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Re: If g(x) = 2^x + x, how many solutions satisfy the equation g(x) = 2?
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07 Feb 2016, 19:20
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TeamGMATIFY wrote:
If g(x) = \(2^x\) + x, how many solutions satisfy the equation g(x) = 2?
A. Zero B. One C. Two D. Three E. Infinite
The number of solutions of 2^x+x = 2 will be equal to number of points of intersection of y=2 and y=2^x+x.
Realize that 2^x+x is exponential in nature and as such once it passes the x=-0.641, the value of y=2^x+x will go on increasing on an exponential basis (note that it is NOT necessary to know this value of 'x' but you do need to know the general nature of an exponential function). Thus, there will only be 1 point of intersection.
B is thus the correct answer.
Refer below for the graph.
Attachment:
2016-02-07_21-19-51.jpg [ 31.64 KiB | Viewed 3169 times ]
Re: If g(x) = 2^x + x, how many solutions satisfy the equation g(x) = 2?
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Updated on: 07 Feb 2016, 20:49
Usually when I see problems like this (graph, number of times a value is hit, exponential equation, etc.), I try to find a key inflection x value. For any a^b question, the inflection x will often be 0. Be careful with this rule of thumb, of course. Don't trust it blindly. Once we've defined our inflection point (x = 0 for this question), we break our equation into three parts: x = 0, x < 0, x > 0. We want to know what the function is doing in each of these ranges. For this specific question, we want to know if each part is upward-trending or downward-trending.
x = 0: 2^0 + 0 = 1. g(1) = 1. Next question: are both x < 0 and x > 0 upward trending? Downward trending? Etc.
x < 0: \(2^x + x, x < 0\) For simplicity, let's flip the signs on everything so we're not dealing with awkward negatives to throw us off: \(\frac{1}{{2^x}} - x, x > 0\). This means that, returning to x < 0, the lesser x is, the more negative y is. The only unsurety is for -1 < x < 0. If you know calculus and limits, \(\frac{1}{{2^x}} + x\) as x approaches 0 will become 1/1 + 0, which is 1. We conclude that the trend is strictly downwards.
x > 0: \(2^x + x, x > 0\) \(2^x\) is a hyperbola when x > 0. We know that it will pass through all positive x points and, for x > 0, all y points above 1. Our trend is strictly upwards.
Since g(1) = 1, g(x<0) < 1, g(x>0) > 1, x>0 is the upward trending half of a hyperbola, and x>0 will cover the case of g(1) = 1, we know that there will be only one point where g(x) crosses 2.
Edit: Added a warning about x = 0 as an inflection point.
Originally posted by Beixi88 on 07 Feb 2016, 20:33.
Last edited by Beixi88 on 07 Feb 2016, 20:49, edited 2 times in total.
Re: If g(x) = 2^x + x, how many solutions satisfy the equation g(x) = 2?
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07 Feb 2016, 20:42
Beixi88 wrote:
Usually when I see problems like this (graph, number of times a value is hit, exponential equation, etc.), I try to find a key inflection x value. For any a^b question, the inflection x will almost always be 0. From here, we break our equation into three parts: x = 0, x < 0, x > 0. We want to know what the function is doing in each of these ranges. For this specific question, we want to know if each part is upward-trending or downward-trending.
x > 0: \(2^x + x, x > 0\) \(2^x\) is a parabola when x > 0. We know that it will pass through all positive x points and, for x > 0, all y points above 1. Our trend is strictly upwards.
IMO, statement above in red is a sweeping statement that is not really true. "Almost always" is a bit misleading.
Secondly, the graph of y=2^x is NOT a parabola but one half of a hyperbola with x axis as the asymptotic axis after \(x \approx -11\)
Re: If g(x) = 2^x + x, how many solutions satisfy the equation g(x) = 2?
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07 Feb 2016, 20:46
Engr2012 wrote:
Beixi88 wrote:
Usually when I see problems like this (graph, number of times a value is hit, exponential equation, etc.), I try to find a key inflection x value. For any a^b question, the inflection x will almost always be 0. From here, we break our equation into three parts: x = 0, x < 0, x > 0. We want to know what the function is doing in each of these ranges. For this specific question, we want to know if each part is upward-trending or downward-trending.
x > 0: \(2^x + x, x > 0\) \(2^x\) is a parabola when x > 0. We know that it will pass through all positive x points and, for x > 0, all y points above 1. Our trend is strictly upwards.
IMO, statement above in red is a sweeping statement that is not really true. "Almost always" is a bit misleading.
Secondly, the graph of y=2^x is NOT a parabola but one half of a hyperbola with x axis as the asymptotic axis after \(x \approx -11\)
Thanks for pointing out those errors. I'll go ahead and amend my post.
Re: If g(x) = 2^x + x, how many solutions satisfy the equation g(x) = 2?
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17 May 2017, 09:38
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TeamGMATIFY wrote:
If g(x) = \(2^x\) + x, how many solutions satisfy the equation g(x) = 2?
A. Zero B. One C. Two D. Three E. Infinite
This question is too hard and unlikely to appear on the GMAT test.
We could draw the graph as Engr2012 did. However, drawing the graph of \(2^x+x\) is too hard without any tool.
Note that \(2^x+x=2 \implies 2^x=2-x\).
We will draw the graph of \(y=2^x\) and \(y=2-x\)
Graph of \(y=2^x\) If \(x=0 \implies y=2^x = 1\). The line go through point (0, 1) If \(x=1 \implies y=2^x = 2\). The line go through point (1, 2) If \(x=2 \implies y=2^x = 4\). The line go through point (2, 4) If \(x=-1 \implies y=2^x = 1/2\). The line go through point (-1, 1/2)
Hence, the graph of \(y=2^x\) looks like the blue line.
Attachment:
graph.png [ 10.67 KiB | Viewed 2069 times ]
Hence, the equation has only 1 root. The answer is B.
We could solve this question using derivative. However, this tool is out of scope.
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If g(x) = 2^x + x, how many solutions satisfy the equation g(x) = 2?
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11 Jul 2017, 09:58
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I agree with the post above that says this question is out of the scope of the GMAT, since you can't solve the resulting equation using conventional GMAT math, so I don't think people should worry about it much. But I suppose you can answer it without using any graphs.
If you think about the function f(x) = 2^x, that is a function that is 'constantly increasing'. If you plug in bigger and bigger values for x, the value of the function gets bigger and bigger. The same is true for the function h(x) = x, and if we add two functions that are constantly increasing, the result will be too. So the function g(x) = 2^x + x is a function that gets bigger and bigger as you plug in larger and larger values of x. Put algebraically,
if a > b, then g(a) > g(b)
and as a logical consequence, if we plug in two different values for x into g(x), we can never get the same answer. So the equation g(x) = 2 can only have at most one solution for x.
Then we just need to confirm there is indeed one solution. This is where things really get outside the scope of the GMAT. If you notice that g(x) can be less than 2 (if you plug in x = 0, say) or greater than 2 (if you plug in x=1, say), then g(x) has to be equal to 2 for some value of x between 0 and 1, because g(x) is constantly and continuously increasing, and has to pass through the value 2 at some point. Technically I'm using something called the 'Intermediate Value Theorem' here though, which is a theorem from calculus that you don't ever need on the GMAT.
The logic above might not make a lot of sense to people who haven't studied calculus (you use this kind of reasoning in a lot of calc problems, but rarely in other kinds of problems), and as I pointed out above, if it doesn't make a lot of sense, it's not something you need to worry about if you're preparing for the GMAT!
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Re: If g(x) = 2^x + x, how many solutions satisfy the equation g(x) = 2?
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