Let $f(x) = x^2 + x \sin x - \cos x$. Then

Let $f(x) = \frac{1-\tan x}{4x-\pi}$ for $x \neq \frac{\pi}{4}$ and $x \in [0, \frac{1}{2}]$. If $f(x)$ is continuous in $[0, \frac{\pi}{2}]$,then $f(\frac{\pi}{4})$ is

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

Define $f: R \rightarrow R$ by $f(x) = \begin{cases} (x-a) \frac{e^{\frac{1}{x-a}}-1}{e^{\frac{1}{x-a}}+1}, & x \neq a \\ 0, & x=a \end{cases}$. Then which one of the following is true?

Given $f(x) = \begin{cases} cx + 1, & x \leq 3 \\ dx + 3, & x > 3 \end{cases}$. If $f$ is continuous at $x = 3$,then $d - c =$ . . . . . . .

If $f(x) = \cos \left[ \frac{\pi}{x} \right] \cos \left( \frac{\pi}{2} (x - 1) \right)$,then $f(x)$ is continuous at: (where $[x]$ is the greatest integer function of $x$)