If $f(x)$ defined as given below is continuous on $R$,then the value of $a+b$ is equal to: $f(x) = \begin{cases} \sin x, & x \leq 0 \\ x^2+a, & 0 < x < 1 \\ b x+3, & 1 \leq x \leq 3 \\ -3, & x > 3 \end{cases}$

If $f(x) = \begin{cases} 1 + x, & \text{when } x \le 2 \\ 5 - x, & \text{when } x > 2 \end{cases}$,then which of the following is true?

If function $f$ is continuous at point $x = \pi$ and $f(x) = \begin{cases} kx+1; & x \leq \pi \\ \cos x; & x > \pi \end{cases}$ then the value of $k$ is $\dots \dots \dots$

If $f(x) = \frac{10^x + 7^x - 14^x - 5^x}{1 - \cos x}$ for $x \neq 0$ is continuous at $x = 0$,then the value of $f(0)$ is:

If the function $f$ defined on $\left(\frac{\pi}{6}, \frac{\pi}{3}\right)$ by $f(x)=\begin{cases} \frac{\sqrt{2} \cos x-1}{\cot x-1}, & x \neq \frac{\pi}{4} \\ k, & x=\frac{\pi}{4} \end{cases}$ is continuous,then $k$ is equal to

If the function $f(x) = \begin{cases} \frac{\log_{e}(1-x+x^{2}) + \log_{e}(1+x+x^{2})}{\sec x - \cos x}, & x \in (-\frac{\pi}{2}, \frac{\pi}{2}) - \{0\} \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then $k$ is equal to.