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