If $f(x) = \begin{cases} x - 1, & x < 0 \\ \frac{1}{4}, & x = 0 \\ x^2, & x > 0 \end{cases}$,then

For the function $f(x) = \begin{cases} \frac{e^{1/x} - 1}{e^{1/x} + 1}, & x \ne 0 \\ 0, & x = 0 \end{cases}$,which of the following is correct?

Let $f(x) = \begin{cases} (x - 1)^{\frac{1}{2 - x}}, & x > 1, x \neq 2 \\ k, & x = 2 \end{cases}$. The value of $k$ for which $f$ is continuous at $x = 2$ is

If for $p \neq q \neq 0$,the function $f(x) = \frac{\sqrt[7]{p(729+x)}-3}{\sqrt[3]{729+qx}-9}$ is continuous at $x=0$,then:

If $f(x) = \begin{cases} \frac{5}{2} - x, & x < 2 \\ 1, & x = 2 \\ x - \frac{3}{2}, & x > 2 \end{cases}$,then:

If the function $f(x)$ is continuous in $0 \leq x \leq \pi$,then the value of $2a+3b$ is where $f(x) = \begin{cases} x+a \sqrt{2} \sin x & \text{if } 0 \leq x < \frac{\pi}{4} \\ 2x \cot x + b & \text{if } \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \\ a \cos 2x - b \sin x & \text{if } \frac{\pi}{2} < x \leq \pi \end{cases}$