Let $f(x) = x^2 + x \sin x - \cos x$. Then

If $f: [-2, 2] \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{\sqrt{1 + cx} - \sqrt{1 - cx}}{x}, & -2 \leq x < 0 \\ \frac{x + 3}{x + 1}, & 0 \leq x \leq 2 \end{cases}$ is continuous on $[-2, 2]$,then $c$ is equal to

If $f(x) = \begin{cases} 1 + \cos x, & x \le 0 \\ a - x, & 0 < x < 2 \\ (x - b)^2, & x \ge 2 \end{cases}$ is continuous at $x=0$ and $x=2$, then find the value of $a^2+b^2$.

Let $f(x) = \begin{cases} 1 + 6x - 3x^2, & x \leq 1 \\ x + \log_2(b^2 + 7), & x > 1 \end{cases}$. Then the set of all possible values of $b$ such that $f(1)$ is the maximum value of $f(x)$ is

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?

If $f(x) = \begin{cases} 1 + x^2, & \text{when } 0 \le x \le 1 \\ 1 - x, & \text{when } x > 1 \end{cases}$,then