Let $f(x) = \begin{cases} 2 - |x^2 + 5x + 6|, & x \neq -2 \\ a^2 + 1, & x = -2 \end{cases}$. Then the range of $a$ such that $f(x)$ has a maximum at $x = -2$ is

Let $f(x) = \min \{x, x^2\}$ for every real number $x$. Then:

If a function defined by $f(x) = \frac{(3^x - 1)^2}{\sin x \log(1 + x)}$,$x \neq 0$,is continuous at $x = 0$,then $f(0) =$

$f$ is continuous at $x=\frac{\pi}{2}$ where,
$f(x)=\begin{cases}\frac{2 k \cos x}{\pi-2 x}, & x \neq \frac{\pi}{2} \\ 2024, & x=\frac{\pi}{2}\end{cases}$ then,the value of $k$ is . . . . . .

If the function $f(x)$,defined below,is continuous everywhere,then $k$ equals: $f(x)=\begin{cases} \frac{2^x-1}{\sqrt{1+x}-1}, & x \neq 0 \\ k, & x=0 \end{cases}$

For $x > 0$,let $h(x) = \begin{cases} \frac{1}{q} & \text{if } x = \frac{p}{q} \text{ (where } p, q \in \mathbb{N} \text{ are relatively prime)} \\ 0 & \text{if } x \text{ is irrational} \end{cases}$. Which of the following statements does not hold true?