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

Question

https://gmatclub.com/forum/download/file.php?id=14499 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 pic.JPG

KarishmaB · Accepted Answer

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. Lines and Slopes.jpg 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.

bagrettin · Answer

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.

zaur2010 · Answer

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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bgpower · Answer

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!

Subanta · Answer

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.

sahilvijay · Answer

Clearly the option E

we can see that g(x) = 3x-2 is the line which will intersect the given y=|2x|+4 at point (6,16)
and refer to the figure .

Kinshook · Answer

Only E. g(x) = 3x - 2 has greater slope = 3 and will intersect the graph of a function f, defined by f(x) = |2x| + 4
Since all other functions has less y-intersect and less than or equal to slope = 2 and will not intersect the graph of a function f, defined by f(x) = |2x| + 4

IMO E

Kinshook · Answer

Given: The figure above shows the graph of a function f, defined by f(x) = |2x| + 4 for all numbers x. Asked: For which of the following functions g defined for all numbers x does the graph of g intersect the graph of f ? For intersection at all values of x, f(x) = g(x) is valid for all x A. g(x) = x - 2 f(x) = |2x| + 4 = g(x) = x-2 |2x| = x - 6 If x>0 ; 2x = x - 6; x = - 6; Not valid since x>0 If x<0; -2x = x- 6; 3x = 6; x = 2; Not valid since x<0 B. g(x) = x + 3 f(x) = |2x| + 4 = g(x) = x + 3 |2x| = x -1 If x>0; 2x = x-1; x = -1; Not valid since x>0 If x<0; -2x = x - 1; 3x = 1; x = 1/3; Not valid since x<0 C. g(x) = 2x - 2 f(x) = |2x| + 4 = g(x) = 2x-2 |2x| = 2x - 6 If x>0; 2x = 2x - 6; 0 = -6; Not valid If x<0; -2x = 2x - 6; 4x = 6; x = 1.5; Not valid since x<0 D. g(x) = 2x + 3 f(x) = |2x| + 4 = g(x) = 2x + 3 |2x| = 2x -1 If x>0; 2x = 2x - 1; 0 = -1; Not valid If x<0; -2x = 2x - 1; 4x = 1; x = 1/4; Not valid since x<0 E. g(x) = 3x - 2 f(x) = |2x| + 4 = g(x) = 3x - 2 |2x| = 3x -6 If x>0; 2x = 3x - 6; x = 6; Valid solution since x>0 If x<0; -2x = 3x -6 ; 5x = 6; x = 6/5 ; Not valid since x<0 Only E. g(x) = 3x -2 has a valid solution x = 6 IMO E

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

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