Which of the following statements are correct ?

(A) If $f : R \to R$ then $f(x) = |x|$ is continuous everywhere.

(B) If $f : R \to R$ then $f(x) = |x|$ is continuous everywhere but not differentiable at $x = 0$.

(C) Let $f : R - \{0\} \to R$ then $f(x) = \frac{1}{x}$ is continuous everywhere.

(D) Let $f : R \to R$ then $f(x) = |x - 1| + |x - 2|$ is continuous everywhere but not differentiable at exactly 2 points.

(E) If $f : R \to R$ then $f(x) = \cot x$ is continuous everywhere.

Choose the correct answer from the options given below :

If $y = \dfrac{\log_e x}{x}$, then $\dfrac{d^2y}{dx^2}$ is :

Let a function $f(x)=\alpha+(\beta^2+5\beta+6)|x|+\gamma|x|^4$, where $\alpha, \beta$ and $\gamma$ are real number. Then the function $f(x)$ is differentiable at $x=0$ if

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

Let $f(x)=\begin{cases}\frac{1-\cos4x}{x^2}, & \text{if } x<0\\2\lambda, & \text{if } x=0\\\frac{\sqrt{x}}{\sqrt{16+\sqrt{x}}-4}, & \text{if } x>0\end{cases}$ such that $f$ continuous at $x=0$. Then, the value of $\lambda$ is equal to:

The graph of a function $f: R \to R$ is shown below, where $R$ is a real numbers

Then which of following statements is correct?

If $x = \frac{a}{1 + t}$ and $y = \frac{a}{(1 + t)^2}$ where $a > 0$ , then $\frac{d^2y}{dx^2}$ at $t = 1$ is