If y=tan−1⁡4x1+5x2+tan−1⁡2+3x3−2x where x∈0,23 if dydx=k1+λx2 where k and λ∈N then λk is - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

Solution:

$y = \tan^{- 1} (\frac{4 x}{1 + 5 x^{2}}) + \tan^{- 1} (\frac{\frac{2}{3} + x}{1 - \frac{2 x}{3}}) = \tan^{- 1} (\frac{5 x - x}{1 + x (5 x)}) + \tan^{- 1} (\frac{x + \frac{2}{3}}{1 - x (\frac{2}{3})})$

$= \tan^{- 1} (5 x) - \tan^{- 1} x + \tan^{- 1} (x) + \tan^{- 1} (\frac{2}{3}) = \tan^{- 1} (5 x) + \tan^{- 1} (\frac{2}{3})$

$\frac{dy}{dx} = \frac{5}{1 + 25 x^{2}} ∴ k = 5, λ = 25, \frac{λ}{k} = 5$

If $y = \tan^{- 1} (\frac{4 x}{1 + 5 x^{2}}) + \tan^{- 1} (\frac{2 + 3 x}{3 - 2 x})$ where $x \in (0, \frac{2}{3})$ if $\frac{dy}{dx} = \frac{k}{1 + {λx}^{2}}$ where k and $λ \in N$ then $\frac{λ}{k}$ is

Solution:

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