JEE MAIN - Mathematics Hindi (2022 - 29th July Evening Shift - No. 7)
यदि अवकल समीकरण $$\frac{d y}{d x}=\frac{x+y-2}{x-y}$$ का हल वक्र बिंदुओं $$(2,1)$$ तथा $$(k+1,2)$$, $$k>0$$ से होकर जाता है, तो
$$2 \tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(k^{2}+1\right)$$
$$\tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(k^{2}+1\right)$$
$$2 \tan ^{-1}\left(\frac{1}{k+1}\right)=\log _{e}\left(k^{2}+2 k+2\right)$$
$$2 \tan ^{-1}\left(\frac{1}{k}\right)=\log _{e}\left(\frac{k^{2}+1}{k^{2}}\right)$$
Comments (0)
