$f(x) = \begin{cases} \frac{(2x^2 - ax + 1) - (ax^2 + 3bx + 2)}{x + 1} & ; x \neq -1 \\ k & ; x = -1 \end{cases}$ is a real-valued function. If $a, b, k \in R$ and $f$ is continuous on $R$,then $k =$

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 ..... .

$f(x) = \begin{cases} [x] + [-x], & \text{when } x \neq 2 \\ \lambda, & \text{when } x = 2 \end{cases}$
If $f(x)$ is continuous at $x = 2$, the value of $\lambda$ will be

If $f(x) = \begin{cases} \frac{1-\sin^3 x}{3 \cos^2 x}, & x < \frac{\pi}{2} \\ \alpha, & x = \frac{\pi}{2} \\ \frac{\beta(1-\sin x)}{(\pi-2 x)^2}, & x > \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then $\alpha \beta =$

If the function $f(x) = \begin{cases} 1+\cos x, & x \leq 0 \\ a-x, & 0 < x \leq 2 \\ x^2-b^2, & x > 2 \end{cases}$ is continuous everywhere,then $a^2+b^2=$

Let $f: R \rightarrow R$ be a continuous function satisfying $f(0)=1$ and $f(2x)-f(x)=x$ for all $x \in R$. If $\lim_{n \rightarrow \infty} \{f(x)-f(\frac{x}{2^n})\} = G(x)$,then $\sum_{r=1}^{10} G(r^2)$ is equal to