The largest interval for which \[{{x}^{12}}-{{x}^{9}}+{{x}^{4}}-x+1>0\] is
A.\[-4 < x < 0\]
B.\[0 < x < 1\]
C.\[-100 < x < 100\]
D.\[-\infty < x < \infty \]

Hint: In this particular problem, the equation which is given that is\[{{x}^{12}}-{{x}^{9}}+{{x}^{4}}-x+1>0\] from that we have to make a three cases and solve each case then you get the three equation by combining these three equation we will get the interval for each cases. So, in this way we have to solve further and get the answer.

Complete step-by-step answer :
In this type of problem, the equation is given that \[{{x}^{12}}-{{x}^{9}}+{{x}^{4}}-x+1>0\]
Case-1:
Consider above equation as \[f(x)\]that is \[f(x)={{x}^{12}}-{{x}^{9}}+{{x}^{4}}-x+1\]
We have to simplify further the above equation we get:
\[f(x)={{x}^{9}}({{x}^{3}}-1)+x({{x}^{3}}-1)+1\]
By simplifying further we get:
\[f(x)=({{x}^{3}}-1)({{x}^{9}}+x)+1\]
But the above equation is greater than 0 that is \[f(x)=({{x}^{3}}-1)({{x}^{9}}+x)+1>0\]
We have to select the value of x in such a way that the above equation should be greater than zero. That means condition should be satisfied that is \[\forall x\ge 1\]
Therefore, \[f(x)=({{x}^{3}}-1)({{x}^{9}}+x)+1>0\,\,\,\,\,\,\,\,\forall x\ge 1---(1)\]
Case-2:
Equation is given as \[f(x)={{x}^{12}}-{{x}^{9}}+{{x}^{4}}-x+1\]
We have to simplify further the above equation we get:
\[f(x)={{x}^{4}}({{x}^{8}}+1)-x({{x}^{8}}+1)+1\]
By simplifying further we get:
\[f(x)=x({{x}^{8}}+1)({{x}^{3}}-1)+1\]
But the above equation is greater than 0 that is \[f(x)=x({{x}^{8}}+1)({{x}^{3}}-1)+1>0\]
We have to select the value of x in such a way that the above equation should be greater than zero. That means condition should be satisfied that is \[\forall x\le 0\]
Therefore, \[f(x)=x({{x}^{8}}+1)({{x}^{3}}-1)+1>0\,\,\,\,\,\,\,\,\forall x\le 0---(2)\]
Equation is given as \[f(x)={{x}^{12}}-{{x}^{9}}+{{x}^{4}}-x+1\]
We have to simplify further the above equation we get:
\[f(x)={{x}^{12}}+{{x}^{4}}(1-{{x}^{5}})+(1-x)\]
But the above equation is greater than 0 that is \[f(x)={{x}^{12}}+{{x}^{4}}(1-{{x}^{5}})+(1-x)>0\]
We have to select the value of x in such a way that the above equation should be greater than zero. That means condition should be satisfied that is \[\text{For}\,\,\,0 < x < 1\]
Therefore, \[f(x)={{x}^{12}}+{{x}^{4}}(1-{{x}^{5}})+(1-x)>0\,\,\,\,\,\,\,\,\text{For}\,\,\,0 < x < 1---(3)\]
By combining the equations (1), (2) and (3) we get them\[f(x)>0\,\,\,\text{for }x\in (-\infty ,\infty )\].
So, the correct option is “option 4”.
So, the correct answer is “Option 4”.

Note : In this type of problem remember that when we are solving such a type of problem select the value of x in such a way that function should be greater than 0. In the given problem we have a polynomial of degree 12, if we solve the given inequality we will have 12 roots or 12 solutions which satisfies the inequality.