If - n 2 - n- then sin -1 9- 2 - 4n 2 6- -4n -

The correct option is A π 6 Ans.(a) n( n+1 ) ( 2n+1 ) #160; 6 =λ n( n+1 ) #160; 2 ∴ 2n+1=3λ ∴ 9λ #160; ^ 2 4n ^ 2 6λ +4n =( ...

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Standard XII

Mathematics

AM,GM,HM Inequality

If ∑ n 2 =λ...

Question

If $\sum n^{2} = λ \sum n$ , then ${sin}^{- 1} (\frac{{9 λ}^{2} - {4 n}^{2}}{6 λ + 4 n}) =$

\frac{π}{6}

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\frac{π}{3}

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\frac{π}{2}

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π

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Solution

The correct option is A $\frac{π}{6}$
Ans.(a)
$\frac{n (n + 1) (2 n + 1)}{6} = λ \frac{n (n + 1)}{2}$
$∴ 2 n + 1 = 3 λ$
$∴ \frac{{9 λ}^{2} - {4 n}^{2}}{6 λ + 4 n} = \frac{(3 λ - 2 n) (3 λ + 2 n)}{2 (3 λ + 2 n)} = \frac{1}{2}$ by (1)
$∴ E = {sin}^{- 1} (\frac{1}{2}) = \frac{π}{6}$

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Similar questions

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f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \begin{matrix} 0, & - \frac{π}{2} < x < - \frac{π}{6} \end{matrix} \begin{matrix} \frac{x}{2}, & x = - \frac{π}{6} \end{matrix} \begin{matrix} x, & - \frac{π}{6} < x < \frac{π}{6} \end{matrix} \begin{matrix} \frac{x}{2}, & x = \frac{π}{6} \end{matrix} \begin{matrix} 0, & \frac{π}{6} < x < \frac{π}{2} \end{matrix} \end{matrix}

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If ∑n2=λ∑n, then sin−1(9λ2−4n26λ+4n)=

The correct option is A π6Ans.(a)n(n+1)(2n+1)6=λn(n+1)2∴2n+1=3λ∴9λ2−4n26λ+4n=(3λ−2n)(3λ+2n)2(3λ+2n)=12 by (1)∴E=sin−1(12)=π6

If $\sum n^{2} = λ \sum n$ , then ${sin}^{- 1} (\frac{{9 λ}^{2} - {4 n}^{2}}{6 λ + 4 n}) =$

The correct option is A $\frac{π}{6}$
Ans.(a)
$\frac{n (n + 1) (2 n + 1)}{6} = λ \frac{n (n + 1)}{2}$
$∴ 2 n + 1 = 3 λ$
$∴ \frac{{9 λ}^{2} - {4 n}^{2}}{6 λ + 4 n} = \frac{(3 λ - 2 n) (3 λ + 2 n)}{2 (3 λ + 2 n)} = \frac{1}{2}$ by (1)
$∴ E = {sin}^{- 1} (\frac{1}{2}) = \frac{π}{6}$