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If f(x) = x(x-p)(x-q), is f(1) > 0? 1) f(2) = 0 2) f(4) = 0 [#permalink]

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15 Nov 2017, 01:08

MathRevolution wrote:

[GMAT math practice question]

If \(f(x) = x(x-p)(x-q)\), is \(f(1) > 0\)?

1) \(f(2) = 0\) 2) \(f(4) = 0\)

\(f(1)=1(1-p)(1-q)\). this will be greater than \(0\) only if both \((1-p)\) & \((1-q)\) are either together positive or negative.

Statement 1: \(f(2)=2(2-p)(2-q)=0\)

=> either \(p=2\) or \(q=2\). So we do not have a unique answer for \(p\) & \(q\). Insufficient (eg. if \(p=2\), then \(q\) can be positive or negative. it will not change the outcome. similarly if \(q=2\) then \(p\) can be anything)

Statement 2: \(f(4)=4(4-p)(4-q)=0\)

=> either \(p=4\) or \(q=4\). So we do not have a unique answer for \(p\) & \(q\). Insufficient

combining 1 & 2 we have \(p=2\) or \(4\) and \(q=2\) or \(4\). but in either case \((1-p)<0\) and \((1-q)<0\)

so we have both \((1-p)\) & \((1-q)\) as negative hence \(f(1)=1(1-p)(1-q)>0\). Sufficient

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since the question includes 2 variables (p and q) and 0 equations, C is most likely to be the answer.

Conditions 1) & 2):

The equations f(2) = 0 and f(4) = 0 tell us that 2 and 4 are roots of x(x-p)(x-q) = 0, and so f(x) = x(x-2)(x-4) Therefore, f(1) = 1(1-2)(1-4) = 1*(-1)*(-3) = 3 > 0, and the answer is ‘yes’. Therefore, conditions 1) and 2), when taken together, are sufficient.

The answer is C, as expected.

Normally, in problems which require 2 or more additional equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E). Answer: C
_________________

Re: If f(x) = x(x-p)(x-q), is f(1) > 0? 1) f(2) = 0 2) f(4) = 0 [#permalink]

Show Tags

17 Nov 2017, 07:48

MathRevolution wrote:

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since the question includes 2 variables (p and q) and 0 equations, C is most likely to be the answer.

Conditions 1) & 2):

The equations f(2) = 0 and f(4) = 0 tell us that 2 and 4 are roots of x(x-p)(x-q) = 0, and so f(x) = x(x-2)(x-4) Therefore, f(1) = 1(1-2)(1-4) = 1*(-1)*(-3) = 3 > 0, and the answer is ‘yes’. Therefore, conditions 1) and 2), when taken together, are sufficient.

The answer is C, as expected.

Normally, in problems which require 2 or more additional equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E). Answer: C

Can you explain how can p & q be variable here? the function has one input x and give an output so the variable should be x. subsequently 2 & 4 should be the roots of x.

Re: If f(x) = x(x-p)(x-q), is f(1) > 0? 1) f(2) = 0 2) f(4) = 0 [#permalink]

Show Tags

17 Nov 2017, 21:26

niks18 wrote:

MathRevolution wrote:

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since the question includes 2 variables (p and q) and 0 equations, C is most likely to be the answer.

Conditions 1) & 2):

The equations f(2) = 0 and f(4) = 0 tell us that 2 and 4 are roots of x(x-p)(x-q) = 0, and so f(x) = x(x-2)(x-4) Therefore, f(1) = 1(1-2)(1-4) = 1*(-1)*(-3) = 3 > 0, and the answer is ‘yes’. Therefore, conditions 1) and 2), when taken together, are sufficient.

The answer is C, as expected.

Normally, in problems which require 2 or more additional equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E). Answer: C

Can you explain how can p & q be variable here? the function has one input x and give an output so the variable should be x. subsequently 2 & 4 should be the roots of x.

it would be great if you could you provide more clarity on the concerns I have pertaining to the definition of variables for this question and solution posted above?

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since the question includes 2 variables (p and q) and 0 equations, C is most likely to be the answer.

