 9, since the denominator cannot be equal to 0.
      o So, the numerator must be ≤ 0, for \(\frac{(x - 2)(x – 9)(x - 4)}{[(x – 9)(x - 4)]^2} ≤ 0\).
      o We need to find the values of x, for which (x - 2) (x – 4)(x - 9) ≤ 0

Approach 1: Wavy-line method

    • The zero points of the inequality are {2, 4, 9} and the wavy-line will be as follows:



    • So, the expression will be equal to zero for x = {2, 4, 9}
      o But, as we already know that, x cannot take the values 4 and 9
      o Thus, x = {2}
    • The expression will be negative in the regions, 4 < x < 9 and x < 2
      o But we know that x is a positive integer, so, the expression will be negative for x = {1, 5, 6, 7, 8}
    • Therefore, the given inequality satisfies for x = {1, 2, 5, 6, 7, 8}

Approach 2: Number-line method

The zero points of the inequality are {2, 4, 9} and highlighting these points on a number line, we get:



    • Thus, (x - 2) (x – 4)(x - 9) will be ≤ 0 in two regions, x ≤ 2 and 4 ≤ x ≤ 9, but we already inferred that x ≠ {4 , 9}
    • And we know that x is a positive integer
    • Therefore, the given inequality satisfies for x = {1, 2, 5, 6, 7, 8}

Hence, the correct answer is option C.

Answer: C

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If x is a positive integer, how many values of x satisfy the.......... [#permalink] New post   31 Oct 2018, 10:52
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How did you get 1 2 5 6 7 and 8 from those 2 intervals?

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If x is a positive integer, how many values of x satisfy the.......... [#permalink] New post   02 Nov 2018, 08:06
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dave13 wrote:

i would appreciate you could take time to answrr my questions below :) thank you! :)

Question 1. Why do we need to multiply the numerator and denominator of LHS with (x – 9)*(x – 4) in which cases do we need to do so ?



Question 2. if the denominator of the above inequality is always greater than 0 and x ≠ 4 or 9, since the denominator cannot be equal to 0. WHY do we remove whole denominator in the end, leave only this (x - 2) (x – 4)(x - 9) ≤ 0 ???


Question 3. And the last question :grin: is it gramatically paralel structure to say "Approach and Working" :) may be "Aprroach and Work" is better or "Approach first and then start Working" :lol: how about "approach while working and by the time you approach it will be solved" :lol:


Hey dave13

The answer to question 1 is that they do so to make the denominator a square.
Remember a square of a number/expression can never be negative. Now, that
we have that in order, for the overall expression to be neagtive or equal to zero,
the numerator must be negative or have a value of zero at maximum.

That's the reason why they remove the denominator and keep only the numerator
Hope that answers your second question. :)

As for your thid question, I am sure generis will have an answer. Unfortunately,
I am not qualified to make that judgement :lol: :lol:
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Re: If x is a positive integer, how many values of x satisfy the.......... [#permalink] New post   25 Apr 2020, 07:57
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EgmatQuantExpert wrote:
If x is a positive integer, how many values of x satisfy the inequality, \(\frac{(x - 2)}{(x^2 – 13x + 36)} ≤ 0\)?

    A. 4
    B. 5
    C. 6
    D. 7
    E. Cannot be determined

To read all our articles: Must read articles to reach Q51


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File comment: You can use this method of plotting transition points on the number line.

This is fast method and with practice one can get used to it.
6226da61-516a-4a8a-b3d5-c724a9240d3f.jpg
6226da61-516a-4a8a-b3d5-c724a9240d3f.jpg [ 81.43 KiB | Viewed 9488 times ]

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Re: If x is a positive integer, how many values of x satisfy the.......... [#permalink] New post   31 Oct 2018, 10:57
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rahulkashyap wrote:
How did you get 1 2 5 6 7 and 8 from those 2 intervals?

Posted from my mobile device


Hi rahulkashyap,
a) x≤2 and x is a +ve integer. hence x could be: 1, 2
b) 4<x<9 and x is a +ve integer. hence x could be: 5,6,7,8
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If x is a positive integer, how many values of x satisfy the.......... [#permalink] New post   02 Nov 2018, 07:22
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EgmatQuantExpert wrote:

Solution


Given:
    • We are given that x is a positive integer
    • And, we are given an inequality, \(\frac{(x - 2)}{(x^2 – 13x + 36)} ≤ 0\)

To find:
    • We need to find out the number of values of x that satisfies the given inequality

Approach and Working:
    • If we observe the given inequality, we can see that the denominator is a quadratic expression.
    • So, let’s try to factorise the quadratic expression, \(x^2 – 13x + 36\)
      o \(x^2 – 13x + 36 = (x – 9) * (x – 4)\)

    • Thus, the inequality can be written as, \(\frac{(x - 2)}{(x – 9)(x - 4)} ≤ 0\)
    • Now, we need to multiply the numerator and denominator of LHS with (x – 9)*(x – 4)
      o \(\frac{(x - 2) (x – 9)(x - 4)}{[(x – 9)(x - 4)]^2} ≤ 0\)
      o The denominator of the above inequality is always greater than 0 and x ≠ 4 or 9, since the denominator cannot be equal to 0.
      o So, the numerator must be ≤ 0, for \(\frac{(x - 2)(x – 9)(x - 4)}{[(x – 9)(x - 4)]^2} ≤ 0\).
      o We need to find the values of x, for which (x - 2) (x – 4)(x - 9) ≤ 0

Approach 1: Wavy-line method

    • The zero points of the inequality are {2, 4, 9} and the wavy-line will be as follows:



    • So, the expression will be equal to zero for x = {2, 4, 9}
      o But, as we already know that, x cannot take the values 4 and 9
      o Thus, x = {2}
    • The expression will be negative in the regions, 4 < x < 9 and x < 2
      o But we know that x is a positive integer, so, the expression will be negative for x = {1, 5, 6, 7, 8}
    • Therefore, the given inequality satisfies for x = {1, 2, 5, 6, 7, 8}

Approach 2: Number-line method

The zero points of the inequality are {2, 4, 9} and highlighting these points on a number line, we get:



    • Thus, (x - 2) (x – 4)(x - 9) will be ≤ 0 in two regions, x ≤ 2 and 4 ≤ x ≤ 9, but we already inferred that x ≠ {4 , 9}
    • And we know that x is a positive integer
    • Therefore, the given inequality satisfies for x = {1, 2, 5, 6, 7, 8}

Hence, the correct answer is option C.

Answer: C





Greetings EgmatQuantExpert , Gladiator59, pushpitkc

i would appreciate you could take time to answrr my questions below :) thank you! :)

Question 1. Why do we need to multiply the numerator and denominator of LHS with (x – 9)*(x – 4) in which cases do we need to do so ?



Question 2. if the denominator of the above inequality is always greater than 0 and x ≠ 4 or 9, since the denominator cannot be equal to 0. WHY do we remove whole denominator in the end, leave only this (x - 2) (x – 4)(x - 9) ≤ 0 ???


Question 3. And the last question :grin: is it gramatically paralel structure to say "Approach and Working" :) may be "Aprroach and Work" is better or "Approach first and then start Working" :lol: how about "approach while working and by the time you approach it will be solved" :lol:
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Re: If x is a positive integer, how many values of x satisfy the.......... [#permalink] New post   02 Nov 2018, 08:38
dave13,

pushpitkc has already answered it perfectly.

I'd like to add that this is a common trick in inequality questions, to make perfect squares and then "ignore them" as the perfect squares of real numbers are greater than or equal to zero.

However, as an advanced step you can go ahead and skip that part. Treat the numerator and denominator the same and use the wavy approach. ( While multiplying by same factor in both num and den we are doing this itself!)

To add a little to q2 raised by you above, -ve * +ve is -ve and +ve * +ve is +ve... Which means multiplying by positive entity does not change the sign... Hence we can ignore the perfect square part in denominator.

Q3: no clue what you're talking about! :-D

Let me know if you wanna followup.

Best,
Gladi

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Re: If x is a positive integer, how many values of x satisfy the.......... [#permalink] New post   28 Mar 2020, 10:25
EgmatQuantExpert wrote:
If x is a positive integer, how many values of x satisfy the inequality, \(\frac{(x - 2)}{(x^2 – 13x + 36)} ≤ 0\)?

    A. 4
    B. 5
    C. 6
    D. 7
    E. Cannot be determined

To read all our articles: Must read articles to reach Q51



Asked: If x is a positive integer, how many values of x satisfy the inequality, \(\frac{(x - 2)}{(x^2 – 13x + 36)} ≤ 0\)?

(x-2)/(x-4)(x-9) <=0

Possible values of x = {1,2,5,6,7,8}

IMO C
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Re: If x is a positive integer, how many values of x satisfy the.......... [#permalink] New post   29 Mar 2020, 06:47
Why can't 0 be one of the integer.

