CSEC Examination: Additional Mathematics: Paper 02: General Proficiency

$ (a.) (i.) \\[3ex] y = 2^x \\[3ex] 3(4^x) - 2^x - 4 = 0 \\[3ex] ................................... \\[3ex] 4^x \\[3ex] = (2^2)^x \\[3ex] = 2^{2x} \quad\dots\text{Law 5 Exp} \\[3ex] = (2^x)^2 \\[3ex] ................................... \\[3ex] 3(2^x)^2 - 2^x - 4 = 0 \\[3ex] \text{Substitute for } y \\[3ex] (ii.) \\[3ex] 3y^2 - y - 4 = 0 \\[3ex] 3y^2(-4) = -12y^2 \\[3ex] \text{Factors are: } 3y \text{ and } -4y \\[3ex] 3y^2 + 3y - 4y - 4 = 0 \\[3ex] 3y(y + 1) - 4(y + 1) = 0 \\[3ex] (y + 1)(3y - 4) = 0 \\[3ex] y + 1 = 0 \text{ or } 3y - 4 = 0 \\[3ex] y = -1 \text{ or } 3y = 4 \\[3ex] \hspace{3em} \text{ or } y = \dfrac{4}{3} \\[5ex] \text{Resubstitute for } y \text{ to solve for }x \\[3ex] \text{When } y = -1 \\[3ex] 2^x = -1 \\[3ex] \log 2^x = \log (-1) \\[3ex] \text{Stop. The logarithm of a negative number does not exist.} \\[5ex] \text{When } y = \dfrac{4}{3} \\[5ex] 2^x = \dfrac{4}{3} \\[5ex] \log 2^x = \log\left(\dfrac{4}{3}\right) \\[5ex] x \log 2 = \log 4 - \log 3 \quad\dots\text{Laws 5 and 2 Log} \\[3ex] x = \dfrac{\log 4 - \log 3}{\log 2} \\[5ex] (b.) \\[3ex] \dfrac{x + 5}{x + 1} \le 0 \\[5ex] \text{Assume: } \\[3ex] \text{Numerator: } x + 5 = 0 \\[3ex] x = -5 \\[5ex] \text{Denominator: } x + 1 = 0 \\[3ex] x = -1 \\[5ex] \underline{\text{Test Intervals are:}} \\[3ex] x \lt -5 \\[3ex] -5 \le x \le -1 \\[3ex] x \gt -1 \\[3ex] $

Let:	x < −5 x = −10	−5 ≤ x ≤ −1 x = −2	x > −1 x = 0
$x + 5$	−	+	+
$x + 1$	−	−	+
$\dfrac{x + 5}{x + 1}$	+	−	+

Less than or equal to zero means nonpositive
So, we are looking at the region that gives a negative value or a zero value
Hence, we focus on the second test interval

$ \text{Testing the end points:} \\[3ex] \text{When } x = -5 \\[3ex] \dfrac{x + 5}{x + 1} = \dfrac{-5 + 5}{-5 + 1} = \dfrac{0}{-4} = 0 \quad\dots\text{okay} \\[5ex] \text{When } x = -1 \\[3ex] \dfrac{x + 5}{x + 1} = \dfrac{-1 + 5}{-1 + 1} = \dfrac{-4}{0} \quad\dots\text{does not exist} \\[5ex] \text{Therefore} \\[3ex] \text{Set Notation: } \{x: -5 \le x \lt -1\} \\[3ex] \text{Interval Notation: } [-5, -1) \\[3ex] $ (c.) (i.)
Let the salaries for each year be shown as:

$ \begin{array}{*{5}{c}} \text{1st Year} & \text{2nd Year} & \text{3rd Year} & \text{4th Year} & \text{5th Year} \\[1pt] \underline{\qquad\qquad} & \underline{\qquad\qquad} & \underline{\qquad\qquad} & \underline{\qquad\qquad} & \underline{\qquad\qquad} \\[1pt] a & b & 56000 & c & 61000 \end{array} $

$ AS_n = a + d(n - 1) \quad\dots\text{nth term of an Arithmetic Sequence} \\[3ex] \text{where} \\[3ex] \text{first term} = a \\[3ex] \text{common difference} = d \\[3ex] \text{number of terms} = n \\[5ex] AS_3 = a + d(3 - 1) \\[3ex] 56000 = a + 2d \\[3ex] a + 2d = 56000 \quad\dots eqn.(1) \\[5ex] AS_5 = a + d(5 - 1) \\[3ex] a + 4d = 61000 \quad\dots eqn.(2) \\[5ex] \text{Elimination and Substitution Method} \\[3ex] eqn.(2) - eqn.(1) \implies \\[3ex] (a + 4d) - (a + 2d) = 61000 - 56000 \\[3ex] a + 4d - a - 2d = 5000 \\[3ex] 2d = 5000 \\[3ex] d = \dfrac{5000}{2} \\[5ex] d = 2500 \\[5ex] \text{Substitute for } d \text{ in } eqn.(1) \\[3ex] a + 2(2500) = 56000 \\[3ex] a = 56000 - 5000 \\[3ex] a = 51000 \\[5ex] SAS_n = \dfrac{n[2a + d(n - 1)]}{2} \quad\dots\text{sum of the n terms of an Arithmetic Sequence} \\[3ex] SAS_5 = \dfrac{5[2(51000) + 2500(5 - 1)]}{2} \\[5ex] = \dfrac{5(102000 + 10000)}{2} \\[5ex] = \dfrac{560000}{2} \\[5ex] = \$280000 \\[5ex] \text{Verify} \\[3ex] \begin{array}{*{5}{c}} \text{1st Year} & \text{2nd Year} & \text{3rd Year} & \text{4th Year} & \text{5th Year} \\[1pt] \underline{\qquad\qquad} & \underline{\qquad\qquad} & \underline{\qquad\qquad} & \underline{\qquad\qquad} & \underline{\qquad\qquad} \\[1pt] 51000 & (51000 + 2500) & 56000 & (56000 + 2500) & 61000 \end{array} $

$ \text{Total amount at the end of the five-year period} \\[3ex] = 51000 + 53500 + 56000 + 58500 + 61000 \\[3ex] = \$280000 \\[5ex] (ii.) \\[3ex] 6 - \dfrac{18}{5} + \dfrac{54}{25} - \dfrac{162}{125} + ... \\[5ex] \text{first term} = a = 6 \\[3ex] \text{common ratio} = r \\[3ex] \text{sum to infinity of a geometric series} = S_{\infty} \\[5ex] r = -\dfrac{18}{5} \div 6 \\[5ex] r = -\dfrac{18}{5} * \dfrac{1}{6} \\[5ex] r = -\dfrac{3}{5} \\[5ex] S_{\infty} = \dfrac{a}{1 - r} \\[5ex] = 6 \div \left[1 - \left(-\dfrac{3}{5}\right)\right] \\[5ex] = 6 \div \left[1 + \dfrac{3}{5}\right] \\[5ex] = 6 \div \left[\dfrac{5}{5} + \dfrac{3}{5}\right] \\[5ex] = 6 \div \dfrac{8}{5} \\[5ex] = 6 \cdot \dfrac{5}{8} \\[5ex] = \dfrac{15}{4} $

