If $f(x) = \begin{cases} 3ax - 2b, & x > 1 \\ ax + b + 1, & x < 1 \end{cases}$ and $\lim_{x \rightarrow 1} f(x)$ exists,then the relation between $a$ and $b$ is

For $t > -1$,let $\alpha_t$ and $\beta_t$ be the roots of the equation $\left((t+2)^{\frac{1}{7}}-1\right) x^2+\left((t+2)^{\frac{1}{6}}-1\right) x+\left((t+2)^{\frac{1}{21}}-1\right)=0$. If $\lim _{t \rightarrow -1^{+}} \alpha_t$ and $\lim _{t \rightarrow -1^{+}} \beta_t$ are the roots of the limiting equation,and $a+b$ is the sum of these roots,then $72(a+b)^2$ is equal to . . . . . . .

If $\lim _{x \rightarrow \infty}\left(\frac{x^2+1}{x+1}-a x-b\right)=0$, where $a, b \in R$, then:

The values of the constants $\alpha$ and $\beta$ such that $\lim_{x \to \infty} \left( \frac{x^2 + 1}{x + 1} - \alpha x - \beta \right) = 0$ are respectively:

If $\lim _{x \rightarrow 1} \frac{x^2-ax+b}{x-1}=5$,then $(a+b)$ is equal to

If $f(x) = \begin{cases} 1+\frac{2x}{a}, & 0 \leq x \leq 1 \\ ax, & 1 < x \leq 2 \end{cases}$,and $\lim_{x \rightarrow 1} f(x)$ exists,then the sum of the cubes of the possible values of $a$ is: