To ensure continuity at \(x = 0\), we require \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \).
Left-hand limit:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1 - \cos 2x}{x^2} = 2 \]
This gives \( f(0) = \alpha = 2 \).
Right-hand limit:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\beta \sqrt{1 - \cos x}}{x} = \frac{\beta}{\sqrt{2}} = 2 \implies \beta = 2\sqrt{2} \]
Calculating \( \alpha^2 + \beta^2 \):
\[ \alpha^2 + \beta^2 = 4 + 8 = 12 \]
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