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The figure above shows the graph of a function f, defined by f(x) = |2 [#permalink]

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28 Mar 2011, 17:17

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A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

59% (01:44) correct 41% (01:55) wrong based on 311 sessions

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The figure above shows the graph of a function f, defined by f(x) = |2x| + 4 for all numbers x. For which of the following functions g defined for all numbers x does the graph of g intersect the graph of f ?

A. g(x) = x - 2 B. g(x) = x + 3 C. g(x) = 2x - 2 D. g(x) = 2x + 3 E. g(x) = 3x - 2

Re: The figure above shows the graph of a function f, defined by f(x) = |2 [#permalink]

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28 Mar 2011, 18:38

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Obviously, this is 3x-2 because it has the slope of 3, so it increases faster than f(x) with its slope of 2. More formally, the right part of f(x) is given by the equation 2x+4. If graphs of 2 functions have the point of intersection then the following equation should have a solution with x>=0: 2x+4=3x-2 x=6 y=2*6+4=16

So, these lines reaaly intersect each other. The answer is E.
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Re: The figure above shows the graph of a function f, defined by f(x) = |2 [#permalink]

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28 Mar 2011, 19:41

I think, I found the solution - ...

if we take the slope as tangent(x) then we notice that 60<x<90, as tan(x)=2 and tan(x)>Sqrt(3)

Now if we take answer choice (E) we get tan(x`)>2 --> angle x`>angle x

the graph g(x)=3x-2 has to cross the graph f at any given time, because of the angle property (x`) and disposition of adjacent (rise in x-coordinate) and opposite (rise in y-coordinate) sides.

all other answer choices could be parallel or placed under the graph {fixed y-coordinate in f(x)} too, they depend on the value of x. We can check simply by plugging in x=0 and x=1 into answer choices A-D.

thanks

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Last edited by zaur2010 on 28 Mar 2011, 19:51, edited 1 time in total.

Re: The figure above shows the graph of a function f, defined by f(x) = |2 [#permalink]

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28 Mar 2011, 19:45

I didn't get the right answer. But I think this what bagrettin interpretation. The x axis has slope of zero, it will never intersect f(x). If you increase the slope of g(x) it might. Howmuch depends on the slope of f(x) -it has to be greater the slope of f(x) otherwise the g(x) being slant will miss the aim.

Re: The figure above shows the graph of a function f, defined by f(x) = |2 [#permalink]

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28 Mar 2011, 20:00

x can have a slope of 0 and y-intercept >=6 what will happen then? the lines should cross ... may be it's better to operate with the slope(x`)>slope(x) for two functions f(x) and g(x`)

gmat1220 wrote:

I didn't get the right answer. But I think this what bagrettin interpretation. The x axis has slope of zero, it will never intersect f(x). If you increase the slope of g(x) it might. Howmuch depends on the slope of f(x) -it has to be greater the slope of f(x) otherwise the g(x) being slant will miss the aim.

Re: The figure above shows the graph of a function f, defined by f(x) = |2 [#permalink]

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30 Dec 2014, 10:33

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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The figure above shows the graph of a function f, defined by f(x) = |2 [#permalink]

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29 Jul 2015, 00:02

Hi guys,

I am actually quite surprised by my capability to even come close to solving this 75% difficulty questions. The only thing I don't understand here is why we're looking for a solution in the 1st quadrant. For example, bagrettin refers to this in his solution.

Answer choices (C) and (D) offer solutions of \(-\frac{1}{2}\) and \(-\frac{7}{4}\) in the x<0 or 2nd quadrant area. Why do we not except these answers here?

Thank you!
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Thank you very much for reading this post till the end! Kudos?

I am actually quite surprised by my capability to even come close to solving this 75% difficulty questions. The only thing I don't understand here is why we're looking for a solution in the 1st quadrant. For example, bagrettin refers to this in his solution.

Answer choices (C) and (D) offer solutions of \(-\frac{1}{2}\) and \(-\frac{7}{4}\) in the x<0 or 2nd quadrant area. Why do we not except these answers here?

Thank you!

We are looking for intersection in the first quadrant because all lines in the given five options have positive slope. A line with positive slope definitely passes through the first and third quadrant. So we are ignoring second quadrant for the time being.

If you understand how lines are drawn on xy axis, you can do this question in 10 seconds.

The graph of f(x) is shown. The line in first quadrant has slope 2.

Let's plot all lines on the xy axis.

Attachment:

Lines and Slopes.jpg [ 1.14 MiB | Viewed 8582 times ]

We know y = x is a line making 45 degree angle with y axis and passing through the centre. Point at it in the given figure. y = x - 2 will be two steps below it passing through (0, -2) y = x + 3 will be two steps above it passing through (0, 3) They both have slope 1 which is less than the slope of f(x). They will not intersect f(x).

Now hold on to y = 2x, a line passing through (0, 0) but with a steeper slope. The slope is the same as that of f(x) so it is parallel to f(x). Both y = 2x - 2 and y = 2x + 3 (blue lines) will have slopes parallel to f(x) and will not intersect it.

The only remaining line is 3x - 2 (green line) which is steeper than f(x) and will intersect it.
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Re: The figure above shows the graph of a function f, defined by f(x) = |2 [#permalink]

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30 Jul 2015, 02:33

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Start by checking the slopes and intercepts of all the lines. On comparison with the reference line provided, we see that the first four lines will never intersect with the given line. Only the last line will intersect because it has a steeper slope.
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Re: The figure above shows the graph of a function f, defined by f(x) = |2 [#permalink]

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01 Aug 2015, 10:09

if two lines intersect we know that we can equate the two linear equation in this 2x+4=g(x). I'm taking only positive values of x since we are looking for a intersection in the first quad. you can easily eliminate options C,D . As for the rest a little calculation shows a,b are out . 2x+4=x-2--> x=-6 which deosnt statisfy the equation. so is B. upon solving E 3x-2=2x+4==>x=6 which satisfies bth equations. @@Bunuuel Can you verfiy this approach and any other concept that can be applied.

