Let $[x]$ denote the greatest integer less than or equal to $x$. Then, the value of $\alpha$ for which the function $f(x)=\begin{cases} \frac{\sin [-x^2]}{[-x^2]}, & x \neq 0 \\ \alpha, & x=0 \end{cases}$ is continuous at $x=0$ is:

The value$(s)$ of $x$ for which the function $f(x) = \begin{cases} 1-x, & x < 1 \\ (1-x)(2-x), & 1 \leq x \leq 2 \\ 3-x, & x > 2 \end{cases}$ fails to be continuous is(are):

If $f(x) = \frac{x+x^2+x^3+\ldots+x^{n}-n}{x-1}$ for $x \neq 1$ is continuous at $x=1$,then $f(1) =$

If the function $f(x) = \begin{cases} \frac{1}{x} \log_{e}\left(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}\right), & x < 0 \\ k, & x = 0 \\ \frac{\cos^{2} x - \sin^{2} x - 1}{\sqrt{x^{2}+1}-1}, & x > 0 \end{cases}$ is continuous at $x = 0$,then $\frac{1}{a} + \frac{1}{b} + \frac{4}{k}$ is equal to:

Let $[x]$ denote the greatest integer less than or equal to $x$. Then $f(x) = \frac{1 + \sin([\cos x])}{\cos([\sin x])}$ is

Discuss the continuity of the function defined by $f(x) = \begin{cases} x + 2, & \text{if } x < 0 \\ -x + 2, & \text{if } x > 0 \end{cases}$