The largest interval containing $\mathrm{x}$ for which…
The largest interval containing $\mathrm{x}$ for which $\mathrm{x}^{12}-\mathrm{x}^9+\mathrm{x}^4-\mathrm{x}+1>0$ is
- $0 < x < 1$
- $-4 < x < 2$
- $-\infty < $ x $ < \infty$
- $-2^{10} < x < 2^{10}$
Solution
We have $f(x)=x^{12}-x^9+x^4-x+1$
where, $\mathrm{f}(\mathrm{x})>0$
case I:- when $x>1$
$\mathrm{x}^{12}+\mathrm{x}^4+1>0$ and $-(\mathrm{x} 9+\mathrm{x})$ is also positive
$
\Rightarrow \mathrm{f}(\mathrm{x})>0 \quad \forall \mathrm{x} " 0
$
Case III :- when $0 < \mathrm{x} < $ !
we have $f(x)=x^{12}-x^9+x^4-x+1$
$
\begin{aligned}
& \Rightarrow \mathrm{x}^4\left(\mathrm{x}^8+1\right)-\mathrm{x}\left(\mathrm{x}^8+1\right)+1 \\
& \Rightarrow\left(\mathrm{x}^8+1\right)\left(\mathrm{x}^4-\mathrm{x}+1\right) \\
& \because \mathrm{x}^8+1>0 \\
& \text { now } \mathrm{x}^4 < \mathrm{x} \\
& \Rightarrow \mathrm{x}^4+1 < \mathrm{x}+1 \\
& \Rightarrow \mathrm{x}^4+1-\mathrm{x} < 1 \text { but } \mathrm{x} 4+1-\mathrm{x}>0 \text { as } \mathrm{x} \in(0,1) \\
& \Rightarrow \mathrm{f}(\mathrm{x})>0
\end{aligned}
$
from (i), (ii), (iii)
Asked in: AP EAMCET 2022 (06 Jul Shift 1)
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