f(x)=x^{2n}+x^n+1, where n is an integer. Is f(x)=1?

[Max@Math Revolution]

[GMAT math practice question]

f(x)=x^{2n}+x^n+1, where n is an integer. Is f(x)=1?

1) x=-1
2) n is a multiple of 5.

[Max@Math Revolution]

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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.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.
f(x)=1
=> x^{2n}+x^n+1=1
=> x^{2n}+x^n=0
=> x^n(x^n +1)=0
=> x^n= 0 or x^n =-1
=> ( x = 0 ) or ( x = -1 and n is odd )

Conditions 1) and 2)
If x = -1 and n = 5, then f(x) = (-1)^{10} + (-1)^5 + 1 = 1 + (-1) + 1 = 1 and the answer is 'yes'.
If x = -1 and n = 10, then f(x) = (-1)^{20} + (-1)^{10} + 1 = 1 + 1 + 1 = 1 and the answer is 'no'.

Thus, both conditions together are not sufficient, since they do not yield a unique solution.

Therefore, E is the answer.
Answer: E

[deloitte247]

$$f\left(x\right)=x^{2n}+x^n+1$$
$$Is\ f\left(x\right)=1?$$

Statement 1
$$x=-1$$
$$f\left(-1\right)=-1^{2n}+\left(-1\right)^n+1$$
If n = 0
$$f\left(-1\right)=-1^{2\left(0\right)}+\left(-1\right)^0+1$$
$$=1+1+1=3$$
If n = 1

$f\left(-1\right)=-1^{\left(2\cdot1\right)}+\left(-1\right)^1-1$$
$$f\left(-1\right)=1-1-1$$
$$f\left(-1\right)=-1$$

Value of n is unknown, hence statement 1 is INSUFFICIENT.

Statement 2
n is a multiple of 5
this means n/5 remains 0 , multiple of 5 includes.
5, 10, 15, 20, 25, 30,.............(n*5)
If n=5 and x=1
$$f\left(x\right)=x^{2n}+x^n+1$$
$$f\left(1\right)=x^{2\cdot5}+x^5+1$$
$$f\left(1\right)=1^{2\cdot5}+1^5+1$$
$$f\left(1\right)=1^{10}+1^5+1=3$$
If n=5 and x=2

$$f\left(x\right)=x^{2n}+x^n+1$$
$$f\left(2\right)=2^{2\cdot5}+2^5+1$$
$$f\left(2\right)=2^{10}+2^5+1$$
$$f\left(2\right)>1$$

Value of x is unknown, hence statement 2 is INSUFFICIENT.

Combining statement 1 and 2 together
x = -1 and n = multiple of 5
If n = 5

$$f\left(x\right)=x^{2n}+x^n+1$$
$$f\left(-1\right)=\left(-1\right)^{2\cdot5}+\left(-1\right)^5+1$$
$$f\left(-1\right)=\left(-1\right)^{2\cdot5}+\left(-1\right)^5+1$$
$$f\left(-1\right)=\left(1\right)+\left(-1\right)+1$$
$$f\left(-1\right)=\left(1\right)+\left(-1\right)+1=1$$

If n = 10

$$f\left(-1\right)=\left(1\right)^{2\cdot10}+\left(-1\right)^{10}+1$$
$$f\left(-1\right)=\left(1\right)+\left(1\right)+1=3$$

Therefore,
If n = odd f(x) = 1, but If n = even f(x) = 3
Answer is not specific, information given is not enough to answer the question , both statement combined together are INSUFFICIENT.

$$answer\ is\ OPTION\ E$$