Functions and Continuity

Relevant wiki: Maclaurin Series

For $f(x)$ to be continuous at $x=0$ , $\implies \displaystyle \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 3$ .

$\begin{aligned} f_-(x) & = \frac{a(1-x\color{#3D99F6}{\sin x}) + b\color{#3D99F6}{\cos x} + 5}{x^2} \quad \quad \small \color{#3D99F6}{\text{By Maclaurin series}} \\ & = \frac{1}{x^2}\left[a\left(1-x\color{#3D99F6}{\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... \right)}\right) + b\color{#3D99F6}{\left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ... \right)} + 5\right] \\ & = \frac{a}{x^2} - a + \frac{ax}{3!} - \frac{ax^3}{5!} - ... + \frac{b}{x^2} - \frac{b}{2!} + \frac{bx^2}{4!} - ... + \frac{5}{x^2} \end{aligned}$

For $f_-(0) = 3$ ,

$\implies \begin{cases} a+b+5 = 0 \\ - a - \dfrac{b}{2} = 3 \end{cases} \color{#D61F06}{\implies b = -4 \implies a = -1}$

$\begin{aligned} f_+(x) & = \left(1+\left(\dfrac{cx + dx^3}{x^2}\right)\right)^{\frac 1 x} \\ & = \left(1+\left(\dfrac{\color{#3D99F6}{\frac{c}{x^2}} + d}{\frac{1}{x}}\right)\right)^{\frac 1 x} \quad \quad \small {\color{#3D99F6}{\text{For }f_+(0) \text{ to converge when } x \to 0^+,}} \ \color{#D61F06}{c = 0} \\ & = \left[ \left(1+\dfrac{1}{\frac{1}{dx}} \right)^{\frac 1 {dx}} \right]^d \end{aligned}$

For $f_+(0) = 3$ ,

$\begin{aligned} \implies f_+(0) & = \lim_{x \to 0^+} \left[ \left(1+\dfrac{1}{\frac{1}{dx}} \right)^{\frac 1 {dx}} \right]^d \\ & = e^d = 3 \\ \color{#D61F06}{\implies d} & \color{#D61F06}{= \ln 3} \end{aligned}$

Therefore, $2e^d + 7c - 3a - 5b = 2e^{\ln 3} + 7(0) - 3(-1) - 5(-4) = 6+0+3+20 = \boxed{29}$