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Re: If x is a positive integer, how many values of x satisfy the.......... [#permalink] New post   12 Apr 2020, 12:09
rahulkashyap wrote:
How did you get 1 2 5 6 7 and 8 from those 2 intervals?

Posted from my mobile device


3 is also satisfying the equation
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Re: If x is a positive integer, how many values of x satisfy the.......... [#permalink] New post   24 Sep 2020, 04:28
Adding to all the above solutions,

the correct option is 6.

The question says Positive integer, hence x cannot take 0 or lesser values. x<=2 - this is a key point in the question.
so X can take 1 and 2 to satisfy the equation and remaining solutions, by following the number line or wavy line method results can be obtained.
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Re: If x is a positive integer, how many values of x satisfy the.......... [#permalink] New post   25 Sep 2020, 07:03
Hi equation holds good at x=4 and 9since it is expression <=0 thus at 4 and 9 the value of expression is 0
thus x=1,2,4,5,6,7,8, and 9.
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Re: If x is a positive integer, how many values of x satisfy the.......... [#permalink] New post   20 Nov 2021, 02:18
Anubha1 wrote:
Hi equation holds good at x=4 and 9since it is expression <=0 thus at 4 and 9 the value of expression is 0
thus x=1,2,4,5,6,7,8, and 9.


At x=4 and x=9, the denominator will become 0 and such an expression where D = 0 is not defined. Hence these situations are to be ignored from solution set
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Re: If x is a positive integer, how many values of x satisfy the.......... [#permalink] New post   03 Jun 2022, 21:31
Here is an alternative solution for those interested:

Factor the denominator to get (x-9)*(x-4) so your overall inequality becomes \(\frac{(x-2)}{(x-9)*(x-4)} ≤ 0\)
We want to find values of x that will return a negative result or zero.

As a first step, you can deduce that x≠9 and x≠4, but x=2 is the first solution. From here, we're only really interested in two ranges: x>2 and x<2.
What if x>2? For any value of x greater than 2, the numerator will be positive. If x=3, the denominator is also positive so it does not satisfy the inequality. But when x = 5, 6, 7, and 8 the denominator is negative, leading to a negative result. So x=5,6,7,8 are four more solutions with a total of 5 at this point. If x>9, the denominator is always positive.

What if x<2? For any value of x less than 2, the numerator will be negative. But since we're given that x is a positive integer, x can only be 1. If x=1, we will get a negative result and this is our final solution, taking the total number of solutions to 6.
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Re: If x is a positive integer, how many values of x satisfy the.......... [#permalink] New post   28 Nov 2023, 18:29
I was wondering, if we did not square the denominator and simply included the points in the denominator on the wave curve line, would that change the answer? With this Q, I tend to get the correct answer with that approach too, so I wonder how essential it is to square the denominator in such questions?

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Re: If x is a positive integer, how many values of x satisfy the.......... [#permalink] New post   29 Nov 2023, 00:46
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TargetMBA007 wrote:
If x is a positive integer, how many values of x satisfy the inequality, \(\frac{(x - 2)}{(x^2 – 13x + 36)} ≤ 0\)?

    A. 4
    B. 5
    C. 6
    D. 7
    E. Cannot be determined

I was wondering, if we did not square the denominator and simply included the points in the denominator on the wave curve line, would that change the answer? With this Q, I tend to get the correct answer with that approach too, so I wonder how essential it is to square the denominator in such questions?

EgmatQuantExpert Bunuel


Here is the easiest way of solving this question:

\(\frac{(x - 2)}{x^2 – 13x + 36} ≤ 0\);

\(\frac{(x - 2)}{(x-4)(x-9)} ≤ 0\).

The "roots", in ascending order, are 2, 4, and 9, which gives us 4 ranges:

    \(x \leq 2\);
    \(2 < x < 4\);
    \(4 < x < 9\);
    \(x>9\).

Next, test an extreme value for \(x\): if \(x\) is some large enough number, say 100, then all three terms will be positive, giving a positive result for the whole expression. So when \(x > 9\), the expression is positive. Now, apply the alternating sign trick: \(\text{(- + - +)}\). So, the ranges when the expression is less than or equal to 0 are: \(x \leq 2\) and \(4 < x < 9\).

Since it is given that x is a positive integer, the possible values for x are 1, 2, 5, 6, 7, and 8, totaling 6 values. Note that while x can be 2, it cannot be 4 or 9, as that would make the denominator 0, and division by zero is not allowed.

Answer: C.
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