First slide

Properties of ITF

Question

$A$ root of the equation $17 x^{2} + 17 x \tan (2 \tan^{- 1} \frac{1}{5} - \frac{π}{4}) - 10 = 0$ , is

Moderate

A

$\frac{10}{17}$
B

$- 1$
C

$\frac{- 7}{17}$
D

$1$

Solution

We have,
$\begin{array}{l} \tan (2 \tan^{- 1} \frac{1}{5} - \frac{π}{4}) \\ = \tan \{\tan^{- 1} \frac{5}{12} - \tan^{- 1} 1\} [∵ 2 \tan^{- 1} x = \tan^{- 1} \frac{2 x}{1 - x^{2}}] \\ = \tan \{\tan^{- 1} (\frac{\frac{5}{12} - 1}{1 + \frac{5}{12}})\} = \frac{- 7}{17} \end{array}$
So, the given equation is
$17 x^{2} - 7 x - 10 = 0 \Rightarrow (x - 1) (17 x + 10) = 0 \Rightarrow x = 1, \frac{- 10}{7}$

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