Further Mathematics (Elective) Essay for Private Candidates

The gradient of the tangent to the curve at a point is the derivative of the curve at that point.

$ (x^2 - y^2 + 16)^2 - 2x = 46 \\[3ex] \text{Differentiating each term with respect to } x \\[3ex] .............................................. \\[3ex] \text{For } (x^2 - y^2 + 16)^2 \\[3ex] \text{Let } p = x^2 - y^2 + 16 \hspace{3em} y = p^2 \\[3ex] \dfrac{dp}{dx} = 2x - 2y\dfrac{dy}{dx} \hspace{3.5em} \dfrac{dy}{dp} = 2p \\[5ex] \dfrac{dy}{dx} = \dfrac{dp}{dx} * \dfrac{dy}{dp}\quad\dots\text{Chain Rule} \\[5ex] = 2x - 2y\dfrac{dy}{dx} * 2p \\[5ex] = \left(2x - 2y\dfrac{dy}{dx}\right)[2(x^2 - y^2 + 16)] \\[5ex] .............................................. \\[3ex] \implies \\[3ex] 2(x^2 - y^2 + 16)\left(2x - 2y\dfrac{dy}{dx}\right) - 2 = 0 \\[5ex] 2(x^2 - y^2 + 16)\left(2x - 2y\dfrac{dy}{dx}\right) = 2 \\[5ex] (x^2 - y^2 + 16)\left(2x - 2y\dfrac{dy}{dx}\right) = 1 \\[5ex] 2x - 2y\dfrac{dy}{dx} = \dfrac{1}{x^2 - y^2 + 16} \\[5ex] 2x - \dfrac{1}{x^2 - y^2 + 16} = 2y\dfrac{dy}{dx} \\[5ex] \dfrac{2x(x^2 - y^2 + 16) - 1}{x^2 - y^2 + 16} = 2y\dfrac{dy}{dx} \\[5ex] 2y\dfrac{dy}{dx} = \dfrac{2x(x^2 - y^2 + 16) - 1}{x^2 - y^2 + 16} \\[5ex] \dfrac{dy}{dx} = \dfrac{2x(x^2 - y^2 + 16) - 1}{2y(x^2 - y^2 + 16)} \\[5ex] \text{At } (1, -2) \\[3ex] x = 1 \hspace{3em} y = -2 \\[3ex] \dfrac{dy}{dx} = \dfrac{2(1)[1^2 - (-2)^2 + 16] - 1}{2(-2)[1^2 - (-2)^2 + 16]} \\[5ex] = \dfrac{2(1 - 4 + 16) - 1}{-4(1 - 4 + 16)} \\[5ex] = \dfrac{2(13) - 1}{-4(13)} \\[5ex] = \dfrac{26 - 1}{-52} \\[5ex] = \dfrac{25}{-52} \\[5ex] = -\dfrac{25}{52} $

Differentiating a function from first principles can also be done by taking the limit of the difference quotient.

$ \underline{\text{Diiference Quotient, DQ}} \\[3ex] DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex] ................................................. \\[3ex] f(x) = 5 - 4x - 3x^2 \\[5ex] f(x + h) = 5 - 4(x + h) - 3(x + h)^2 \\[3ex] = 5 - 4x - 4h - 3[(x + h)(x + h)] \\[3ex] = 5 - 4x - 4h - 3(x^2 + hx + hx + h^2) \\[3ex] = 5 - 4x - 4h - 3(x^2 + 2hx + h^2) \\[3ex] = 5 - 4x - 4h - 3x^2 - 6hx - 3h^2 \\[5ex] f(x + h) - f(x) \\[3ex] = (5 - 4x - 4h - 3x^2 - 6hx - 3h^2) - (5 - 4x - 3x^2) \\[3ex] = 5 - 4x - 4h - 3x^2 - 6hx - 3h^2 - 5 + 4x + 3x^2 \\[3ex] = -4h - 6hx - 3h^2 \\[3ex] = h(-4 - 6x - 3h) \\[5ex] DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex] = \dfrac{h(-4 - 6x - 3h)}{h} \\[5ex] = -4 - 6x - 3h \\[3ex] ................................................. \\[3ex] \underline{\text{Derivative}} \\[3ex] \dfrac{dy}{dx} = \displaystyle{\lim_{h \to \infty}} DQ \\[5ex] = \displaystyle{\lim_{h \to \infty}} (-4 - 6x - 3h) \\[5ex] = -4 - 6x - 3(0) \\[3ex] = -4 - 6x $

