For which of the following functions is g(c-d) = g(c) - g(d)

In such questions these are the steps I find useful!!

1) The first step is to check whether you could solve the expression stated in the question( in case the expression is too long) or substitute the expression into the solutions in some manner so as to solve it quickly. If so, this is the quickest way out and usually with a little bit of practice, you can start judging whether this step would work or not before you place your pen on the paper to solve the question.

2) The next method is to find whether the solutions can be substituted into the original expression in some way that is fast and quick. This usually takes some time and hence, mostly does not work. However, if the answer options involve numbers instead of expressions, this might work better.

This method is only for those questions in which the answer options and the questions both contain expressions.

3) You could try substitution of base values to confirm.

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Why is option C Wrong
\sqrt{3(c-d)} = \sqrt{(3c)}-\sqrt{(3d)}

Option C says g(x) = \(\sqrt{3x}\)

So, g(c-d) = \(\sqrt{c-d}\)
g(c) = \(\sqrt{c}\)
g(d) = \(\sqrt{d}\)

We cannot say that \(\sqrt{c-d}\) = \(\sqrt{c}\) - \(\sqrt{d}\) for all values of c and d
So we cannot say that g(c-d) = g(c) - g(d) for all values of c and d.

Thus, option c is wrong.

put c = 3 and d = 2 or directly substitute c and d

we get option D,

A g(x) = x^3

Power of "x" is changed, so IGNORE

B g(x) = x+5

"x" is added by constant, so IGNORE

C g(x) = \sqrt{3x}

Power of "x" is changed, so IGNORE

D g(x) = 5x

"x" is multiplied by constant 5, so required result is true

E g(x) 15/x

\(\frac{15}{x} = 15 * x^{-1}\)

Again power of "x" is changed, so IGNORE

Answer = D

Plug in values as c=4 and d=2
which will bring down the funtion in problem statement to g(2)=g(4)-g(2).
Now substituting the values in the option available as per the above equation LHS=RHS is option D.

For which of the following functions is g(c-d) = g(c) - g(d) for all positive numbers c and d?

A g(x) = x^3
B g(x) = x+5
C g(x) = 3x−−√3x
D g(x) = 5x
E g(x) 15/x


well its jus asking if they are the same or not, for that just substitute the value and we can find out.
Option D is the corect answer.

We can let c = 2 and d = 1, and thus we are solving for both sides being equal when:

g(1) = g(2) - g(1)

Starting with answer choice E, we have 15/x:

g(1) = 15/1 = 15

g(2) = 15/2 = 7.5

g(2) - F(1) = 7.5 - 15 = -7.5

Since 15 does not equal -7.5, E is not correct.

Moving to D, we have 5x:

g(1) = 5 x 1 = 5

g(2) = 5 x 2 = 10

g(2) - g(1) = 10 - 5 = 5

Since 5 does equal 10 - 5, D might be correct.

For answer choice C, we have √3x:

g(1) = √(3*1) = √3

g(2) = √(3*2) = √6

Since √3 does not equal √6 - √3 = √3(√2 - 1), C is not correct.

For answer choice B, we have x + 5:

g(1) = 1 + 5 = 6

g(2) = 2 + 5 = 7

Since 6 does not equal 7 - 6 = 1, B is not correct.

For answer choice A, we have x^3:

g(1) = 1

g(2) = 8

Since 1 does not equal 8 - 1 = 7, A is not correct.

Since the only answer choice that is definitely not incorrect is D, it must be the correct answer.

Answer: D

ScottTargetTestPrep EMPOWERgmatRichC

Would you mind explaining your reasoning behind picking the numbers you picked for c and d?

Does one need to be strategic with the number-picking or will any small integers work for such questions?

Thanks in advance!

Hi achloes,

In most situations in which you can TEST VALUES to get to the correct answer, you can use whichever value(s) you like (as long as the values fit whatever 'restrictions' the prompt defines). There are absolutely situations in which choosing specific numbers can make the work easier; in this prompt, we are told that the values of C and D and POSITIVE numbers - so selecting a couple of small positive integers (for example, 2 and 1, respectively) makes working through the five answer choices really easy.

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: Rich.C@empowergmat.com

Just pick numbers that are easy to work with. For example, here, 2 and 1 seemed like they would not make the math too difficult.