If $f(x)$ is continuous and $f\left( \frac{9}{2} \right) = \frac{2}{9}$,then $\lim_{x \to 0} f \left( \frac{1 - \cos 3x}{x^2} \right)$ is equal to:

If $f$ is defined by $f(x) = \begin{cases} \frac{1-\cos ax}{x \sin x}, & x \neq 0 \\ \frac{1}{2}, & x = 0 \end{cases}$ and $f$ is continuous at $x=0$,then $a^{2} =$ . . . . . . .

If $f(x) = \begin{cases} \frac{\cos(ax) - \cos(bx)}{x^2}, & x \neq 0 \\ \frac{1}{2}(b^2 - a^2), & x = 0 \end{cases}$ where $a$ and $b$ are real and distinct constants,then:

Let $[\bullet]$ denote the greatest integer function,and let $f(x) = \min \{\sqrt{2}x, x^2\}$. Let $S = \{x \in (-2, 2) : \text{the function } g(x) = |x|[x^2] \text{ is discontinuous at } x\}$. Then $\sum_{x \in S} f(x)$ equals:

If $f(x) = \begin{cases} \frac{\sqrt{\pi} - \sqrt{\cos^{-1} x}}{\sqrt{x+1}}, & x \neq -1 \\ \frac{1}{\sqrt{\lambda \pi}}, & x = -1 \end{cases}$ is right continuous at $x = -1$,then $\lambda = $

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