For \( f(x) \) to be continuous at \( x = 0 \), we require:
\(\lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x)\).
Given: \(f(0) = \frac{0}{3} = 0\).
Right-hand limit:
Consider:
\(\lim_{x \to 0^+} \frac{\sqrt{ax + b^2x^2} - \sqrt{ax}}{b \sqrt{ax}}\).
To simplify, multiply the numerator and the denominator by the conjugate of the numerator:
\(\lim_{x \to 0^+} \frac{(\sqrt{ax + b^2x^2} - \sqrt{ax}) \times (\sqrt{ax + b^2x^2} + \sqrt{ax})}{b \sqrt{ax} \times (\sqrt{ax + b^2x^2} + \sqrt{ax})}\).
This simplifies to:
\(\lim_{x \to 0^+} \frac{ax + b^2x^2 - ax}{b \sqrt{ax} \times (\sqrt{ax + b^2x^2} + \sqrt{ax})}\).
Simplifying further:
\(\lim_{x \to 0^+} \frac{b^2x^2}{b \sqrt{ax} \times (\sqrt{ax + b^2x^2} + \sqrt{ax})}\).
Canceling terms and simplifying:
\(\lim_{x \to 0^+} \frac{bx}{\sqrt{a} \times 2 \sqrt{ax}} = \frac{b}{2a}\).
For continuity, we equate this limit to the value of \( f(0) \):
\(\frac{b}{2a} = 3 \implies \frac{b}{a} = 6\).
Therefore:
\(\boxed{6}\).
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