Let circle \( C \) be the image of
\[ x^2 + y^2 - 2x + 4y - 4 = 0 \]
in the line
\[ 2x - 3y + 5 = 0 \]
and \( A \) be the point on \( C \) such that \( OA \) is parallel to the x-axis and \( A \) lies on the right-hand side of the centre \( O \) of \( C \).
If \( B(\alpha, \beta) \), with \( \beta < 4 \), lies on \( C \) such that the length of the arc \( AB \) is \( \frac{1}{6} \) of the perimeter of \( C \), then \( \beta - \sqrt{3}\alpha \) is equal to:
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32