If $x = a\left(t - \frac{1}{t}\right)$, $y = b\left(t + \frac{1}{t}\right)$, then $\frac{dy}{dx} =$

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

The given function $f(x) = [x]$ is discontinuous at :

If $y^2 = ax + b, a,b \in \mathbb{R}$ where $\mathbb{R}$ is the set of real numbers, then $y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2$ is equal to:

Match List-I with List-II

$\begin{array}{|l|l|} \hline \rule{0pt}{2.8ex}\text{List-I} & \text{List-II} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(A) } f(x) = |x| & \text{(I) Not differentiable at } x=-2 \text{ only} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(B) } f(x) = |x+2| & \text{(II) Not differentiable at } x=0 \text{ only} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(C) } f(x) = |x^2-4| & \text{(III) Not differentiable at } x=2 \text{ only} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(D) } f(x) = |x-2| & \text{(IV) Not differentiable at } x=2,-2 \text{ only} \\[1.2ex] \hline \end{array}$

Choose the correct answer from the options given below:

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 $\frac{d}{dx}[ax^3 + ax^2 + ax + 1] = 9x^2 + 6x + 3$, then $a$ is equal to