If x i -1 - i - N- thenlog x2 x1-log x3 x2-log xn xn-1-log x1 xnhas the minimum valueA- 1B- 2C- 0D- None

The correct option is D NoneOn applying A.M. ≥ G.M. E ≥ n (log_x_2 x_1. log_x_3 x_2 …… log_x_1 x_n)^1/n=n Hence the minimum value is n. ...

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Inequality

If x i >1 ∀ i...

Question

If $x_{i} > 1 \forall i ϵ N$ , then
$l o g_{x_{2}} x_{1} + l o g_{x_{3}} x_{2} + \dots \dots + l o g_{x_{n}} x_{n - 1} + l o g_{x_{1}} x_{n}$
has the minimum value

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x_{i} > 1 \forall i ϵ N

l o g_{x_{2}} x_{1} + l o g_{x_{3}} x_{2} + \dots \dots + l o g_{x_{n}} x_{n - 1} + l o g_{x_{1}} x_{n}

If $f (x) = c o s (l o g x)$ , then $f (x) f (y) - \frac{1}{2} {f (\frac{x}{y}) + f (x y)}$ has the value

S_{n} = (_{0}^{n}) - (_{1}^{n - 1}) + (_{2}^{n - 2}) - \dots

^{p} C_{q} = 0

p < q

S_{n}

z = (\frac{1 + i}{1 - i})

\frac{1}{l o g_{2} n} + \frac{1}{l o g_{3} n} + \frac{1}{l o g_{4} n} + \dots \frac{1}{l o g_{1983} n}

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