Algebra - find the largest interval in

1708

$$\textbf{Let $\mathbf{f(x) = x^{12}-x^9+x^4-x+1}$}\\\\ \textbf{Let $\mathbf{x\geq 1,}$ Then $\mathbf{f(x)=x^9(x^3-1)+x(x^3-1)+1<0}$ }\\\\ \textbf{Now for $\mathbf{x>1}$, Then $\mathbf{f(x)=x^{12}+x^4(1-x^5)+(1-x)}<0$ }\\\\$

$\textbf{So Largest Interval for which $\mathbf{f(x)<0}$ is $\mathbf{x>1}$ }$\\\\ \boxed{\boxed{\mathbf{x\in\left(-\infty,1\right)}}}$

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Lokesh Verma ·May 17 '11 at 3:36

So the final answer should be R....

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paresh baruah ·May 23 '11 at 8:52

yes ans is R. plz help

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Lokesh Verma ·May 24 '11 at 3:35

arrey the answer is given by jagdish in #2...

he has proved it for both the cases... when x>1 and when x<1

so effectively he has proved it for all real values of x..

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find the largest interval in

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