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Logarithms

If | cos -1...

Question

If $∣ ∣ ∣ {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) ∣ ∣ ∣ < \frac{π}{3}$ , then $x$ belongs to the interval

A

(- \sqrt{3}, \sqrt{3})

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B

(\frac{- 1}{\sqrt{3}}, \frac{1}{\sqrt{3}})

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C

(0, 1 \sqrt{3})

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D

none of these

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Solution

The correct option is B $(\frac{- 1}{\sqrt{3}}, \frac{1}{\sqrt{3}})$

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3 s i n^{- 1} \frac{2 x}{1 + x^{2}} - 4 c o s^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 t a n^{- 1} \frac{2 x}{1 + x^{2}} = \frac{π}{3}

∣ ∣ ∣ {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}} ∣ ∣ ∣ < \frac{π}{3}

3 s i n^{- 1} \frac{2 x}{1 + x^{2}} - 4 c o s^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 t a n^{- 1} \frac{2 x}{1 + x^{2}} = \frac{π}{3}

log ({log}_{(1 / 3)} (x^{2} - \frac{10 x}{3} + 1))

is defined, then the interval in which

x

∣ ∣ ∣ {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) ∣ ∣ ∣ < \frac{π}{3}

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