If $\mathop {\lim }\limits_{x \to 1} {\sin ^{ - 1}}\left( {\frac{k}{{\ln x}} - \frac{k}{{x - 1}}} \right)$ exists, then the number of integers in the range of $k$ is:

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:

If $\mathop {\lim }\limits_{x \to 1} \frac{{{x^4} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to k} \frac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}}$,then $k$ is

Let $\alpha, \beta \in R$ be such that $\lim _{x \rightarrow 0} \frac{x^2 \sin (\beta x)}{\alpha x-\sin x}=1$. Then $6(\alpha+\beta)$ equals

If $\mathop {\lim }\limits_{x \to 2} \frac{{\tan (x - 2)({x^2} + (a - 2)x - 2a)}}{{({x^2} - 4x + 4)}} = 7$,then $a$ is -

If $\lim _{x \rightarrow 1^{+}} \frac{(x-1)(6+\lambda \cos (x-1))+\mu \sin (1-x)}{(x-1)^3}=-1$,where $\lambda, \mu \in \mathbb{R}$,then $\lambda+\mu$ is equal to