Conditions 1) & 2):

The equations f(2) = 0 and f(4) = 0 tell us that 2 and 4 are roots of x(x-p)(x-q) = 0, and so f(x) = x(x-2)(x-4) Therefore, f(1) = 1(1-2)(1-4) = 1*(-1)*(-3) = 3 > 0, and the answer is ‘yes’. Therefore, conditions 1) and 2), when taken together, are sufficient.

The answer is C, as expected.

Normally, in problems which require 2 or more additional equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E). Answer: C

Can you explain how can p & q be variable here? the function has one input x and give an output so the variable should be x. subsequently 2 & 4 should be the roots of x.

If p and q are determined, it means the function f is determined. Since f is determined, we can figure out f(1). That's why we have 2 variable to identify the function f(x).
_________________

Can you explain how can p & q be variable here? the function has one input x and give an output so the variable should be x. subsequently 2 & 4 should be the roots of x.

If p and q are determined, it means the function f is determined. Since f is determined, we can figure out f(1). That's why we have 2 variable to identify the function f(x).

Thanks for your reply. While i understand that to calculate the value of the function, we need the values of p & q. but should this make p & q a variable or are they simply unknown constants. The value of p & q will not change for any other input, for eg. f(3)=3(3-p)(3-q), f(5)=5(5-p)(5-q) and so on....

In my opinion this function has only one independent variable x and the dependent variable is f(x). 2 & 4 are the roots of f(x), which in turn leads to the values of p & q. this is not a multivariate function i hope.

Finally even if we assume that p & q are variables, then how can we simply assume that p=2 & q=4? why can't this be other way round, I mean p=4 & q=2 ?

Can you explain how can p & q be variable here? the function has one input x and give an output so the variable should be x. subsequently 2 & 4 should be the roots of x.

If p and q are determined, it means the function f is determined. Since f is determined, we can figure out f(1). That's why we have 2 variable to identify the function f(x).

Thanks for your reply. While i understand that to calculate the value of the function, we need the values of p & q. but should this make p & q a variable or are they simply unknown constants. The value of p & q will not change for any other input, for eg. f(3)=3(3-p)(3-q), f(5)=5(5-p)(5-q) and so on....

In my opinion this function has only one independent variable x and the dependent variable is f(x). 2 & 4 are the roots of f(x), which in turn leads to the values of p & q. this is not a multivariate function i hope.

Finally even if we assume that p & q are variables, then how can we simply assume that p=2 & q=4? why can't this be other way round, I mean p=4 & q=2 ?

Let's define a variable as an unknown value we need to identify, which means an unknown constant is a variable we should find. To determine the function we need to identify the values of p and q. Thus p and q are variables we need to get in order to determine the function f(x), because we can solve this question after figuring out the function f(x)
_________________

Can you explain how can p & q be variable here? the function has one input x and give an output so the variable should be x. subsequently 2 & 4 should be the roots of x.

If p and q are determined, it means the function f is determined. Since f is determined, we can figure out f(1). That's why we have 2 variable to identify the function f(x).

Thanks for your reply. While i understand that to calculate the value of the function, we need the values of p & q. but should this make p & q a variable or are they simply unknown constants. The value of p & q will not change for any other input, for eg. f(3)=3(3-p)(3-q), f(5)=5(5-p)(5-q) and so on....

In my opinion this function has only one independent variable x and the dependent variable is f(x). 2 & 4 are the roots of f(x), which in turn leads to the values of p & q. this is not a multivariate function i hope.

Finally even if we assume that p & q are variables, then how can we simply assume that p=2 & q=4? why can't this be other way round, I mean p=4 & q=2 ?

Let's define a variable as an unknown value we need to identify, which means an unknown constant is a variable we should find. To determine the function we need to identify the values of p and q. Thus p and q are variables we need to get in order to determine the function f(x), because we can solve this question after figuring out the function f(x)

Again thanks for your reply. Even if for the sake of argument we assume that an unknown "constant" can be termed as variable (which I am not very sure of), this does not explain the assumption that p=2 & q=4. it could be other way round as well i.e p=4 & q=2

as f(2)=0 & f(4)=0 so 2 & 4 are definitely roots of f(x) but shouldn't these be the values of x which will further lead to the values of p & q, which in this case happens to be either 2 or 4?