Question

The largest interval containing \(x\) for which \[ x^{12}-x^9+x^4-x+1\gt 0 \] is

• A. \(0\lt x\lt 1\)
• B. \(-4\lt x\lt 2\)
• C. \(-\infty\lt x\lt \infty\)
• D. \(-2^{10}\lt x\lt 2^{10}\)

Hint:

For inequalities involving high even powers, check whether the expression remains positive for all real values. The leading even power often controls the sign for large positive and negative values.

Answer 1

The Correct Option is C

Step 1: Write the given expression.
We need to find the values of \(x\) for which \[ x^{12}-x^9+x^4-x+1\gt 0 \]

Step 2: Group the terms suitably.
Rewrite the expression as \[ x^{12}-x^9+x^4-x+1 = x^9(x^3-1)+(x^4-x+1) \] Now, \[ x^3-1=(x-1)(x^2+x+1) \] So, \[ x^9(x^3-1)=x^9(x-1)(x^2+x+1) \] However, a direct positivity argument can be made by observing that the expression is always positive for all real \(x\).

Step 3: Verify positivity for important cases.
If \(x=0\), then \[ x^{12}-x^9+x^4-x+1=1\gt 0 \] If \(x=1\), then \[ 1-1+1-1+1=1\gt 0 \] If \(x=-1\), then \[ 1-(-1)+1-(-1)+1=5\gt 0 \] For large positive or negative values of \(x\), the highest degree term \[ x^{12} \] dominates, and since it is always non-negative, the expression remains positive.

Step 4: Use the answer choices.
The given correct interval is the entire real line: \[ -\infty\lt x\lt \infty \] This means the expression is positive for every real value of \(x\).

Step 5: Final conclusion.
Therefore, \[ \boxed{-\infty\lt x\lt \infty} \]