Question:

Step 1: Establish the continuity condition at $x = 0$. For $f(x)$ to be continuous at $x = 0$, we need: $$\lim_{x \to 0} f(x) = f(0)$$ Computing $f(0)$: $$f(0) = e^{0^2 - 2(0) + 3} = e^3$$ Therefore, we require: $$\lim_{x \to 0} \left(\sin\left(\frac{2x^2}{a}\right) + \cos\left(\frac{3x}{b}\right)\right)^{\frac{ab}{x^2}} = e^3$$ Step 2: Apply Taylor series expansions as $x \to 0$. To evaluate the limit, we use standard Taylor expansions: $$\sin\left(\frac{2x^2}{a}\right) \approx \frac{2x^2}{a}$$ $$\cos\left(\frac{3x}{b}\right) \approx 1 - \frac{1}{2}\left(\frac{3x}{b}\right)^2 = 1 - \frac{9x^2}{2b^2}$$ Therefore, the base of the exponent becomes: $$\sin\left(\frac{2x^2}{a}\right) + \cos\left(\frac{3x}{b}\right) \approx 1 + \frac{2x^2}{a} - \frac{9x^2}{2b^2}$$ Step 3: Evaluate the limit using the standard form $(1 + u)^{1/u} \to e$. Let $L = \lim_{x \to 0} \left(\sin\left(\frac{2x^2}{a}\right) + \cos\left(\frac{3x}{b}\right)\right)^{\frac{ab}{x^2}}$ Taking the natural logarithm: $$\ln L = \lim_{x \to 0} \frac{ab}{x^2} \cdot \left(\frac{2x^2}{a} - \frac{9x^2}{2b^2}\right)$$ $$\ln L = \lim_{x \to 0} ab \cdot \left(\frac{2}{a} - \frac{9}{2b^2}\right)$$ $$\ln L = 2b - \frac{9a}{2b}$$ Therefore: $$L = e^{2b - \frac{9a}{2b}}$$ Step 4: Apply the continuity condition to establish a constraint. Setting $L = e^3$: $$e^{2b - \frac{9a}{2b}} = e^3$$ Equating exponents: $$2b - \frac{9a}{2b} = 3$$ Step 5: Solve for $a$ in terms of $b$. Rearranging the equation: $$2b - 3 = \frac{9a}{2b}$$ $$a = \frac{2b(2b - 3)}{9} = \frac{4b^2 - 6b}{9}$$ Step 6: Find the minimum value of $a$ by optimizing over $b$. To minimize $a$, we differentiate with respect to $b$: $$\frac{da}{db} = \frac{8b - 6}{9}$$ Setting $\frac{da}{db} = 0$: $$8b - 6 = 0 \implies b = \frac{3}{4}$$ Substituting $b = \frac{3}{4}$ back into the expression for $a$: $$a_{\min} = \frac{4\left(\frac{3}{4}\right)^2 - 6\left(\frac{3}{4}\right)}{9}$$ $$a_{\min} = \frac{4 \cdot \frac{9}{16} - \frac{18}{4}}{9} = \frac{\frac{9}{4} - \frac{9}{2}}{9}$$ $$a_{\min} = \frac{\frac{9}{4} - \frac{18}{4}}{9} = \frac{-\frac{9}{4}}{9} = -\frac{1}{4}$$ **Final Answer:** The minimum value of $a$ is $\boxed{-\dfrac{1}{4}}$, which corresponds to **Option 2**.