$ (a.) (i.) \\[3ex] (x - h)^2 + (y - k)^2 = r^2 \quad\dots\text{Standard Equation of a Circle} \\[3ex] \text{where } \\[3ex] (h, k) = \text{center of the circle} \\[3ex] r = \text{radius of the circle} \\[5ex] (x - 2)^2 + (y - 1)^2 = 5 \\[3ex] \text{Compare to the standard equation of a circle} \\[3ex] r^2 = 5 \\[3ex] r = \sqrt{5} \\[5ex] (ii.) \\[3ex] \underline{\text{Finding the points of intersection}} \\[3ex] x-\text{intercept} \\[3ex] \text{set } y = 0 \text{ and solve for } x \\[3ex] (x - 2)^2 + (0 - 1)^2 = 5 \\[3ex] (x - 2)^2 + (-1)^2 = 5 \\[3ex] (x - 2)^2 + 1 = 5 \\[3ex] (x - 2)^2 = 5 - 1 \\[3ex] (x - 2)^2 = 4 \\[3ex] x - 2 = \pm\sqrt{4} \\[3ex] x = 2 \pm 2 \\[3ex] x = 2 + 2 \text{ or } 2 - 2 \\[3ex] x = 4 \text{ or } 0 \\[3ex] x-\text{intercepts are: } (0, 0) \text{ and } (4, 0) \\[5ex] y-\text{intercept} \\[3ex] \text{set } x = 0 \text{ and solve for } y \\[3ex] (0 - 2)^2 + (y - 1)^2 = 5 \\[3ex] (-2)^2 + (y - 1)^2 = 5 \\[3ex] 4 + (y - 1)^2 = 5 \\[3ex] (y - 1)^2 = 5 - 4 \\[3ex] (y - 1)^2 = 1 \\[3ex] y - 1 = \pm\sqrt{1} \\[3ex] y = 1 \pm 1 \\[3ex] y = 1 + 1 \text{ or } 1 - 1 \\[3ex] y = 2 \text{ or } 0 \\[3ex] y-\text{intercepts are: } (0, 0) \text{ and } (0, 2) \\[5ex] \text{Points of Intersection are:} \\[3ex] (0, 0) \\[3ex] (0, 2) \\[3ex] (4, 0) \\[5ex] (b.) \\[3ex] |BC| = |AC| = 20\;m \\[3ex] \angle ACB = \theta \quad\dots\text{Diagram} \\[3ex] \text{Area of the building space} = \text{Area of sector ACB} \\[3ex] (i.) \\[3ex] \text{Area of the building space} = \dfrac{200\pi}{3}\;m^2 \quad\dots\text{Given} \\[5ex] \text{Area of sector ACB} = \dfrac{\pi r^2\theta}{360} \\[5ex] \implies \\[3ex] \dfrac{\pi r^2\theta}{360} = \dfrac{200\pi}{3} \\[5ex] \theta = \dfrac{200\pi}{3} \cdot 360 \cdot \dfrac{1}{\pi r^2} \\[5ex] = 200 \cdot 120 \cdot \dfrac{1}{20^2} \\[5ex] = \dfrac{200 \cdot 120}{400} \\[5ex] = 20 \cdot 3 \\[3ex] = 60^\circ \\[3ex] \text{Convert to radians using the Unity Fraction Method} \\[3ex] 180^\circ = \pi \;RAD \\[3ex] 60^\circ \cdot \dfrac{\pi\;RAD}{180^\circ} \\[5ex] \theta = \dfrac{\pi}{3}\;RAD \\[5ex] (ii.) \\[3ex] \underline{\text{Based on the Diagram}} \\[3ex] \angle BCD = 90^\circ \\[3ex] \angle BCD = \angle ACB + \angle ACD \\[3ex] 90 = \theta + \angle ACD \\[3ex] \angle ACD = 90 - \theta \\[3ex] = 90 - 60 \\[3ex] = 30^\circ \\[5ex] \text{Area of the land used for farming} = \text{Area of Right } \triangle ACD \\[3ex] \underline{\text{Right} \triangle ACD} \\[3ex] |AC| = \text{radius of the circle} = \text{hypotenuse of the right triangle} = 20\;m \\[3ex] |AD| = \perp\text{height} \\[3ex] |DC| = \text{base} \\[5ex] \underline{\text{SOHCAHTOA}} \\[3ex] \sin\angle ACD = \dfrac{|AD|}{AC|} \\[5ex] \sin 30^\circ = \dfrac{|AD|}{20} \\[5ex] |AD| = 20\sin 30 \\[3ex] = 20 \cdot \dfrac{1}{2} \\[5ex] = 10\;m \\[5ex] \cos\angle ACD = \dfrac{|DC|}{AC|} \\[5ex] \cos 30^\circ = \dfrac{|DC|}{20} \\[5ex] |DC| = 20\cos 30 \\[3ex] = 20 \cdot \dfrac{\sqrt{3}}{2} \\[5ex] = 10\sqrt{3}\;m \\[5ex] \text{Area of Right } \triangle ACD = \dfrac{1}{2} \cdot \text{base } \cdot \perp\text{height} \\[3ex] = \dfrac{1}{2} \cdot |DC| \cdot |AD| \\[5ex] = \dfrac{1}{2} \cdot 10\sqrt{3} \cdot 10 \\[5ex] = 50\sqrt{3}\;m^2 \\[5ex] (c.) \\[3ex] \overrightarrow{OB} = 3i - 4j \\[3ex] \overrightarrow{BC} = -5i + 13j \\[3ex] (i.) \\[3ex] \overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{BC} \\[3ex] = (3i - 4j) + (-5i + 13j) \\[3ex] = 3i - 4j - 5i + 13j \\[3ex] = -2i + 9j \\[5ex] (ii.) \\[3ex] \langle Vector \rangle = x i + y j \quad\dots\text{vector in component form} \\[3ex] \overrightarrow{OB} = 3i - 4j = x_1 i + y_1 j \\[3ex] x_1 = 3 \\[3ex] y_1 = -4 \\[5ex] \overrightarrow{OC} = -2i + 9j = x_2 i + y_2 j \\[3ex] x_2 = -2 \\[3ex] y_2 = 9 \\[5ex] \overrightarrow{OB} \cdot \overrightarrow{OC} \\[3ex] = (x_1 \cdot x_2) + (y_1 \cdot y_2) \\[3ex] = 3(-2) + -4(9) \\[3ex] = -6 - 36 \\[3ex] = -52 \\[5ex] (d.) \\[3ex] c^2 = a^2 + b^2 \quad\dots\text{Pythagorean Theorem} \\[5ex] \underline{\text{SOHCAHTOA}} \\[3ex] \sin x = \dfrac{a}{c} \\[5ex] \cos x = \dfrac{b}{c} \\[5ex] \text{Prove the identity } \sin^2 x + \cos^2 x = 1 \\[3ex] \text{From the LHS} \\[3ex] \sin^2 x = \left(\dfrac{a}{c}\right) = \dfrac{a^2}{c^2} \\[5ex] \cos^2 x = \left(\dfrac{b}{c}\right) = \dfrac{b^2}{c^2} \\[5ex] \sin^2 x + \cos^2 x \\[3ex] = \dfrac{a^2}{c^2} + \dfrac{b^2}{c^2} \\[5ex] = \dfrac{a^2 + b^2}{c^2} \\[5ex] = \dfrac{c^2}{c^2} \\[5ex] = 1 \\[3ex] = \text{RHS} $