$ 2^{3x + 1} - 3(2^{2x}) + 2^{x + 1} = 2^x \\[3ex] (2^{3x} \cdot 2^1) - 3(2^x)^2 + (2^x \cdot 2^1) = 2^x \quad\dots\text{Law 5...Exp} \\[3ex] [(2^x)^3 \cdot 2] - 3(2^x)^2 + (2^x \cdot 2) = 2^x \\[3ex] \text{Let } 2^x = p \\[3ex] (p^3 \cdot 2) - 3p^2 + (p \cdot 2) = p \\[3ex] 2p^3 - 3p^2 + 2p - p = 0 \\[3ex] 2p^3 - 3p^2 + p = 0 \\[3ex] p(2p^2 - 3p + 1) = 0 \\[3ex] p = 0 \text{ OR } 2p^2 - 3p + 1 = 0 \\[3ex] ......................................... \\[3ex] 2p^2 - 3p + 1 = 0 \\[3ex] 2p^2(1) = 2p^2 \\[3ex] \text{Factors are: } -2p \text{ and } -p \\[3ex] 2p^2 - 2p - p + 1 = 0 \\[3ex] 2p(p - 1) - 1(p - 1) = 0 \\[3ex] (p - 1)(2p - 1) = 0 \\[3ex] p - 1 = 0 \text{ OR } 2p - 1 = 0 \\[3ex] p = 1 \text{ OR } 2p = 1 \\[3ex] p = 1 \text{ OR } p = \dfrac{1}{2} \\[5ex] p = 0, \dfrac{1}{2}, 1 \\[5ex] ......................................... \\[3ex] \text{But } 2^x = p \\[3ex] \text{When } p = 0 \\[3ex] 2^x = 0 \\[3ex] \text{Introducing Logarithms to both sides} \\[3ex] \log 2^x = \log 0 \\[3ex] x \log 2 = undefined \quad\dots\text{The argument of a logarithm must be positive} \\[5ex] \text{When } p = \dfrac{1}{2} \\[5ex] 2^x = \dfrac{1}{2} \\[5ex] 2^x = 2^{-1} \\[3ex] x = -1 \\[5ex] \text{When } p = 1 \\[3ex] 2^x = 1 \\[3ex] 2^x = 2^{0} \quad\dots\text{Law 3...Exp} \\[3ex] x = 0 \\[5ex] x = -1, 0 \\[3ex] $ Check (If you have time)

$x = -1, 0$

LHS

RHS

$ 2^{3x + 1} - 3(2^{2x}) + 2^{x + 1} \\[3ex] \text{When } x = -1 \\[3ex] 2^{3(-1) + 1} - 3(2^{2(-1)}) + 2^{-1 + 1} \\[3ex] 2^{-2} - 3(2^{-2}) + 2^0 \\[3ex] \dfrac{1}{2^2} - 3\left(\dfrac{1}{2^2}\right) + 1 \\[5ex] \dfrac{1}{4} - \dfrac{3}{4} + \dfrac{4}{4} \\[5ex] \dfrac{2}{4} \\[5ex] \dfrac{1}{2} $ $ \text{When } x = 0 \\[3ex] 2^{3(0) + 1} - 3(2^{2(0)}) + 2^{0 + 1} \\[3ex] 2^{1} - 3(2^0) + 2^(1) \\[3ex] 2 - 3(1) + 2 \\[3ex] 1 $

$ 2^x \\[3ex] \text{When } x = -1 \\[3ex] 2^{-1} \\[3ex] \dfrac{1}{2} $ $ \text{When } x = 0 \\[3ex] 2^0 \\[3ex] 1 $

$ \text{Dataset 1} \\[3ex] 2, 9, 7m, 3n, 21, 24 \\[3ex] \text{Median} = \dfrac{7m + 3n}{2} = 16 \\[5ex] 7m + 3n = 2(16) \\[3ex] 7m + 3n = 32...eqn.(1) \\[5ex] \text{Dataset 2} \\[3ex] 4, 8, 6, 4m, 19, 2n \\[3ex] \text{Mean} = \dfrac{4 + 8 + 6 + 4m + 19 + 2n}{6} = 9\dfrac{1}{2} \\[5ex] \dfrac{4m + 2n + 37}{6} = \dfrac{19}{2} \\[5ex] 4m + 2n + 37 = 6\left(\dfrac{19}{2}\right) \\[5ex] 4m + 2n = 3(19) - 37 \\[3ex] 4m + 2n = 20 \\[3ex] 2m + n = 10 ...eqn.(2) \\[5ex] \text{From } eqn.(2) \\[3ex] n = 10 - 2m...eqn.(3) \\[3ex] \text{Substitute for } n \text{ in } eqn.(1) \\[3ex] 7m + 3(10 - 2m) = 32 \\[3ex] 7m + 30 - 6m = 32 \\[3ex] m = 32 - 30 \\[3ex] m = 2 \\[5ex] \text{Substitute for } m \text{ in } eqn.(3) \\[3ex] n = 10 - 2(2) \\[3ex] n = 6 $

