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Given that:

$\lim_{x \to 2} \frac{\sin(x^3 - 5x^2 + ax + b)}{(\sqrt{x-1} - 1)\log_e(x-1)} = m$

Calculation:

$\because$ Denominator = 0 at $x = 2$

$\Rightarrow$ Numerator = 0 at $x = 2$

$\Rightarrow 2^3 - 5(2)^2 + a(2) + b = 0$

$\Rightarrow 2a + b = 12$ ... (1)

$m = \lim_{x \to 2} \frac{\frac{\sin(x^3 - 5x^2 + ax + b)}{(x^3 - 5x^2 + ax + b)} \cdot (x^3 - 5x^2 + ax + b)}{\frac{\sqrt{x-1} - 1}{\sqrt{x-1} + 1} \cdot \frac{\log_e(1 + (x-2))}{(x-2)} \cdot (x-2)}$

$\Rightarrow m = \lim_{x \to 2} 2 \frac{x^3 - 5x^2 + ax + b}{(x-2)^2}$

$\Rightarrow m = \lim_{x \to 2} 2 \frac{3x^2 - 10x + a}{2(x-2)}$

$\because$ Denominator = 0 at $x = 2$

$\Rightarrow$ Numerator = 0 at $x = 2$

$\Rightarrow 3(2)^2 - 10(2) + a = 0$

$\Rightarrow a = 8$

Substituting $a = 8$ in (1):

$2(8) + b = 12$

$\Rightarrow b = -4$

$\therefore m = \lim_{x \to 2} 2 \left( \frac{6x - 10}{2} \right)$

$\Rightarrow m = 2$

$\therefore a + b + m = 8 - 4 + 2 = 6$

∴ The correct answer is option (2).

Given that:

$\frac{dy}{dx} = y(2 + \ln x)$

Calculation:

$\int \frac{1}{y} , dy = \int (2 + \ln x) , dx$

⇒ $\ln y = 2x + x \ln x - x + C$

⇒ $\ln y = x + x \ln x + C$

Given point (1, e): $\ln e = 1 + 1 \ln 1 + C$

⇒ $1 = 1 + 0 + C \Rightarrow C = 0$

⇒ $\ln y = x + x \ln x$

⇒ $f(x) = y = e^{x + x \ln x}$

⇒ $f(e) = e^{e + e} = e^{2e}$

∴ The correct answer is option (3).

Given that:

$f(x) = \begin{cases} \left| \frac{\sin x}{x} \right| & , \quad x \neq 0 \ 1 & , \quad x = 0 \end{cases}$, $x \in (-2\pi, 2\pi)$

Calculation:

$\lim_{x \to 0} \left| \frac{\sin x}{x} \right| = 1 = f(0)$, hence $f(x)$ is continuous.

$f'(x) = \frac{d}{dx} \left| \frac{\sin x}{x} \right| = \frac{x \cos x - \sin x}{x^2}$

$f'(x) = 0 \Rightarrow \tan x = x$

From the graph, there are 3 solutions to $\tan x = x$ in $(-2\pi, 2\pi)$.

Additionally, $f'(x)$ does not exist at $x = -\pi, \pi$.

Total critical points = 3 (from $\tan x = x$) + 2 (where $f'(x)$ does not exist) = 5.

∴ The correct answer is option (3).

Given that:

$\int_{0}^{3}\dfrac{e^x+e^{-x}}{[x]!},dx$

Calculation:

$\int_{0}^{3}\dfrac{e^x+e^{-x}}{[x]!},dx$

⇒ $\int_{0}^{1}(e^x+e^{-x}),dx+\int_{1}^{2}(e^x+e^{-x}),dx+\dfrac{1}{2}\int_{2}^{3}(e^x+e^{-x}),dx$

⇒ \(\left[e^x-e^{-x}\right]{0}^{1}+\left[e^x-e^{-x}\right]{1}^{2}+\dfrac{1}{2}\left[e^x-e^{-x}\right]_{2}^{3}\)

⇒ $\left(e-\dfrac{1}{e}\right)+\left(e^2-e-\dfrac{1}{e^2}+\dfrac{1}{e}\right)+\dfrac{1}{2}\left(e^3-e^2-\dfrac{1}{e^3}+\dfrac{1}{e^2}\right)$

⇒ $\dfrac{1}{2}\left(e^2+e^3-\dfrac{1}{e^2}-\dfrac{1}{e^3}\right)$

∴ The correct answer is option (2).

Given that:

$(1 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0, ; y(0) = 0$

Calculation:

$\int \frac{dy}{y+1} = -\int \frac{\cos x}{1+\sin x} dx$

⇒ $\ln(y+1) = -\ln(1+\sin x) + C$

⇒ $y+1 = \frac{e^{C}}{1+\sin x}$

Using $y(0) = 0$:

⇒ $1+1 = \frac{e^{C}}{1+\sin 0}$

⇒ $C = 0$

⇒ $y+1 = \frac{1}{1+\sin x}$

Given $y = -\frac{1}{2}, ; x = \alpha$:

⇒ $-\frac{1}{2} + 1 = \frac{1}{1+\sin \alpha}$

⇒ $\frac{1}{2} = \frac{1}{1+\sin \alpha}$

⇒ $1+\sin \alpha = 2$

⇒ $\sin \alpha = 1$

⇒ $\alpha = \frac{\pi}{2}$

∴ The correct answer is option (4).

Given that:

$\log_{0.6} \left| \frac{2x - 5}{x^2 - 4} \right| \geq 0$

$\left| \frac{2x - 5}{x^2 - 4} \right| \leq 1$

$x \neq \frac{5}{2}$

$-1 \leq \frac{2x - 5}{x^2 - 4} \leq 1$

$\frac{(x+1)^2 - 10}{(x-2)(x+2)} \geq 0$

$x \in (-\infty, -1-\sqrt{10}] \cup (-2, 2) \cup [-1+\sqrt{10}, \infty)$

$\frac{(x-1)^2}{(x-2)(x+2)} \geq 0$

$x \in (-\infty, -2) \cup (2, \infty) \cup {1}$

Calculation:

$(1) \cap (2) \cap (3)$

$x \in (-\infty, -1-\sqrt{10}] \cup {1} \cup \left[ -1+\sqrt{10}, \frac{5}{2} \right) \cup \left( \frac{5}{2}, \infty \right)$

$a + b + c + d + e = -2 + 1 + 5 = 4$

Given that:

$a_1 + a_2 + \ldots + a_n = 6n^3$

$a_1 + a_2 + \ldots + a_n + a_{n+1} = 6(n+1)^3$

Calculation:

$a_{n+1} = 6(n+1)^3 - 6n^3$

⇒ $a_{n+1} = 6((n+1)-n)((n+1)^2 + n^2 + n(n+1))$

⇒ $a_{n+1} = 6(1)(3n^2 + 3n + 1)$

⇒ $a_n = 6(3(n-1)^2 + 3(n-1) + 1)$

⇒ $a_n = 6(3n^2 - 3n + 1)$

⇒ $\sum_{k=1}^6 \left( \frac{6(3k^2 + 3k + 1) - 6(3k^2 - 3k + 1)}{36} \right)^2$

⇒ $\sum_{k=1}^6 \left( \frac{36k}{36} \right)^2 = \sum_{k=1}^6 k^2 = \frac{6 × 7 × 13}{6} = 91$

Given that:

$8b^2 - 4ac < 0 \Rightarrow 2b^2 < ac$

Calculation:

Total combinations of $b, a, c$:

$1 \times 4 + 1 \times 3 + 1 \times 4 + 1 \times 4 + 1 \times 2 + 1 \times 2 = 19$

Valid combinations satisfying $2b^2 < ac$:

$2 + 3 + 4 + 4 + 2 + 2 = 17$

Required probability:

$\frac{17}{64} = \frac{m}{n}$, where $m = 17, n = 64$

$m + n = 17 + 64 = 81$

Given that:

$r^2 = \left( \frac{\sqrt{14}}{2} \right)^2 + \left( r - \frac{1}{\sqrt{2}} \right)^2$

Calculation:

$r^2 = \frac{7}{2} + \left( r - \frac{1}{\sqrt{2}} \right)^2$

⇒ $r^2 = \frac{7}{2} + r^2 - \sqrt{2}r + \frac{1}{2}$

⇒ $\sqrt{2}r = 4$

⇒ $r = 2\sqrt{2}$

⇒ $r^2 = 8$

Given that:

$\int_a^b f(x) dx + \int_{f(a)}^{f(b)} f^{-1}(x) dx = bf(b) - af(a)$

$\alpha = (2\sqrt{3})(4) - (0)(2)$

Calculation:

$\alpha = 8\sqrt{3}$

$\alpha^2 = 192$