$ (a.) (i.) \\[3ex] f(x) = \sqrt{x} - \cos 4x \\[3ex] ........................................................................ \\[3ex] y = \sqrt{x} \\[3ex] = x^{\dfrac{1}{2}} \quad\dots\text{Law 7 Exp} \\[5ex] y' = \dfrac{1}{2} \cdot x^{\dfrac{1}{2} - 1} \quad\dots\text{Power Rule} \\[7ex] = \dfrac{1}{2} \cdot x^{-\dfrac{1}{2}} \\[7ex] = \dfrac{1}{2} \cdot \dfrac{1}{x^{\dfrac{1}{2}}} \quad\dots\text{Law 6 Exp} \\[9ex] = \dfrac{1}{2} \cdot \dfrac{1}{\sqrt{x}} \quad\dots\text{Law 7 Exp} \\[5ex] = \dfrac{1}{2\sqrt{x}} \\[5ex] y = \cos 4x \\[3ex] \text{Let } p = 4x \hspace{3em} \implies y = \cos p \\[3ex] \dfrac{dp}{dx} = 4 \hspace{3.5em} \dfrac{dy}{dp} = -\sin p \\[5ex] \dfrac{dy}{dx} = \dfrac{dp}{dx} \cdot \dfrac{dy}{dp} \quad\dots\text{Chain Rule} \\[5ex] = 4 \cdot -\sin p \\[3ex] = -4\sin p \\[3ex] = -4\sin 4x \\[3ex] ........................................................................ \\[3ex] f'(x) = \dfrac{1}{2\sqrt{x}} - (-4\sin4x) \quad\dots\text{Difference Rule} \\[5ex] = \dfrac{1}{2\sqrt{x}} + 4\sin4x \\[5ex] (ii.) \\[3ex] g(x) = (x + 1)^4 \\[3ex] \text{Let } p = x + 1 \hspace{3em} \implies g = p^4 \\[3ex] \dfrac{dp}{dx} = 1 \hspace{4em} \dfrac{dg}{dp} = 4p^3 \quad\dots\text{Power Rule} \\[5ex] g'(x) = \dfrac{dp}{dx} \cdot \dfrac{dg}{dp} \quad\dots\text{Chain Rule} \\[5ex] = 1 \cdot 4p^3 \\[3ex] = 4(x + 1)^3 \\[5ex] \text{For } 4(x + 1)^3 \\[3ex] \text{Using the Power Rule and Chain Rule again} \\[3ex] g''(x) = 3 \cdot 4 \cdot 1 \cdot (x + 1)^{3 - 1} \\[3ex] = 12(x + 1)^2 \\[5ex] (b.) \\[3ex] \text{side of the block of ice} = \text{side of a cube} = k \\[3ex] \text{surface area} = A \\[3ex] \underline{\text{Given:}} \\[3ex] \dfrac{dk}{dt} = -\dfrac{1}{3}\;cm/min \quad\dots\text{because the ice block is melting} \\[5ex] k = 150\;cm \\[3ex] A = 6k^2 \quad\dots\text{Surface Area of a Cube} \\[3ex] \dfrac{dA}{dk} = 2 \cdot 6 \cdot k^{2 - 1} \quad\dots\text{Power Rule} \\[5ex] = 12k \\[3ex] = 12(150) \\[3ex] = 180\;cm^2/cm \\[5ex] \dfrac{dA}{dt} = \dfrac{dA}{dk} \cdot \dfrac{dk}{dt} \\[5ex] = 180 \cdot -\dfrac{1}{3} \\[5ex] = -60\;cm^2/min \\[3ex] $ The surface area of the ice block is melting at the rate of 60 square centimeters per minute.

