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The general solution of $x\sqrt{1+y^{2}}\,dx + y\sqrt{1+x^{2}}\,dy = 0$ is

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Standard integral: $\displaystyle\int \frac{x}{\sqrt{1+x^2}}\,dx = \sqrt{1+x^2} + C$. Recognise this pattern in separation-of-variables problems.

Updated On: Apr 8, 2026

$\sin^{-1}x + \sin^{-1}y = c$
$x^{2} + y^{2} = \sqrt{1+x^{2}} + \sqrt{1+y^{2}} + c$
$\sqrt{1+x^{2}} + \sqrt{1+y^{2}} = c$
$\tan^{-1}x - \tan^{-1}y = c$

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The Correct Option is C

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If $f(x) = \begin{cases} x, & 0 \le x \le 1 \\ 2x - 1, & 1<x \end{cases}$, then

MET - 2019
Continuity

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