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Re: Inequality - Absolute value [#permalink]
10 Apr 2011, 20:39
I don't know how to do graphically, but the idea is that we have to maximize the values of both the terms in the expression. In that case, the last option is the biggest one with -2 as the value :
x2– x = (-2)^2 -(-2) = 4 + 2 = 6
For the other options :
3x – 1 -> Max value in the given range is 3*2 - 1 = 5 x2 + 1 -> Max value in the given range is (2)^2 + 1 = 5 3 – x -> Max value in the given range is 3 - (-2) = 5 x – 3 -> Max value in the given range is 2 - 3 = -1
Answer - E _________________
Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)
Re: Inequality - Absolute value [#permalink]
10 Apr 2011, 22:15
There is trick used by designers. more often when you have rhe choice of negative and positive values and the argument is dependent on the sign of the number- negative number will be decisive. The people will overlook the negative value more than the positive the designer is going to use this trap. Since we should look in the opposite way first exhaust all the negative values to ensure there is no trap. And you are good!
IEsailor wrote:
for those of you who havent already understood, options B and E are X^2 + 1 X^2 - X
Re: Inequality - Absolute value [#permalink]
11 Apr 2011, 13:17
2
This post received KUDOS
IEsailor wrote:
If x is a number such that –2 ≤ x ≤ 2, which of the following has the largest possible absolute value?
3x – 1 x2 + 1 3 – x x – 3 x2– x
Can this be done Algebrically / graphically.
I got the answer using substitution but found that time consuming.
If you know how to plot graphs, this is very straight forward, most of the expressions are very simple to plot.
In case you want to do algebraically, here are a couple of simple rules you need to know: (a) For linear expressions, the maxima or minima will always be at the extreme ends of a range (b) For quadratic expressions, the maxima or minima will always either be the extreme end of a range of the global max/min of the function
3x – 1 : At ends, value is 5 and -7 x2 + 1 : At ends, value is 5 and global min value is at x=0, where it is 1 3 – x : At ends, value is 1, 5 x – 3 : At ends, value is -5,-1 x2– x : At ends, value is 2,6. Global min value is at x=0.5, where it is -0.25
(Its easy to tell what the global min for these simple quadratic functions is knowing where there roots are, it is half way between the roots)
Re: Inequality - Absolute value [#permalink]
11 Apr 2011, 22:47
1
This post received KUDOS
gmat1220 wrote:
hello shrouded1 I don't understand your last statement. Pls can you elaborate.
(Its easy to tell what the global min for these simple quadratic functions is knowing where there roots are, it is half way between the roots)
On way to get the global min is applying the calculus. - y = x^2+1 dy/dx = 0 or 2x = 0. So the min occurs at x=0
Similarly y = x^2-x dy/dx = 0 or 2x - 1 = 0 . So the min occurs at x = 1/2 = 0.5
Thats implicitly what I did. But I used a little trick so I dont have to do the calculations. For any quadratic function with real roots, the minima/maxima (depending on wether the coefficient of x^2 is positive or negative) will occur at the mean of the roots. (This is very easy to prove, all I am saying is that the minima is -b/2a, which you can show by differentiation).
So for x^2-x = x(x-1) the roots are 0,1 and the minima will be at 0.5
For x^2+1, the roots are not real, but this function is simply a vertically translated version of x^2=0, hence it also minimizes/maximizes at the same level which is x=0.
Generalizing, you can always use the fornula x=(-b/2a) for the min or max valuation _________________
If x is a number such that –2 ≤ x ≤ 2 which of the following [#permalink]
02 Jul 2013, 09:37
Expert's post
1
This post was BOOKMARKED
emailmkarthik wrote:
Hey Bunuel,
Do you have any alternate easier method to tackle this problem?
If x is a number such that –2 ≤ x ≤ 2, which of the following has the largest possible absolute value? A. 3x – 1 B. x^2 + 1 C. 3 – x D. x – 3 E. x^2 – x
First of all notice that we can eliminate options C and D right away: \(|3-x|=|x-3|\), so these two options will have the same maximum value and since we cannot have two correct answers in PS questions then none of them is correct.
Evaluate each option by plugging min and max possible values of x:
A. 3x – 1 --> max for x=-2 --> |3*(-2)-1|=7.
B. x^2 + 1 --> max for x=-2 or x=2 --> |2^2 + 1|=5.
E. x^2 – x --> max for x=-2 --> |(-2)^2 - (-2)|=6.
Re: Inequality - Absolute value [#permalink]
02 Jul 2013, 09:50
Bunuel wrote:
emailmkarthik wrote:
Hey Bunuel,
Do you have any alternate easier method to tackle this problem?
