If -1 n2 - 10n - 21-6- - -6- n - N terms- then n can be

The correct options are A 4 B 3Taking cot on both the sides, we get n^2 10n+21.6 lt;π cot(π6) ...( inequality sign gets reversed, since cot(x) is a ...

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Standard XII

Mathematics

Inverse Function

If -1 n2 - ...

Question

If ${cot}^{- 1} (\frac{n^{2} - 10 n + 21.6}{π}) > \frac{π}{6}, n ϵ N$ terms, then $n$ can be

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Solution

The correct options are
A 4
B 3
Taking $c o t$ on both the sides, we get
$n^{2} - 10 n + 21.6 < π c o t (\frac{π}{6})$ ...( inequality sign gets reversed, since cot(x) is a decreasing function).
$n^{2} - 10 n + 21.6 < π (\sqrt{3})$
$n^{2} - 10 n + 21.6 < 5.44$
$n^{2} - 10 n + 16.19 < 0$
Consider
$g (n) = n^{2} - 10 n + 16.19$
$g (2) = 4 - 20 + 16.19$
$= 0.19$ ...(i)
$g (3) = 9 - 30 + 16.19$
$= - 4.81 < 0$ ...(ii)
$g (4) = 16 - 40 + 16.19$
$= - 7.81 < 0$ ...(iii)
$g (8) = 64 - 80 + 16.19$
$= 0.19$ ...(iv)
Hence only 3, and 4 satisfies the above inequality,

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