Match List-I with List-II | List-I | List-II | | --- | --- | | Function f(x) | Points of Non-Differentiability | | --- | --- | | (A) $f(x) = \vert x\vert + 1$ | (I) Not differentiable at $x = 3$ only | | (B) $f(x) = \vert x - 3\vert $ | (II) Not differentiable at $x = -3$ only | | (C) $f(x) = \vert x + 3\vert $ | (III) Not differentiable at $x = 3, -3$ only | | (D) $f(x) = \vert x^2 - 9\vert $ | (IV) Not differentiable at $x = 0$ only | Choose the correct answer from the options given below:

Match List-I with List-II

Where $\mathbb{R}$ is set of real numbers

Choose the correct answer from the options given below:

If $f(x) = \begin{cases} \frac{x^2 - 9}{x - 3}, & x \neq 3 \\ 5, & x = 3 \end{cases}$ then $f(x)$ :

If $x^2 - y^2 = t - \frac{1}{t}$, and $x^4 + y^4 = t^2 + \frac{1}{t^2}$, then which of the following is correct?

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 derivative of $|x|$ with respect to $x$, where $x \in R \sim \{0\}$ is

If $y=\tan^{-1}(x+y)$, then $\frac{d^2y}{dx^2}$ is equal to: