If the function
$ f(x) = \begin{cases} \frac{\cos ax - \cos 9x}{x^2}, & \text{if } x \neq 0 \\ 16, & \text{if } x = 0 \end{cases} $
is continuous at $ x = 0 $, then $ a = ? $
$ \lim_{x \to -\frac{3}{2}} \frac{(4x^2 - 6x)(4x^2 + 6x + 9)}{\sqrt{2x - \sqrt{3}}} $
\[ f(x) = \begin{cases} \frac{(4^x - 1)^4 \cot(x \log 4)}{\sin(x \log 4) \log(1 + x^2 \log 4)}, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases} \]
Find \( e^k \) if \( f(x) \) is continuous at \( x = 0 \).
If
\[ A = \{ P(\alpha, \beta) \mid \text{the tangent drawn at P to the curve } y^3 - 3xy + 2 = 0 \text{ is a horizontal line} \} \]
and
\[ B = \{ Q(a, b) \mid \text{the tangent drawn at Q to the curve } y^3 - 3xy + 2 = 0 \text{ is a vertical line} \} \]
then \( n(A) + n(B) = \)
Evaluate the integral: \[ \int \frac{\sec x}{3(\sec x + \tan x) + 2} \,dx \]
\[ y = \sqrt{\sin(\log(2x)) + \sqrt{\sin(\log(2x)) + \sqrt{\sin(\log(2x))} + \dots \infty}} \]
The area of the region (in square units) enclosed by the curves \( y = 8x^3 - 1 \), \( y = 0 \), \( x = -1 \), and \( x = 1 \) is: