Let $e^y(x+1) = 1$. Then which of the following are TRUE? (A) $\frac{d^2y}{dx^2} = -\frac{1}{(x+1)^2}$ (B) $\frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2$ (C) $\left.\frac{d^2y}{dx^2}\right|_{x=0} = -1$ (D) $\left.\frac{d^2y}{dx^2}\right|_{x=0} = 1$ (E) $\left.\frac{d^2y}{dx^2}\right|_{x=1} = \frac{1}{4}$ Choose the correct answer from the options given below:

If $x\sqrt{1+y}+y\sqrt{1+x}=0, x\neq y$, then $\frac{dy}{dx}$ is equal to

If $y = 3e^{2x} + 2e^{3x}$, then $\frac{d^2y}{dx^2} + 6y$ is equal to

If $\sqrt{y+x} + \sqrt{y-x} = a$, $a > 1$, $\frac{d^2y}{dx^2}$ is equal to :

If $x = 3t^2$ and $y = \sqrt{t}$, then the value of $\frac{d^2y}{dx^2}$ is

If $f(x+y) = f(x) + f(y)$ for all $x, y \in R$ (set of real numbers) and $f(x)$ is continuous at $x = 0$, then identify the correct statement ?

The function $f(x) = |x - 1|$ is