(c.) Let:
velocity = v
time = t
acceleration = a

Graphical Solution:


At t = 4, v = 4 (From the graph)
So, we draw a tangent line to the curve at Point 1 (4, 4)
This is because we were asked to Then, we look for another point, preferable perfectly aligned point at both axis and call it Point 2 (0, 2)

$ \text{Point 1} (4, 4) \hspace{3em} \text{Point 2} (0, 2) \\[3ex] t_1 = 4 \hspace{6em} t_2 = 0 \\[3ex] v_1 = 4 \hspace{6em} v_2 = 2 \\[5ex] a = \dfrac{v_2 - v_1}{t_2 - t_1} \\[5ex] = \dfrac{2 - 4}{0 - 4} \\[5ex] = \dfrac{-2}{-4} \\[5ex] = 0.5\;m/s^2 \\[3ex] $ Student: SamDom For Peace
Teacher: What's good?
Student: Can we do it algebraically?
I mean...using your graph does not restrict us to using a graphical approach only.
Does it?
Teacher: Based on the wording of the question, I think we can use an algebraic approach.
We just need to use the graph.
Student: Okay, can we use an algebraic approach?
Teacher: Sure, let's do it.

Algebraic Solution:
An observation of the curve in the graph shows a positive square root function.
Let us verify it using well-aligned points.

$ (t, v) \\[3ex] (0, 0) \\[3ex] (1, 2) \\[3ex] \sqrt{1} \cdot 2 = 1 \cdot 2 = 2 \\[5ex] (4, 4) \\[3ex] \sqrt{4} \cdot 2 = 2 \cdot 2 = 4 \\[5ex] (9, 6) \\[3ex] \sqrt{9} \cdot 2 = 3 \cdot 2 = 6 \\[5ex] (t, v) \\[3ex] v(t) = \sqrt{t} \cdot 2 \\[3ex] v(t) = 2\sqrt{t} \\[5ex] v = 2t^{\dfrac{1}{2}} \quad\dots\text{Law 7 Exp} \\[5ex] a = \dfrac{dv}{dt} = \dfrac{1}{2} \cdot 2 \cdot t^{\dfrac{1}{2} - 1} \\[7ex] = t^{-\dfrac{1}{2}} \\[5ex] = \dfrac{1}{t^{\dfrac{1}{2}}} \quad\dots\text{Law 6 Exp} \\[7ex] = \dfrac{1}{\sqrt{t}} \quad\dots\text{Law 6 Exp} \\[5ex] a|_{t = 4} = \dfrac{1}{\sqrt{t}} \\[5ex] = \dfrac{1}{\sqrt{4}} \\[5ex] = \dfrac{1}{2} \\[5ex] = 0.5\;m/s^2 $

