For x (0, ^ -1 5 2 ) , the function f (x) = cot –1 ( 2 x+ 5 x 7 )
Mathematics · JEE Main · NTA Exams — Limit, Continuity and Differentiability
For \(x \in\left(0, \tan ^{-1} \sqrt{\frac{5}{2}}\right)\), the function
f (x) = cot–1 \(\left(\frac{\sqrt{2} \sin x+\sqrt{5} \cos x}{\sqrt{7}}\right)\)
f (x) = cot–1 \(\left(\frac{\sqrt{2} \sin x+\sqrt{5} \cos x}{\sqrt{7}}\right)\)
- increases in
- decreases in \(\left(0, \tan ^{-1} \sqrt{\frac{5}{2}}\right)\)
- increases in \(\left(0, \tan ^{-1} \sqrt{\frac{2}{5}}\right)\) and decreases in
\(\left(\tan ^{-1} \sqrt{\frac{2}{5}}, \tan ^{-1} \sqrt{\frac{5}{2}}\right)\) - increases in \(\left(\tan ^{-1} \sqrt{\frac{2}{5}}, \tan ^{-1} \sqrt{\frac{5}{2}}\right)\) and decreases in \(\left(0, \tan ^{-1} \sqrt{\frac{2}{5}}\right)\)
Answer
(D) increases in ( ^ -1 2 5 , ^ -1 5 2 ) and decreases in (0, ^ -1 2 5 )
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