The problem consists of four independent questions: d) Newton's Law of Cooling: The temperature of a cold drink is initially $3^\circ C$. After 30 minutes in a $15^\circ C$ room, the temperature increases to $10^\circ C$. We need to find a formula for the temperature of the cold drink after $t$ hours, and determine when the temperature will be $14^\circ C$. a) Function Analysis: Given a function $f: \{1, 2, 3\} \rightarrow \{0, 4\}$ such that $f(1) = 0$, $f(2) = 4$, and $f(3) = 4$, we need to sketch the graph of the function, find where $f$ has a local minimum value, find where $f$ has a local minimum value but not a global minimum, and find where $f$ has a global maximum value. b) Definite Integral: Given that $f$ is continuous on $R$ and $\int_0^1 f(x) dx = 3$, we need to find $\int_\pi^{\frac{5\pi}{4}} \cos(2x) f(\sin(2x)) dx$. c) Fundamental Theorem of Calculus: Given $f(x) = \int_{x^2}^{2025} (e^{y^2+1}) dy$, we need to find $f'(x)$ using the Fundamental Theorem of Calculus part 1 and then find $f'(1)$.

1. Problem Description

The problem consists of four independent questions:

d) Newton's Law of Cooling: The temperature of a cold drink is initially $3^\circ C$. After 30 minutes in a $15^\circ C$ room, the temperature increases to $10^\circ C$. We need to find a formula for the temperature of the cold drink after $t$ hours, and determine when the temperature will be $14^\circ C$.

a) Function Analysis: Given a function $f: \{1, 2, 3\} \rightarrow \{0, 4\}$ such that $f(1) = 0$, $f(2) = 4$, and $f(3) = 4$, we need to sketch the graph of the function, find where $f$ has a local minimum value, find where $f$ has a local minimum value but not a global minimum, and find where $f$ has a global maximum value.

b) Definite Integral: Given that $f$ is continuous on $R$ and $\int_0^1 f(x) dx = 3$, we need to find $\int_\pi^{\frac{5\pi}{4}} \cos(2x) f(\sin(2x)) dx$.

c) Fundamental Theorem of Calculus: Given $f(x) = \int_{x^2}^{2025} (e^{y^2+1}) dy$, we need to find $f'(x)$ using the Fundamental Theorem of Calculus part 1 and then find $f'(1)$.

2. Solution Steps

d) (i) Newton's Law of Cooling states that the rate of change of temperature is proportional to the difference between the object's temperature and the ambient temperature. Let $T(t)$ be the temperature of the cold drink at time $t$ (in hours). Then

$\frac{dT}{dt} = k(T - 15)$

where $k$ is a constant. The solution to this differential equation is

$T(t) = 15 + (T_0 - 15)e^{kt}$

where $T_0$ is the initial temperature. We are given that $T_0 = 3$, so

$T(t) = 15 + (3 - 15)e^{kt} = 15 - 12e^{kt}$.

We are also given that $T(\frac{1}{2}) = 10$. Plugging this in, we get

$10 = 15 - 12e^{k/2}$

$-5 = -12e^{k/2}$

$e^{k/2} = \frac{5}{12}$

$\frac{k}{2} = \ln(\frac{5}{12})$

$k = 2\ln(\frac{5}{12})$

Thus, $T(t) = 15 - 12e^{2\ln(\frac{5}{12})t} = 15 - 12(e^{\ln(\frac{5}{12})})^{2t} = 15 - 12(\frac{5}{12})^{2t}$.

(ii) We want to find $t$ such that $T(t) = 14$. So,

$14 = 15 - 12(\frac{5}{12})^{2t}$

$-1 = -12(\frac{5}{12})^{2t}$

$(\frac{5}{12})^{2t} = \frac{1}{12}$

$2t \ln(\frac{5}{12}) = \ln(\frac{1}{12})$

$t = \frac{\ln(\frac{1}{12})}{2\ln(\frac{5}{12})} = \frac{-\ln(12)}{2(\ln(5) - \ln(12))} \approx \frac{-2.4849}{2(-0.8755)} \approx \frac{-2.4849}{-1.751} \approx 1.419$ hours.

To the nearest hour, $t = 1$ hour.

a) (i) The graph consists of three points: $(1, 0)$, $(2, 4)$, and $(3, 4)$.

(ii) A local minimum occurs at $x = 1$ because $f(1) = 0$ is less than $f(2) = 4$.

(iii) A local minimum that is not a global minimum doesn't exist in this function, since the minimum at $x=1$ is the global minimum as well. Thus, the answer is None.

(iv) Global maximum occurs at $x = 2$ and $x = 3$ since $f(2) = f(3) = 4$ are the largest function values.

b) Let $u = \sin(2x)$. Then $du = 2\cos(2x) dx$, so $\cos(2x) dx = \frac{1}{2} du$.

When $x = \pi$, $u = \sin(2\pi) = 0$. When $x = \frac{5\pi}{4}$, $u = \sin(\frac{5\pi}{2}) = \sin(\frac{\pi}{2}) = 1$.

$\int_\pi^{\frac{5\pi}{4}} \cos(2x) f(\sin(2x)) dx = \int_0^1 f(u) \frac{1}{2} du = \frac{1}{2} \int_0^1 f(u) du = \frac{1}{2} (3) = \frac{3}{2}$.

c) By the Fundamental Theorem of Calculus,

$f'(x) = \frac{d}{dx} \int_{x^2}^{2025} (e^{y^2+1}) dy = -e^{(x^2)^2+1} \frac{d}{dx} x^2 = -2x e^{x^4+1}$.

Therefore, $f'(1) = -2(1)e^{1^4+1} = -2e^2$.

3. Final Answer

d) (i) $T(t) = 15 - 12(\frac{5}{12})^{2t}$

(ii) 1 hour

a) (i) Graph with points (1,0), (2,4), (3,4)

(ii) $x = 1$

(iii) None

(iv) $x = 2, 3$

b) $\frac{3}{2}$

c) $f'(1) = -2e^2$