First slide

Properties of ITF

Question

$If f (x) = a x + b, a > 0 such that f : [- 1, 1] \to [0, 2], then$ $\cot [\cot^{- 1} 7 + \cot^{- 1} 8 + \cot^{- 1} 18] is equal to$

Moderate

A

$f (- 1)$
B

$f (0)$
C

$f (1)$
D

$f (2)$

Solution

$Given that f (x) = a x + b$
$Differentiating with respect to x on both sides$
$\begin{array}{l} \Rightarrow f^{'} (x) = a > 0 \\ ∴ f (x) is an increasing function. \end{array}$
$\begin{array}{l} \Rightarrow f (- 1) = 0 and f (1) = 2 \\ - a + b = 0 and a + b = 2 \\ \Rightarrow a = b = 1 \\ ∴ f (x) = x + 1 \\ Now, \cot [\cot^{- 1} 7 + \cot^{- 1} 8 + \cot^{- 1} 18] \\ = \cot \{\tan^{- 1} (\frac{1}{7}) + \tan^{- 1} (\frac{1}{8}) + \tan^{- 1} (\frac{1}{18})\} \\ = \cot \{\tan^{- 1} (\frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7} \cdot \frac{1}{8}}) + \tan^{- 1} (\frac{1}{18})\} \\ = \cot \{\tan^{- 1} (\frac{15}{55}) + \tan^{- 1} (\frac{1}{18})\} \end{array}$
$\begin{array}{l} = \cot \{\tan^{- 1} (\frac{3}{11}) + \tan^{- 1} (\frac{1}{18})\} \\ = \cot \{\tan^{- 1} (\frac{\frac{3}{11} + \frac{1}{18}}{1 - \frac{3}{11} \cdot \frac{1}{18}})\} \\ = \cot \{\tan^{- 1} (\frac{65}{195})\} \end{array}$
$\begin{array}{l} = \cot \{\tan^{- 1} (\frac{1}{3})\} \\ = \cot (\cot^{- 1} 3) \\ = 3 \\ = 1 + 2 \\ = f (2) \end{array}$

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