$ C \text{ is the constant of integration} \\[5ex] (a.) \\[3ex] y = -2x^2 + 7x \text{ and the lines: } x = 2 \text{ and } x = 1 \\[5ex] \text{When } x = 2 \\[3ex] y = -2(2)^2 + 7(2) \\[3ex] = -2(4) + 14 \\[3ex] = -8 + 14 \\[3ex] = 6 \\[3ex] 6 \gt 0 \quad\dots\text{above the } x-\text{axis} \\[5ex] \text{When } x = 1 \\[3ex] y = -2(1)^2 + 7(1) \\[3ex] = -2(1) + 7 \\[3ex] = -2 + 7 \\[3ex] = 5 \\[3ex] 5 \gt 0 \quad\dots\text{above the } x-\text{axis} \\[5ex] y \gt 0 \text{ for all } x \in [1, 2] \\[5ex] \text{Area} = \displaystyle\int_1^2 (-2x^2 + 7x)\;dx \\[5ex] = \left[-\dfrac{2x^3}{3} + \dfrac{7x^2}{2}\right]_1^2 \\[5ex] = \left[-\dfrac{2(2)^3}{3} + \dfrac{7(2)^2}{2}\right] - \left[-\dfrac{2(1)^3}{3} + \dfrac{7(1)^2}{2}\right] \\[5ex] = \left(-\dfrac{16}{3} + 14 \right) - \left(-\dfrac{2}{3} + \dfrac{7}{2}\right) \\[5ex] = \left(\dfrac{-16 + 42}{3}\right) - \left(\dfrac{-4 + 21}{6}\right) \\[5ex] = \dfrac{26}{3} - \dfrac{17}{6} \\[5ex] = \dfrac{52 - 17}{6} \\[5ex] = \dfrac{35}{6}\text{ square units} \\[5ex] (b.) \\[3ex] x_1 = 0 \\[3ex] x_2 = 1 \\[3ex] y = 2x^2 - 3 \\[3ex] \text{Volume} = \displaystyle\int_{x_1}^{x_2} \pi y^2 \; dx \quad\dots\text{rotation about the x-axis} \\[5ex] = \pi \cdot \displaystyle\int_0^1 (2x^2 - 3)^2\; dx \\[5ex] ................................................... \\[3ex] (2x^2 - 3)^2 \\[3ex] = (2x^2 - 3)(2x^2 - 3) \\[3ex] = 4x^4 - 6x^2 - 6x^2 + 9 \\[3ex] = 4x^4 - 12x^2 + 9 \\[5ex] \displaystyle\int (4x^4 - 12x^2 + 9)\; dx \\[5ex] = \dfrac{4x^5}{5} - \dfrac{12x^3}{3} + 9x \\[5ex] = \dfrac{4x^5}{5} - 4x^3 + 9x \\[5ex] ................................................... \\[3ex] = \pi \cdot \left[\dfrac{4x^5}{5} - 4x^3 + 9x\right]_0^1 \\[5ex] = \pi \cdot \left[\left(\dfrac{4(1)^5}{5} - 4(1)^3 + 9(1)\right) - \left(\dfrac{4(0)^5}{5} - 4(0)^3 + 9(0)\right)\right] \\[5ex] = \pi \cdot \left[\left(\dfrac{4}{5} - 4 + 9\right) - (0 - 0 + 0)\right] \\[5ex] = \pi \cdot \left[\dfrac{4 - 20 + 45}{5}\right] \\[5ex] = \dfrac{29\pi}{5}\text{ cubic units} \\[5ex] (c.) \\[3ex] y = \displaystyle\int \dfrac{dy}{dx}\;dx \\[5ex] = \displaystyle\int \left[\dfrac{10(5x + 1)^2}{3}\right]\;dx \\[5ex] = \dfrac{10}{3}\displaystyle\int (5x + 1)^2\;dx \\[5ex] ................................................... \\[3ex] (5x + 1)^2 \\[3ex] = (5x + 1)(5x + 1) \\[3ex] = 25x^2 + 5x + 5x + 1 \\[3ex] = 25x^2 + 10x + 1 \\[3ex] ................................................... \\[3ex] = \dfrac{10}{3}\displaystyle\int (25x^2 + 10x + 1)\;dx \\[5ex] = \dfrac{10}{3} \left[\dfrac{25x^3}{3} + \dfrac{10x^2}{2} + x \right] + C \\[5ex] = \dfrac{10}{3} \left[\dfrac{25x^3}{3} + 5x^2 + x\right] + C \\[5ex] \text{At } \left(0, \dfrac{7}{9}\right) \\[5ex] x = 0 \hspace{3em} y = \dfrac{7}{9} \\[5ex] \implies \\[3ex] \dfrac{7}{9} = \dfrac{10}{3} \left[\dfrac{25(0)^3}{3} + 5(0)^2 + 0\right] + C \\[5ex] \dfrac{7}{9} = \dfrac{10}{3}(0 + 0 + 0) + C \\[5ex] \dfrac{7}{9} = 0 + C \\[5ex] C = \dfrac{7}{9} \\[5ex] \text{So, the equation of the curve is: } \\[3ex] y = \dfrac{10}{3} \left[\dfrac{25x^3}{3} + 5x^2 + x\right] + \dfrac{7}{9} \\[5ex] y = \dfrac{250x^3}{9} + \dfrac{50x^2}{3} + \dfrac{10x}{3} + \dfrac{7}{9} \\[5ex] (d.) \\[3ex] \text{Let the displacement} = p \\[3ex] v = (5t - 6)^{\dfrac{1}{2}} \\[7ex] v = \dfrac{dp}{dt} \\[5ex] dp = v\;dt \\[3ex] p = \displaystyle\int v\;dt \\[3ex] = \displaystyle\int (5t - 6)^{\dfrac{1}{2}} \; dt \\[7ex] \underline{\text{Integration by Algebraic Substitution}} \\[3ex] \text{Let } k = 5t - 6 \\[3ex] \dfrac{dk}{dt} = 5 \\[5ex] \dfrac{dt}{dk} = \dfrac{1}{5} \\[5ex] dt = \dfrac{dk}{5} \\[5ex] \implies \\[3ex] p = \displaystyle\int k^{\dfrac{1}{5}} \cdot \dfrac{dk}{5} \\[7ex] = \dfrac{1}{5} \displaystyle\int k^{\dfrac{1}{2}}\;dk \\[7ex] = \dfrac{1}{5} \cdot \dfrac{k^{\dfrac{3}{2}}}{\dfrac{3}{2}} + C \\[9ex] = \dfrac{1}{5} \cdot k^{\dfrac{3}{2}} \cdot \dfrac{2}{3} + C \\[7ex] = \dfrac{2k^{\dfrac{3}{2}}}{15} + C \\[7ex] \text{Substitute back} \\[3ex] = \dfrac{2(5t - 6)^{\dfrac{3}{2}}}{15} + C \\[7ex] \text{When } t = 2\;s \hspace{3em} p = 3\;m \\[3ex] 3 = \dfrac{2[5(2) - 6]^{\dfrac{3}{2}}}{15} + C \\[7ex] 3 = \dfrac{2[5(2) - 6]^{\dfrac{3}{2}} + 15C}{15} \\[7ex] 15(3) = 2(10 - 6)^{\dfrac{3}{2}} + 15C \\[7ex] 45 = 2(4)^{\dfrac{3}{2}} + 15C \\[7ex] 15C = 45 - 2(4)^{\dfrac{3}{2}} \\[7ex] C = \dfrac{45 - 2(\sqrt{4})^3}{15} \\[5ex] = \dfrac{45 - 2(2)^3}{15} \\[5ex] = \dfrac{45 - 16}{15} \\[5ex] = \dfrac{29}{15} \\[5ex] \implies \\[3ex] p = \dfrac{2(5t - 6)^{\dfrac{3}{2}}}{15} + \dfrac{29}{15} \\[5ex] p = \dfrac{2(5t - 6)^{\dfrac{3}{2}} + 29}{15} $

