If y=tan−1⁡4×1+5×2+tan−1⁡2+3×3−2x where x∈0,23 if dydx=k1+λx2 where k and λ∈N then λk is

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

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### Solution:

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

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

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