$ \underline{\text{Binomial Probability Distribution}} \\[3ex] S = \text{Sample Space} \\[3ex] n(S) = 10 \\[3ex] n(good) = 3 \\[3ex] n(bad) = 10 - 3 = 7\quad\dots\text{word of reference} \\[3ex] p = P(bad) = \dfrac{7}{10} \\[5ex] q = P(good) = \dfrac{3}{10} \\[5ex] P(x) = C(n, x) * p^x * q^{n - x} \\[5ex] n = 4 \quad\dots\text{four mangoes are selected} \\[3ex] (a.) \\[3ex] \text{one is bad} \implies x = 1 P(x = 1) \\[3ex] = C(4, 1) * \left(\dfrac{7}{10}\right)^1 * \left(\dfrac{3}{10}\right)^{4 - 1} \\[5ex] = \dfrac{189}{2500} \\[5ex] = 0.0756 \\[5ex] (b.) \\[3ex] \text{not more than two are bad} \implies \text{at most 2} \implies \le 2 \\[3ex] P(x \le 2) \\[3ex] = P(x = 0) + P(x = 1) + P(x = 2) \\[3ex] ................................................................. \\[3ex] P(x = 0) \\[3ex] = C(4, 0) * \left(\dfrac{7}{10}\right)^0 * \left(\dfrac{3}{10}\right)^{4 - 0} \\[5ex] = 0.0081 \\[5ex] P(x = 1) = 0.0756 \\[5ex] P(x = 2) \\[3ex] = C(4, 2) * \left(\dfrac{7}{10}\right)^2 * \left(\dfrac{3}{10}\right)^{4 - 2} \\[5ex] = \dfrac{1323}{5000} \\[5ex] = 0.2646 \\[3ex] ................................................................. \\[3ex] = 0.0081 + 0.0756 + 0.2646 \\[3ex] = 0.3483 $

$ velocity = v \\[3ex] acceleration = a \\[5ex] r = t^4 + 3t^3 - t^2 + 2t \\[3ex] (a.) \\[3ex] v = \dfrac{dr}{dt} = 4t^3 + 9t^2 - 2t + 2 \quad\dots\text{Power Rule} \\[5ex] \text{At } t = 3 \\[3ex] v(3) = 4(3)^3 + 9(3)^2 - 2(3) + 2 \\[3ex] v(3) = 185\;m/s \\[5ex] (b.) \\[3ex] a = \dfrac{dv}{dt} = 12t^2 + 18t - 2 \quad\dots\text{Power Rule} \\[5ex] \text{At } t = 3 \\[3ex] a(3) = 12(3)^2 + 18(3) - 2 \\[3ex] a(3) = 160\;m/s^2 $

$ F_1 = (6\;N, 030^\circ) \\[3ex] F_{1x} = 6\cos 30^\circ = 5.196152423\;N \\[3ex] F_{1y} = 6\sin 30^\circ = 3\;N \\[5ex] F_2 = (8\;N, 075^\circ) \\[3ex] F_{2x} = 8\cos 75^\circ = 2.070552361\;N \\[3ex] F_{2y} = 8\sin 75^\circ = 7.72740661\;N \\[5ex] F_3 = (x\;N, \theta^\circ) \\[5ex] \underline{\text{At Equilibrium}} \\[3ex] \Sigma F_x = 0 \\[3ex] F_{1x} + F_{2x} + F_{3x} = 0 \\[3ex] 5.196152423 + 2.070552361 + F_{3x} = 0 \\[3ex] F_{3x} = 0 - 5.196152423 - 2.070552361 \\[3ex] F_{3x} = -7.266704784\;N \\[5ex] \Sigma F_y = 0 \\[3ex] F_{1y} + F_{2y} + F_{3y} = 0 \\[3ex] 3 + 7.72740661 + F_{3y} = 0 \\[3ex] F_{3y} = 0 - 3 - 7.72740661 \\[3ex] F_{3y} = -10.72740661\;N \\[5ex] \underline{\text{Magnitude of } F_3} \\[3ex] F_3 = \sqrt{F_{3x}^2 + F_{3y}^2} \\[3ex] F_3 = \sqrt{(-7.266704784)^2 + (-10.72740661)^2} \\[3ex] = 12.95693833 \\[3ex] \approx 13\;N\quad\dots\text{to the nearst whole number} \\[5ex] \underline{\text{Direction of } F_3, \theta} \\[3ex] \theta = \tan^{-1}\left(\dfrac{F_y}{F_x}\right) \\[5ex] \theta = \tan^{-1}\left(\dfrac{F_{3y}}{F_{3x}}\right) \\[5ex] = \tan^{-1}\left(\dfrac{-10.72740661}{-7.266704784}\right) \\[5ex] = 55.8864353^\circ \\[3ex] \text{But: } \\[3ex] F_{3x} = -7.266704784\;N\quad\dots\text{negative} \\[3ex] F_{3y} = -10.72740661\;N\quad\dots\text{negative} \\[3ex] \implies F_3 \text{ is in the 3rd Quadrant} \\[3ex] \therefore \theta = 180^\circ + 55.8864353^\circ \\[3ex] = 235.8864353 \\[3ex] \approx 236^\circ \quad\dots\text{to the nearest whole number} $