If x is a number such that –2 ≤ x ≤ 2, which of the following has the largest possible absolute value? A. 3x – 1 B. x^2 + 1 C. 3 – x D. x – 3 E. x^2 – x
First of all notice that we can eliminate options C and D right away: \(|3-x|=|x-3|\), so these two options will have the same maximum value and since we cannot have two correct answers in PS questions then none of them is correct.
Evaluate each option by plugging min and max possible values of x:
A. 3x – 1 --> max for x=-2 --> |3*(-2)-1|=7.
Attachment:
MSP40111b1fh8bi1bf48ff500001bihb6c5hie997cf.gif
B. x^2 + 1 --> max for x=-2 or x=2 --> |2^2 + 1|=5.
Attachment:
MSP566209hcf53gca1831500002g3b78339ag7a6gh.gif
E. x^2 – x --> max for x=-2 --> |(-2)^2 - (-2)|=6.
Attachment:
MSP5861i671ha12e2ie48c00005gbb4caa2heiag53.gif
Answer: A.
Nice. Makes sense. Thanks.
Would it possible to use basic calculus to solve this? something like \(\frac{d}{dx}(3x-1) = 0\)? This results in max value of 3. Something on these lines.
Re: Inequality - Absolute value [#permalink]
02 Jul 2013, 09:58
Expert's post
emailmkarthik wrote:
Bunuel wrote:
emailmkarthik wrote:
Hey Bunuel,
Do you have any alternate easier method to tackle this problem?
If x is a number such that –2 ≤ x ≤ 2, which of the following has the largest possible absolute value? A. 3x – 1 B. x^2 + 1 C. 3 – x D. x – 3 E. x^2 – x
First of all notice that we can eliminate options C and D right away: \(|3-x|=|x-3|\), so these two options will have the same maximum value and since we cannot have two correct answers in PS questions then none of them is correct.
Evaluate each option by plugging min and max possible values of x:
A. 3x – 1 --> max for x=-2 --> |3*(-2)-1|=7.
Attachment:
MSP40111b1fh8bi1bf48ff500001bihb6c5hie997cf.gif
B. x^2 + 1 --> max for x=-2 or x=2 --> |2^2 + 1|=5.
Attachment:
MSP566209hcf53gca1831500002g3b78339ag7a6gh.gif
E. x^2 – x --> max for x=-2 --> |(-2)^2 - (-2)|=6.
Attachment:
MSP5861i671ha12e2ie48c00005gbb4caa2heiag53.gif
Answer: A.
Nice. Makes sense. Thanks.
Would it possible to use basic calculus to solve this? something like \(\frac{d}{dx}(3x-1) = 0\)? This results in max value of 3. Something on these lines.
Correct me if I'm wrong.
I don't quite understand what you mean here. Also, don't think that it's a good idea to use derivative to solve this question. _________________
Re: Inequality - Absolute value [#permalink]
02 Jul 2013, 10:09
Bunuel wrote:
emailmkarthik wrote:
Bunuel wrote:
Hey Bunuel,
Do you have any alternate easier method to tackle this problem?
If x is a number such that –2 ≤ x ≤ 2, which of the following has the largest possible absolute value? A. 3x – 1 B. x^2 + 1 C. 3 – x D. x – 3 E. x^2 – x
First of all notice that we can eliminate options C and D right away: \(|3-x|=|x-3|\), so these two options will have the same maximum value and since we cannot have two correct answers in PS questions then none of them is correct.
Evaluate each option by plugging min and max possible values of x:
A. 3x – 1 --> max for x=-2 --> |3*(-2)-1|=7.
Attachment:
MSP40111b1fh8bi1bf48ff500001bihb6c5hie997cf.gif
B. x^2 + 1 --> max for x=-2 or x=2 --> |2^2 + 1|=5.
Attachment:
MSP566209hcf53gca1831500002g3b78339ag7a6gh.gif
E. x^2 – x --> max for x=-2 --> |(-2)^2 - (-2)|=6.
Attachment:
MSP5861i671ha12e2ie48c00005gbb4caa2heiag53.gif
Answer: A.
Nice. Makes sense. Thanks.
Would it possible to use basic calculus to solve this? something like \(\frac{d}{dx}(3x-1) = 0\)? This results in max value of 3. Something on these lines.
Correct me if I'm wrong.
I don't quite understand what you mean here. Also, don't think that it's a good idea to use derivative to solve this question.
What I meant was: Generally to find the maxima/minima of an equation, we take the derivative of the equation and equate it to 0 and then solve the equation. This results in the max/min value. I was wondering if it is possible to do something on these lines here( taking derivative of each of the options, equate it to 0, solve to get max and min value)
But yes, even if this method results in the right answer, it might be cumbersome.
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