The point('s) at which the function $f$ given by $f(x) = \begin{cases} \dfrac{|x|}{2x}, & x<0 \\ -\dfrac{1}{2}, & x\geq0 \end{cases}$ is continuous is (are)

If $e^y(x + 1) = 1$, then

If $y = x^{\sin x}$, then value of $x^{-\sin x} \frac{dy}{dx} - \cos x \log x$ is :

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 $e^y(1+x)=1$, then $\dfrac{d^2y}{dx^2}$ is:

If $x^m y^n = (x + y)^{m+n}$, then $\frac{d^2y}{dx^2}$ is equal to:

Consider the following statements for the curve $f(x) = x|x|$ and find that which of the following(s) are/is correct :

(A) $f(x) = x|x|$ is differentiable at $x = 0$.

(B) $f(x) = x|x|$ is continuous at $x = 0$ but not differentiable at $x = 0$.

(C) $f(x) = x|x|$ has point of infection at $x = 0$.

(D) $f(x) = x|x|$ is symmetrical about y-axis.

Choose the correct answer from the options given below :