Question
The largest interval for which $${x^{12}} - {x^9} + {x^4} - x + 1 > 0{\text{ is}}$$
A.
$$ - 4 < x \leqslant 0$$
B.
$$0 < x < 1$$
C.
$$ - 100 < x < 100$$
D.
$$ - \infty < x < \infty $$
Answer :
$$ - \infty < x < \infty $$
Solution :
$$\eqalign{
& {\text{Given expression }}{x^{12}} - {x^9} + {x^4} - x + 1 = f\left( x \right)\,\left( {{\text{say}}} \right) \cr
& {\text{For }}x < 0\,{\text{ put}}\,\,x = - y\,\,{\text{where}}\,y > 0{\text{ }} \cr
& {\text{then we get }}\,f\left( x \right) = {y^{12}} + {y^9} + {y^4} + y + 1 > 0\,\,{\text{for }}\,y > 0 \cr
& {\text{For }}0 < x < 1,\,\,{x^9} < {x^4} \cr
& \Rightarrow \, - {x^9} + {x^4} > 0 \cr
& {\text{Also }}1 - x > 0\,\,{\text{and }}{x^{12}} > 0 \cr
& \Rightarrow \,\,{x^{12}} - {x^9} + {x^4} + 1 - x > 0 \cr
& \Rightarrow \,\,f\left( x \right) > 0 \cr
& {\text{For }}x > 1 \cr
& f\left( x \right) = x\left( {{x^3} - 1} \right)\left( {{x^8} + 1} \right) + 1 > 0 \cr
& {\text{So}}\,\,f\left( x \right) > 0\,\,{\text{for}}\,\, - \infty < x < \infty . \cr} $$