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प्रश्न
Find the maximum and minimum values of y = tan \[x - 2x\] .
उत्तर
\[\text { Given }: f\left( x \right) = y = \tan x - 2x\]
\[ \Rightarrow f'\left( x \right) = \sec^2 x - 2\]
\[\text { For a local maxima or local minima, we must have }\]
\[ f'\left( x \right) = 0\]
\[ \Rightarrow \sec^2 x - 2 = 0\]
\[ \Rightarrow \sec^2 x = 2\]
\[ \Rightarrow \sec x = \pm \sqrt{2}\]
\[ \Rightarrow x = \frac{\pi}{4} \text { and } \frac{3\pi}{4}\]
\[\text { Thus, x }= \frac{\pi}{4} \text { and }x = \frac{3\pi}{4}\text { are the possible points of local maxima or a local minima } . \]
\[\text { Now,} \]
\[f''\left( x \right) = 2 \sec^2 x \tan x\]
\[\text { At }x = \frac{\pi}{4}: \]
\[ f''\left( \frac{\pi}{4} \right) = 2 \sec^2 \left( \frac{\pi}{4} \right) \tan \left( \frac{\pi}{4} \right) = 4 > 0\]
\[\text { So, }x = \frac{\pi}{4} \text { is a point of local minimum } . \]
\[\text { The local minimum value is given by }\]
\[f\left( \frac{\pi}{4} \right) = \tan\left( \frac{\pi}{4} \right) - 2 \times \frac{\pi}{4} = 1 - \frac{\pi}{2}\]
\[\text { At x} = \frac{3\pi}{4}: \]
\[ f''\left( \frac{3\pi}{4} \right) = 2 \sec^2 \left( \frac{3\pi}{4} \right) \tan \left( \frac{3\pi}{4} \right) = - 4 < 0\]
\[\text{ So,} x = \frac{3\pi}{4}\text { is a point of local maximum }. \]
\[\text { The local maximum value is given by }\]
\[f\left( \frac{3\pi}{4} \right) = \tan \left( \frac{3\pi}{4} \right) - 2 \times \frac{3\pi}{4} = - 1 - \frac{3\pi}{2}\]
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