$ \text{Let: } \\[3ex] \text{Male} = M \\[3ex] \text{Female} = F \\[3ex] \text{Instagram} = I \\[3ex] \text{TikTok} = T \\[3ex] \text{Sample Space} = S \\[5ex] (a.) \\[3ex] n(I) = 120 \\[3ex] n(T) = 150 \\[3ex] n(M) = 110 \\[3ex] n(F) = 160 \\[3ex] n(S) = 270 \\[5ex] (i.) \\[3ex] P(I \cap F) \\[3ex] \\[3ex] = \dfrac{70}{270} \\[5ex] = \dfrac{7}{27} \\[5ex] (ii.) \\[3ex] P(T | M) = \dfrac{P(T \cap M)}{P(M)} \quad\dots\text{Conditional Probability} \\[5ex] = \dfrac{60}{110} \\[5ex] = \dfrac{6}{11} \\[5ex] (iii.) \\[3ex] P(M \cup I) = P(M) + P(I) - P(M \cap I) \quad\dots\text{General Addition Rule} \\[3ex] = \dfrac{110}{270} + \dfrac{120}{270} - \dfrac{50}{270} \\[5ex] = \dfrac{180}{270} \\[5ex] = \dfrac{2}{3} \\[5ex] $ (b.) Let us arrange the data in ascending order.

