If a + frac{pi}{2}

Question

(A) Given the inequality: $ a + \frac{\pi}{2} < 2 	an^{-1} x + 3 \cot^{-1} x < b $.We know that $ 	an^{-1} x + \cot^{-1} x = \frac{\pi}{2} $,so $

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$ \frac{\pi}{2} $ and $ 2 \pi $

Vedclass · Answer

(A) Given the inequality: $ a + \frac{\pi}{2} We know that $ 	an^{-1} x + \cot^{-1} x = \frac{\pi}{2} $,so $ 	an^{-1} x = \frac{\pi}{2} - \cot^{-1} x $.Substituting this into the expression: $ 2(\frac{\pi}{2} - \cot^{-1} x) + 3 \cot^{-1} x = \pi - 2 \cot^{-1} x + 3 \cot^{-1} x = \pi + \cot^{-1} x $.Now the inequality becomes: $ a + \frac{\pi}{2} Subtracting $ \pi $ from all parts: $ a - \frac{\pi}{2} Since the range of $ \cot^{-1} x $ is $ (0, \pi) $,we have $ 0 Comparing $ a - \frac{\pi}{2} $ a - \frac{\pi}{2} = 0 \Rightarrow a = \frac{\pi}{2} $.$ b - \pi = \pi \Rightarrow b = 2\pi $.Thus,$ a = \frac{\pi}{2} $ and $ b = 2\pi $.

If $ a + \frac{\pi}{2} < 2 \tan^{-1} x + 3 \cot^{-1} x < b $,then find the values of $ a $ and $ b $.

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