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.

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

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.

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