If $f(x) = \int_{-1}^x |t| \, dt$,$x \ge -1$,then

If $f(x) = \frac{e^{x^2} - \cos x}{x^2}$ for $x \neq 0$ is continuous at $x = 0$,then $f(0) = $.

The function $f(x) = \cot x$ is discontinuous at every point of the set

Assertion $(A)$: $f(x)=|x-a|+|x-b|$ is continuous on $R$. Reason $(R)$: $\frac{|x-\alpha|}{x-\alpha}$ is continuous at $x \in R-\{\alpha\}$. The correct option among the following is:

Let $a, b \in R, b \neq 0$. Define a function $f(x) = \begin{cases} a \sin \frac{\pi}{2}(x-1), & \text{for } x \leq 0 \\ \frac{\tan 2x - \sin 2x}{bx^3}, & \text{for } x > 0 \end{cases}$. If $f$ is continuous at $x = 0$,then $10 - ab$ is equal to ...... .

Let $f: R \rightarrow R$ be defined as $f(x) = \begin{cases} 2 \sin \left(-\frac{\pi x}{2}\right), & \text{if } x < -1 \\ |ax^2 + x + b|, & \text{if } -1 \leq x \leq 1 \\ \sin(\pi x), & \text{if } x > 1 \end{cases}$. If $f(x)$ is continuous on $R$,then $a + b$ equals ..... .