The smallest and the largest values of tan-1 1-x1-x - 0- x- 1 are-

The correct option is C 0,π4 tan ^ 1(1 x/1+x) 0 ≤ x≤ 1 1≤ x+1 ≤ 2 1≥1/1+x≥1/2 #160; 0 ≤1 x/1+x≤ 1 ∴ xϵ [ 1,1] by #160; ...

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Odd Function

The smallest ...

Question

The smallest and the largest values of
${tan}^{- 1} (\frac{1 - x}{1 + x})$ , $0 \leq x \leq 1$ are.

0, π

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0, \frac{π}{4}

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- \frac{π}{4}, \frac{π}{4}

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\frac{π}{4}, \frac{π}{2}

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Solution

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tan x > cot x

x \in (0, π) - {\frac{π}{2}}

{tan}^{- 1} (\frac{3}{4}) + {tan}^{- 1} (\frac{1}{7}) = \frac{π}{4}

x > 0, y > 0, {tan}^{- 1} (\frac{x}{y}) + {tan}^{- 1} (\frac{y - x}{y + x}) = \frac{π}{4}

f (x) = \frac{1 - tan x}{4 x - π}, x \neq \frac{π}{4}, x \in [0, \frac{π}{4}]

f (x)

[0, \frac{π}{2}]

f (\frac{π}{4})

Set of values of $x$ in $(0, π)$ satisfying $1 + {log}_{2} sin x + {log}_{2} sin 3 x \geq 0$ is

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} x + a^{2} \sqrt{2} s i n x, & 0 \leq x < \frac{π}{4} x c o t x + b, & \frac{π}{4} \leq x < \frac{π}{2} b s i n 2 x - a c o s 2 x, & \frac{π}{2} \leq x \leq π \end{matrix}

[0, π]

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