The figure above shows the graph of a function f, defined by f(x) = |2 [#permalink]

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01 Aug 2015, 10:26

zaur2010 wrote:

Attachment:

pic.JPG

The figure above shows the graph of a function f, defined by f(x) = |2x| + 4 for all numbers x. For which of the following functions g defined for all numbers x does the graph of g intersect the graph of f ?

A. g(x) = x - 2 B. g(x) = x + 3 C. g(x) = 2x - 2 D. g(x) = 2x + 3 E. g(x) = 3x - 2

My quick analysis-

Lines A and B have slopes which are less than the slope of f(x) and will always be below the graph.

Lines C and D are parallel to F(x).

We are left with E which has more positive slope and will definitely cut the graph at some point.

I donot mind Kudos _________________

Our greatest weakness lies in giving up. The most certain way to succeed is always to try just one more time.

I hated every minute of training, but I said, 'Don't quit. Suffer now and live the rest of your life as a champion.-Mohammad Ali

Last edited by samichange on 01 Aug 2015, 11:16, edited 1 time in total.

The figure above shows the graph of a function f, defined by f(x) = |2 [#permalink]

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01 Aug 2015, 10:52

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PriyankaKabbina wrote:

if two lines intersect we know that we can equate the two linear equation in this 2x+4=g(x). I'm taking only positive values of x since we are looking for a intersection in the first quad. you can easily eliminate options C,D . As for the rest a little calculation shows a,b are out . 2x+4=x-2--> x=-6 which deosnt statisfy the equation. so is B. upon solving E 3x-2=2x+4==>x=6 which satisfies bth equations. @@Bunuuel Can you verfiy this approach and any other concept that can be applied.

Couple of concepts tested in this question:

1. |x| =x for x \(\geq\) 0 . This is used to find the equation of |2x|+4 in the 1st quadrant.

2. Slope of lines y = mx+c and y = mx + b are same. 2.1: A greater value of the slope indicates a greater angle that the line makes with the positive direction of the x-axis.

Point 1 is used to find that the equation of the line in the first quadrant will be , y =2x+4 and thus its slope will be 2.

Points 2 and 2.1 are used to eliminate options A and B. Slope of x \(\pm\) a will be 1 and will thus be diverging away from line y =2x+4 (as slope of 1 < slope of 2, in the 1st quadrant).

Point 2 shows that lines in options C and D will be parallel to y =2x+4 ( as the slopes of [y = 2x \(\pm\) a] = slope of [y =2x+4]). 2 parallel lines can never intersect and thus these 2 options are eliminated.

Thus E is the only remaining answer choice.

Your method is fine for this question but the method mentioned above by me can be applied to any question related to lines in xy coordinate plane.

Re: The figure above shows the graph of a function f, defined by f(x) = |2 [#permalink]

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01 Aug 2015, 10:53

samichange wrote:

zaur2010 wrote:

Attachment:

pic.JPG

The figure above shows the graph of a function f, defined by f(x) = |2x| + 4 for all numbers x. For which of the following functions g defined for all numbers x does the graph of g intersect the graph of f ?

A. g(x) = x - 2 B. g(x) = x + 3 C. g(x) = 2x - 2 D. g(x) = 2x + 3 E. g(x) = 3x - 2

My quick analysis-

Lines A and B have slopes which are less than the slope of f(x) and will always be below the graph.

Lines C and D are parallel to F(x).

We are left with D which has more positive slope and will definitely cut the graph at some point.

Re: The figure above shows the graph of a function f, defined by f(x) = |2 [#permalink]

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01 Aug 2015, 11:15

Engr2012 wrote:

samichange wrote:

zaur2010 wrote:

Attachment:

pic.JPG

The figure above shows the graph of a function f, defined by f(x) = |2x| + 4 for all numbers x. For which of the following functions g defined for all numbers x does the graph of g intersect the graph of f ?

A. g(x) = x - 2 B. g(x) = x + 3 C. g(x) = 2x - 2 D. g(x) = 2x + 3 E. g(x) = 3x - 2

My quick analysis-

Lines A and B have slopes which are less than the slope of f(x) and will always be below the graph.

Lines C and D are parallel to F(x).

We are left with D which has more positive slope and will definitely cut the graph at some point.

I donot mind Kudos

I think you wanted to mention E instead of D.

Thanks for pointing out
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Re: The figure above shows the graph of a function f, defined by f(x) = |2 [#permalink]

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15 Sep 2015, 23:58

The figure above shows the graph of a function f, defined by f(x) = |2x| + 4 for all numbers x. For which of the following functions g defined for all numbers x does the graph of g intersect the graph of f ?

Understanding slope would be helpful in this question .

Standard slope eq : y=mx+b

as the slope of f(x) is 2, for g(x) to intersect, the slope of g(x) needs to be bigger than 2. (If slope is equal to 2, then it would be parallel)

Only E has the slope bigger than 2.

A. g(x) = x - 2 --> Wrong, as slope is less than 2 B. g(x) = x + 3 --> Wrong, as slope is less than 2 C. g(x) = 2x - 2 --> Wrong, as slope is not great than 2, which makes it parallel of f(x) D. g(x) = 2x + 3 --> Wrong, as slope is not great than 2, which makes it parallel of f(x) E. g(x) = 3x - 2 --> CORRECT, as slope is bigger than 2.

Re: The figure above shows the graph of a function f, defined by f(x) = |2 [#permalink]

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18 Oct 2016, 19:41

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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