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Solving functions ques

Expert replies
by greatchap » Fri Nov 19, 2010 11:20 pm
Hi Guys,

Inspite of some search, I am unable to determine how to solve questions with functions ?

For e.g.

If function f(x) satisfies f(x)=f(x^2) for all x, which of the following must be true?

(A) f(4)=f(2) f(2)
(B) f(16)-f(2)=0
(C) f(-2)+f(4)=0
(D) f(3)=3f(3)
(E) f(0)=0

I don't know where to start ?

Please help me out here.

Thanks,

GR
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Source: — Problem Solving |

by shovan85 » Sat Nov 20, 2010 12:09 am
greatchap wrote: If function f(x) satisfies f(x)=f(x^2) for all x, which of the following must be true?

(A) f(4)=f(2) f(2)
(B) f(16)-f(2)=0
(C) f(-2)+f(4)=0
(D) f(3)=3f(3)
(E) f(0)=0
IMO B

f(x)=f(x^2) for all x

A: Incorrect
f(2) = f(2^2) = f(4)
And f(2) * f(2) = [f(2)]^2

Thus f(4) = f (2) and not f(4) = f(2)*f(2)

B: Correct
f(2) = f(2^2) = f(4)
f(4) = f(4^2) = f(16)
Thus f(2) = f(16) => f(16)-f(-2) = 0

C: Incorrect
f(-2) = f([-2]^2) = f(4)
Thus, f(-2) - f(4) = 0 Not f(-2)+f(4)=0

D: Incorrect
f(3) = f(3^2) = f(9)
Thus f(3) = f(9)
But there is no way to say f(9) = 3f(3)

E: Incorrect
f(0) = f(0^2) but we do not know what is f(0) from the Question.
So we cannot say f(0) = 0

As per solving, see the question what is given as the function. Here f(x) = f(x^2) for all x.
Then go to each option take one side either LHS or RHS of the eqn and expand it as per the condition given in Q.
Once you reach or cross the other side after expansion then compare with the other side and make a decision.
See here after B I would not have bothered about rest as B is concretely correct.
If the problem is Easy Respect it, if the problem is tough Attack it
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by greatchap » Sat Nov 20, 2010 4:02 am
Thanks for your help buddy.

Though I got a hang of it but to understand it better, can I illustrate -

B: Correct
f(2) = f(2^2) = f(4)
f(4) = f(4^2) = f(16)
Thus f(2) = f(16) => f(16)-f(-2) = 0

So here f(16)=f(4)=f(2) so thats why result is 0 right?

but if we solve last choice f(0)=0 then isnt it true- as f(0)=f(0^2) which is 0.
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by GMATGuruNY » Sat Nov 20, 2010 4:09 am
greatchap wrote:Hi Guys,

Inspite of some search, I am unable to determine how to solve questions with functions ?

For e.g.

If function f(x) satisfies f(x)=f(x^2) for all x, which of the following must be true?

(A) f(4)=f(2) f(2)
(B) f(16)-f(2)=0
(C) f(-2)+f(4)=0
(D) f(3)=3f(3)
(E) f(0)=0

I don't know where to start ?

Please help me out here.

Thanks,

GR
Scan the answers, looking for one that's easy to prove.

Answer choices A and B discuss f(2). Let's start there.

The problem stipulates that f(x) = f(x^2) for all values of x. This means that:

f(2) = f(2^2) = f(4)
So f(2) = f(4)

f(4) = f(4^4) = f(16)
So f(4) = f(16)

Thus, f(2) = f(4) = f(16).

So we can prove answer choice B: f(16) - f(2) = 0.

The correct answer is B.
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by GMATGuruNY » Sat Nov 20, 2010 4:17 am
greatchap wrote:Thanks for your help buddy.

Though I got a hang of it but to understand it better, can I illustrate -

B: Correct
f(2) = f(2^2) = f(4)
f(4) = f(4^2) = f(16)
Thus f(2) = f(16) => f(16)-f(-2) = 0

So here f(16)=f(4)=f(2) so thats why result is 0 right?

but if we solve last choice f(0)=0 then isnt it true- as f(0)=f(0^2) which is 0.
We don't know the actual function itself. We know only the relationship between x values of the function: that, for all values of x, f(x) = f(x^2).

If x=0, we know that f(0) = f(0^2) = f(0), but we don't know the actual value of f(0).
Last edited by GMATGuruNY on Sat Nov 20, 2010 4:22 am, edited 2 times in total.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
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by shovan85 » Sat Nov 20, 2010 4:17 am
greatchap wrote:
So here f(16)=f(4)=f(2) so thats why result is 0 right?
Yes. f(16) = f(2) => f(16) - f(2) = 0.
greatchap wrote: but if we solve last choice f(0)=0 then isnt it true- as f(0)=f(0^2) which is 0.
No. How do you know f(0) = 0?
A function can be defined in many ways. Unless the value f(0) is nmentioned we cannot say f(0) = 0
If the problem is Easy Respect it, if the problem is tough Attack it
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by greatchap » Sat Nov 20, 2010 4:28 am
Thanks. I got this one. Appreciate your help guys.
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