For which of the following functions is f(a+b) = F(a) + f(b)

[Guest]

For which of the following functions is f(a+b) = F(a) + f(b) for all positive numbers a and b?

f(x) = X^2

f(x) = X+1

f(x) = square root of x

F(x) = 2/x

F(x) = -3x*

[rohit801]

we want f(a+b) = F(a) + f(b). let's look at each option:
1. f(x) = X^2 : F(a) = a^2; f(b)= b^2 and f(a+b) = (a+b)^2; clearly (a+b)^2 will not equal a^2 + b^2.

2. f(a) = a+1; f(b) = b+1 so, f(a+b) = a+b+1 which does not equal Fa + Fb, which is a+b +2.

3 f(a) = square root of a; f(b)= square root of b and f(a+b)=SQRT [a+b]. here also, the equality doesn't hold [ u get it, right?]

4. F(a+b) = 2/a; F(b) = 2/b. so, F(a+b) = 2/(a+ b). Now, does 2/a + 2/b = 2/(a+b) - Nope!

5. F(a) = -3a; F(b)= -3b; so, F(a+b) = -3(a+b). Now, -3a-3b= -3 (a+b) = F(a+b) - GOOD.

let me know whether there is confusion.

thanks

[RonPurewal]

here's probably the easiest way to handle problems like this:

plug in your own numbers.

this problem is formulated in such a way that it's supposed to work for all choices of a and b. that means that, if you pick particular values of a and b, it must work for those values.

so:
let
a = 2
b = 3
therefore a + b = 5
(nb: there's nothing special about these values; they're random choices)

(a)
f(a + b) = f(5) = 5^2 = 25
f(a) + f(b) = f(2) + f(3) = 2^2 + 3^2 = 13
nope

(b)
f(a + b) = f(5) = 5 + 1 = 6
f(a) + f(b) = f(2) + f(3) = (2 + 1) + (3 + 1) = 7
nope

(c)
f(a + b) = f(5) = rad(5)
f(a) + f(b) = f(2) + f(3) = rad(2) + rad(3) - NOTE: can't simplify this!
nope

(d)
f(a + b) = f(5) = 2/5
f(a) + f(b) = f(2) + f(3) = 2/2 + 2/3 = 5/3
nope

(e)
f(a + b) = f(5) = -3(5) = -15
f(a) + f(b) = f(2) + f(3) = -3(2) + -3(3) = -15
works

answer = e

--

note that, for some unfortunate choices of numbers (such as picking 0 for one of the numbers), some of the other choices will appear to 'work' as well. if this is the case, simply pick new numbers and start again, but only with the choices that worked the first time. (remember, 0 and 1 are generally bad choices for picking numbers)

[Guest]

is there a shortcut to do this problem? I can solve it by plugging in number....but I'm timed myself and it took me exactly two and 1/2 minutes to write the solution NOT including reading the problem and trying to comprehend it. if i confront a problem like this on the real gmat, should i spend extra seconds on this? stacy said not to waste more than 2 min on solving a problem...

[RonPurewal]

is there a shortcut to do this problem? I can solve it by plugging in number....but I'm timed myself and it took me exactly two and 1/2 minutes to write the solution NOT including reading the problem and trying to comprehend it. if i confront a problem like this on the real gmat, should i spend extra seconds on this? stacy said not to waste more than 2 min on solving a problem...

2.5 minutes is okay, as long as you don't make a habit of it. remember that there are two parts to time management:

1) micromanagement, which roughly means limiting yourself to two minutes ON AVERAGE per problem.
remember what "average" means!! it doesn't mean that you have to cut the rope at exactly two minutes; there are certain problem types (most notably word translations and geometry) that routinely demand 3+ minutes even from very able students. conversely, there are also problem types (most notably fdp's and short algebra problems) that should take substantially less than two minutes per problem, often less than 1 minute. the idea is to get this to balance out.

2) macro-management, which means checking in on the timing guidelines periodically (= DO NOT stare at the timer in the corner of the screen after every single problem). this is where you figure out EVERY SO OFTEN whether you're in a hole, and whether you need to throw away questions.
hide the timer when you are not looking at it. look at it once every 5-10 problems.

