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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote Function This topic has 1 expert reply and 3 member replies ?? Attachments This post contains an attachment. You must be logged in to download/view this file. Please login or register as a user. Legendary Member Joined 15 Apr 2011 Posted: 1085 messages Followed by: 21 members Upvotes: 158 if we take the slope as tangent(x) then we notice that 60Sqrt(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. Attachments This post contains an attachment. You must be logged in to download/view this file. Please login or register as a user. _________________ Success doesn't come overnight! Master | Next Rank: 500 Posts Joined 08 Feb 2011 Posted: 436 messages Followed by: 6 members Upvotes: 72 N:Dure wrote: ?? f(x)=|2x|+4; now we have to find the equation of function g, whose graph will intersect the function f(x); we must know that when two curves intersect each other, then their point of intersection lies on both curves, lets utilize this principle while solving this question..!! lets work with option A) g(x)=x-2; to find the point of intersection of this function with function, let equation both the equations; f(x)=g(x); |2x|+4=x-2; when x>0; we have; 2x+4=x-2; x=-6; which is not possible as x>0; when x<0; we have; -2x+4=x-2; 3x=6; x=2; which again is not possible as x<0; hence curve g(x)=x-2; will not intersect the curve f(x)=|2x|+4 we can similarly with all the remaining options, lets consider option E)g(x)=3x-2; |2x|+4=3x-2; when x>0; we have; 2x+4=3x-2; x=6; which is possible as x>0; when x<0; we have; -2x+4=3x-2; 5x=6; x=6/5 which is not possible as x<0; hence g(x)=3x-2; will intersect the curve at only point x=6; hence answer should be option E _________________ O Excellence... my search for you is on... you can be far.. but not beyond my reach! GMAT/MBA Expert GMAT Instructor Joined 08 Jan 2008 Posted: 3225 messages Followed by: 611 members Upvotes: 1710 GMAT Score: 800 What's the source of this question? An old GMAT paper test? Or is it not even from the GMAT? For those of you reading this question and incredibly confused, I wouldn't sweat it - it's unlikely you'll see anything similar on test day. You certainly aren't expected to know any trig for the GMAT, so the first solution presented, while elegant, isn't one you'd ever use on test day. Realistically, the only way to solve this problem is via a combination of picking numbers and logic/common sense. We can see from the graph that f(x) gets bigger much more quickly than does x. So, we're going to need a function with a fairly large multiplier of the x term in order to intersect the f(x) function. I'm immediately drawn to (E), the only choice with "3x" in it. Let's compare the two functions: f(x) = |2x| + 4 and g(x) = 3x - 2 Ignoring negative values of x (since that will make g(x) negative and f(x) has a minimum value of +4), let's do some plug 'n play. If x=1, then f(x)= 6 and g(x)= 1... way too small. If x=5, then f(x) = 14 and g(x) = 13... close, still too small - but we got closer! Without even plugging in more numbers, I'm now convinced that (E) will work. If I really felt the need to keep going, then: If x=6, then f(x)= 16 and g(x) = 16... ding ding ding... choose (E)! If you saw the trends (f(x) has a min value of 4 and gets bigger as x either increases or decreases from 0; A through D don't increase as quickly as f(x)), then the fact that (E) is the only answer with a 3x multiplier should have been enough to pick it without actually plugging in any numbers. _________________ Stuart Kovinsky | Kaplan GMAT Faculty | Toronto Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount Free GMAT Practice Test under Proctored Conditions! - Find a practice test near you or live and online in Kaplan's Classroom Anywhere environment. Register today! Master | Next Rank: 500 Posts Joined 22 Mar 2011 Posted: 135 messages Followed by: 3 members Upvotes: 21 Test Date: 25th June 2011 GMAT Score: 720 To me, this appears to be a slope question. for g(x) to intersect f(x), the slope of g(x) has to be greater than f(x). In addition, the question provides a big help by stating that slope is equally steep on both quadrant 1 & 2 Now if the slope of g(x) is greater than f(x) than it will have to intersect with f(x) irrespective of the y intercept of g(x) i.e if y intercept of g(x) is +ve the line will intersect with f(x) in the second quadrant and if y intercept of g(x) is -ve the line will intersect with f(x) in the first quadrant. With this in mind, we simply have to scan the choices and find the equation with greatest slope i.e choice E • Free Practice Test & Review How would you score if you took the GMAT Available with Beat the GMAT members only code • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code • Magoosh Study with Magoosh GMAT prep Available with Beat the GMAT members only code • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code • FREE GMAT Exam Know how you'd score today for$0

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