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If $\beta$ satisfies the inequation $x^{2}-x-6>0$, then a value exists for
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1907 Upvotes
Verified Answer
The correct answer is:
$\sec \left(\sec ^{-1} \beta\right)$
Since, $\beta$ satisfy the inequation
$$
\begin{array}{rlr}
& & x^{2}-x-6>0 \\
\Rightarrow & & \beta^{2}-\beta-6>0 \\
\Rightarrow & & (\beta-3)(\beta+2)>0 \\
\Rightarrow & & \beta>3, \beta < -2
\end{array}
$$
$$
\begin{array}{rlr}
& & x^{2}-x-6>0 \\
\Rightarrow & & \beta^{2}-\beta-6>0 \\
\Rightarrow & & (\beta-3)(\beta+2)>0 \\
\Rightarrow & & \beta>3, \beta < -2
\end{array}
$$
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