(1) and (2) complement each other. DO NOT PRACTICE ONE OF THESE TWO FORMS OF TIME MANAGEMENT TO THE EXCLUSION OF THE OTHER ONE. in other words:
1) don't concentrate so much on spending exactly two minutes per problem that you wind up cutting off lots of productive solutions just because you're obsessed with the clock. this applies here; if you know you are solidly and certainly on the way to a solution, then it's worthwhile to spend the extra minute or whatever.
2) at the same time, don't completely neglect your per-problem timing responsibility. you don't want to wait until the next big "checkpoint" only to discover that you're behind by 5 questions and you have no idea what happened.

[ronaldramlan]

is there a shortcut to do this problem? I can solve it by plugging in number....but I'm timed myself and it took me exactly two and 1/2 minutes to write the solution NOT including reading the problem and trying to comprehend it. if i confront a problem like this on the real gmat, should i spend extra seconds on this? stacy said not to waste more than 2 min on solving a problem...

Yes, there is ... if you know the conditions for a function to be perfectly linear. There are 2 conditions for a function to be perfectly linear :
1. f(x+y) must equal f(x)+f(y)
2. f(kx) must equal kf(x)

Now, I believe that at a quick glance you can tell that choice A, C and D are not linear at all. So, ignore them. Thus, you're left with choice B and E to test and you should be able to quickly identify that for f(x) = x + 1, f(x+y) will not equal f(x) + f(y).
This is interesting because although we understand that f(x) = x + c (c = constant) is a linear function, yet it's not a perfect linear, at least according to the conditions mentioned above.

[mschwrtz]

If I were considering ronaldramlan's approach under test conditions, my first concern would be that the some nonlinear functions might meet at least one of the conditions for perfect linearity, and so that the correct answer might in fact not be linear. With no clock running, I'd be happy to check this out (turns out only linear functions can satisfy that first condition) but under test conditions I know that I at least would be much more secure trying values.

If you want to treat the conditions for perfect linearity as suggesting likely answers to test first, that'll actually work quite well.

(If you've already thought very deeply about functions independent of your prep for the GMAT, and happen to know with certainty that fact that I had to take some time to think about, that's great too, but I don't think that many students will fall under that description.)

Also, I'm sure Stacey will correct me if I'm wrong, but I'd be very surprised if she said quite that you should never take more than 2:00 on a question. In my experience, students who have trouble managing time don't take a little over 2:00 on each question, but rather well over 2:00 on several. I wouldn't worry about an occasional 2:30. Too put it differently, ditto Ron's remarks.

[johnhillescobar]

Regarding time management:

I really agree with RonPurewal and mschwrtz since you don't have to be worried about spending more than 2 minutes on certain questions occasionally. In my opinion, the problem is to be obsessed with time while taking the actual test.

I think that it is really dangerous checking out the clock regularly during the test; on the contrary, you must check the clock a couple of times, no more than that. However, prior training in time management is crucial to balance time when you are taking the test.

My recommendation is to create "spare time" for the test. It means to work thoroughly on time issues depending on the question level during your preparation.

I divided the official guide into three levels (easy, medium and hard) in order to determine my average time in each level so that I could set feasible time improvement goals. As a result of applying this strategy, I discovered that my average time on easy questions was 1:02 minutes (primarily my questions went from 14 to 47 seconds), on medium level questions was 1:31 minutes (mainly my questions timed from 14 to 47 seconds) and on hard questions 1:57 minutes (sometimes I spent 3 minutes per question). Once you get your averages, I study each problem in which I spent over 2 minutes in order to find out a better way to solve it. Remember, these are average time so you don't have to spend exactly 2 minutes per question.

This strategy allowed me to create "spare time" to challenging questions by taking time from easier questions. Also, the strategy allowed me to finish my practice test 8 minutes before.

Hope this helps,

[StaceyKoprince]

Just to echo - on quant problems, we need to average 2m. Sometimes we'll spend more than that, sometimes less.

As a general rule, it's not a good idea to spend more than 30sec longer than the desired average. In other words, for quant questions (on which we want to avg 2min), it's generally not a good idea to spend more than 2m30s. I tell my students that I'll let them get away with a couple of questions in the 2m30s to 3m range, but nothing above 3m ever. Plus, they have to justify to me with the hard data that, when they do choose to get into the 2m30s to 3m range, they've made a wise choice.

A "wise choice" means they get such questions right 60+ percent of the time AND they don't have to rush and get some other question wrong as a result.

