Question

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

• A. $ \frac{10}{17} $
• B. $ -1 $
• C. $ -\frac{7}{17} $
• D. $ 1 $

Answer 1

Answer: (D)

We have, $ \tan \left( 2{{\tan }^{-1}}\frac{1}{5}-\frac{\pi }{4} \right) $ $ =\tan \left( {{\tan }^{-1}}\frac{5}{12}-{{\tan }^{-1}}1 \right) $ $ \left[ \because 2{{\tan }^{-1}}x={{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}} \right] $ $ =\tan \left\{ {{\tan }^{-1}}\left( \frac{\frac{5}{12}-1}{1+\frac{5}{12}} \right) \right\}=\frac{-7}{17} $ So, the given equation is $ 17{{x}^{2}}-7x-10=0 $ $ \Rightarrow $ $ (x-1)(17x+10)=0 $ $ \Rightarrow $ $ x=1,\frac{-10}{17} $