TABLE 2: STUDENTS' SCORES ON A POP QUIZ (SORTED IN ASCENDING ORDER)

42	42	51	52	55
60	60	63	66	70
71	71	72	73	73
75	76	76	76	77
78	79	80	84	84
84	84	88	92	93
94	96	97	98	99

(i.) minimum = 42
maximum = 99
To construct the stem-and-leaf diagram, each score is separated into a stem (the tens digit) and a leaf (the units digit).
The stems are arranged in ascending order vertically, and the corresponding leaves are listed in ascending order horizontally.

$ \begin{array}{r|l} \text{Stem} & \text{Leaf} \\ \hline 4 & 2 \quad 2 \\ 5 & 1 \quad 2 \quad 5 \\ 6 & 0 \quad 0 \quad 3 \quad 6 \\ 7 & 0 \quad 1 \quad 1 \quad 2 \quad 3 \quad 3 \quad 5 \quad 6 \quad 6 \quad 6 \quad 7 \quad 8 \quad 9 \\ 8 & 0 \quad 4 \quad 4 \quad 4 \quad 4 \quad 8 \\ 9 & 2 \quad 3 \quad 4 \quad 6 \quad 7 \quad 8 \quad 9 \end{array} \\[10ex] \text{Key: } 4 \mid 2 \text{ represents a score of } 42 \\[5ex] $ (ii.) The distribution of the data is left-skewed because fewer students earned scores in the 40s and 50s while the majority of students earned scores in the 70s, 80s, and 90s.

$ (iii.) \\[3ex] \text{sample size, } N = 35 \\[3ex] \text{25th Percentile Position} = \dfrac{25N}{100}th \\[5ex] = \dfrac{35}{4}th \\[5ex] = 8.75th \\[3ex] \lceil 8.75 \rceil = 9th \\[3ex] \text{25th Percentile, } Q_1 = 9th\text{ value} = 66 \\[5ex] \text{50th Percentile Position} = \dfrac{50N}{100}th \\[5ex] = \dfrac{35}{2}th \\[5ex] = 17.5th \\[3ex] \lceil 17.5 \rceil = 18th \text{50th Percentile, } Q_2 = 18th\text{ value} = 76 \\[5ex] \text{75th Percentile Position} = \dfrac{75N}{100}th \\[5ex] = \dfrac{3 \cdot 35}{4}th \\[5ex] = 26.25th \\[3ex] \lceil 26.25 \rceil = 27th \\[3ex] \text{75th Percentile, } Q_3 = 27th\text{ value} = 84 \\[5ex] (iv.) \\[3ex] \text{Semi-Interquartile Range, } SIQR = \dfrac{Q_3 - Q_1}{2} \\[5ex] = \dfrac{84 - 66}{2} \\[5ex] = \dfrac{18}{2} \\[5ex] = 9 \\[3ex] $ (c.) A stem‑and‑leaf plot is better for displaying students’ quiz scores because it shows every individual value, making the distribution clearer than a box‑and‑whisker plot, which only summarizes the data.