[agha79]

we want f(a+b) = F(a) + f(b). let's look at each option:
1. f(x) = X^2 : F(a) = a^2; f(b)= b^2 and f(a+b) = (a+b)^2; clearly (a+b)^2 will not equal a^2 + b^2.

2. f(a) = a+1; f(b) = b+1 so, f(a+b) = a+b+1 which does not equal Fa + Fb, which is a+b +2.

3 f(a) = square root of a; f(b)= square root of b and f(a+b)=SQRT [a+b]. here also, the equality doesn't hold [ u get it, right?]

4. F(a+b) = 2/a; F(b) = 2/b. so, F(a+b) = 2/(a+ b). Now, does 2/a + 2/b = 2/(a+b) - Nope!

5. F(a) = -3a; F(b)= -3b; so, F(a+b) = -3(a+b). Now, -3a-3b= -3 (a+b) = F(a+b) - GOOD.

let me know whether there is confusion.

thanks

hey Ron,
in your explanation to this question, you told that plugging in values of a and b would be most probably the easiest way to solve such question. so is the above method flawed by any means? and can it be applied to similar questions of function?

[jnelson0612]

we want f(a+b) = F(a) + f(b). let's look at each option:
1. f(x) = X^2 : F(a) = a^2; f(b)= b^2 and f(a+b) = (a+b)^2; clearly (a+b)^2 will not equal a^2 + b^2.

2. f(a) = a+1; f(b) = b+1 so, f(a+b) = a+b+1 which does not equal Fa + Fb, which is a+b +2.

3 f(a) = square root of a; f(b)= square root of b and f(a+b)=SQRT [a+b]. here also, the equality doesn't hold [ u get it, right?]

4. F(a+b) = 2/a; F(b) = 2/b. so, F(a+b) = 2/(a+ b). Now, does 2/a + 2/b = 2/(a+b) - Nope!

5. F(a) = -3a; F(b)= -3b; so, F(a+b) = -3(a+b). Now, -3a-3b= -3 (a+b) = F(a+b) - GOOD.

let me know whether there is confusion.

thanks

hey Ron,
in your explanation to this question, you told that plugging in values of a and b would be most probably the easiest way to solve such question. so is the above method flawed by any means? and can it be applied to similar questions of function?

I'm going to weigh in here. Every method has possible drawbacks, but plugging in real numbers is one of the most useful methods you should have at hand when you take the GMAT. It is invaluable over and over and works on a variety of problems. If you have a problem similar to this one plugging in numbers should work just fine (and be a very efficient method).

[rhasija19]

Hi Ron,

I was going through the last few sentences of your post(copied below). I see that you are suggesting 0/1 will get some options to appear correct. However, i have a doubt that since it is given that a & b are positive numbers, why would you choose 0 in the first place as a test number. Since it is neither positive nor negative,

Thanks
-Ravik

here's probably the easiest way to handle problems like this:

plug in your own numbers.

this problem is formulated in such a way that it's supposed to work for all choices of a and b. that means that, if you pick particular values of a and b, it must work for those values.

so:
let
a = 2
b = 3
therefore a + b = 5
(nb: there's nothing special about these values; they're random choices)

(a)
f(a + b) = f(5) = 5^2 = 25
f(a) + f(b) = f(2) + f(3) = 2^2 + 3^2 = 13
nope

(b)
f(a + b) = f(5) = 5 + 1 = 6
f(a) + f(b) = f(2) + f(3) = (2 + 1) + (3 + 1) = 7
nope

(c)
f(a + b) = f(5) = rad(5)
f(a) + f(b) = f(2) + f(3) = rad(2) + rad(3) - NOTE: can't simplify this!
nope

(d)
f(a + b) = f(5) = 2/5
f(a) + f(b) = f(2) + f(3) = 2/2 + 2/3 = 5/3
nope

(e)
f(a + b) = f(5) = -3(5) = -15
f(a) + f(b) = f(2) + f(3) = -3(2) + -3(3) = -15
works

answer = e

--

note that, for some unfortunate choices of numbers (such as picking 0 for one of the numbers), some of the other choices will appear to 'work' as well. if this is the case, simply pick new numbers and start again, but only with the choices that worked the first time. (remember, 0 and 1 are generally bad choices for picking numbers)

[RonPurewal]

• the advice about 0 and 1 is general advice (= not just for this specific problem)

• go back and read that part again—i'm advising people NOT to pick those kinds of numbers!