$ (a.) \\[3ex] \underline{\text{Given Equation}} \\[3ex] 5x^2 - 9x - 4 = 0 \\[3ex] \text{Compare to: } ax^2 + bx + c = 0 \\[3ex] a = 5 \\[3ex] b = -9 \\[3ex] c = -4 \\[3ex] \text{Roots} = \alpha \text{ and } \beta \\[3ex] \text{Sum of roots, } \alpha + \beta \\[3ex] = -\dfrac{b}{a} \\[5ex] = \dfrac{-(-9)}{5} \\[5ex] = \dfrac{9}{5} \\[5ex] \text{Product of roots, } \alpha\beta \\[3ex] = \dfrac{c}{a} \\[5ex] = -\dfrac{4}{5} \\[5ex] \underline{\text{New Equation}} \\[3ex] \text{Roots} = (\alpha^2 - 3) \text{ and } (\beta^2 - 3) \\[3ex] \text{Sum of roots} \\[3ex] = (\alpha^2 - 3) + (\beta^2 - 3) \\[3ex] = \alpha^2 - 3 + \beta^2 - 3 \\[3ex] = \alpha^2 + \beta^2 - 6 \\[3ex] ................................................ \\[3ex] \text{But: } (\alpha + \beta)^2 \\[3ex] = (\alpha + \beta)(\alpha + \beta) \\[3ex] = \alpha^2 + \alpha\beta + \alpha\beta + \beta^2 \\[3ex] = \alpha^2 + 2\alpha\beta + \beta^2 \\[3ex] \text{So, } (\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta \\[3ex] \implies \\[3ex] \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \\[3ex] = \left(\dfrac{9}{5}\right)^2 - 2\left(-\dfrac{4}{5}\right) \\[5ex] = \dfrac{81}{25} + \dfrac{8}{5} \\[5ex] = \dfrac{81 + 40}{25} \\[5ex] \therefore \alpha^2 + \beta^2 = \dfrac{121}{25} \\[5ex] ................................................ \\[3ex] = \dfrac{121}{25} - 6 \\[5ex] = \dfrac{121 - 150}{25} \\[5ex] = -\dfrac{29}{25} \\[5ex] \text{Product of roots} \\[3ex] = (\alpha^2 - 3)(\beta^2 - 3) \\[3ex] = \alpha^2\beta^2 - 3\alpha^2 - 3\beta^2 + 9 \\[3ex] = (\alpha\beta)^2 - 3(\alpha^2 + \beta^2) + 9 \\[3ex] = \left(-\dfrac{4}{5}\right)^2 - 3\left(\dfrac{121}{25}\right) + 9 \\[5ex] = \dfrac{16}{25} - \dfrac{363}{25} + \dfrac{225}{25} \\[5ex] = -\dfrac{122}{25} \\[5ex] \underline{\text{Quadratic Equation if the Sum and Product of Roots is given}} \\[3ex] x^2 - \text{(Sum of roots) } x + \text{Product of roots} = 0 \\[3ex] x^2 - \left(-\dfrac{29}{25}\right)x + -\dfrac{122}{25} = 0 \\[5ex] x^2 + \dfrac{29}{25}x -\dfrac{122}{25} = 0 \\[5ex] LCD = 25 \\[3ex] 25x^2 + 29x - 122 = 0 $

$ (b.) \\[3ex] f(x) = \dfrac{x}{5} - 3 \hspace{3em} g(x) = \dfrac{7x - 5}{2} \\[5ex] \underline{\text{Inverse of} f(x)} \\[3ex] y = \dfrac{x}{5} - 3 \\[5ex] x = \dfrac{y}{5} - 3 \\[5ex] x + 3 = \dfrac{y}{5} \\[5ex] y = 5(x + 3) \\[3ex] f^{-1}(x) = 5(x + 3) \\[5ex] (i.) \\[3ex] f^{-1}(x) \circ g(x) \\[3ex] = f^{-1}[g(x)] \\[3ex] = f^{-1}\left(\dfrac{7x - 5}{2}\right) \\[5ex] = 5\left(\dfrac{7x - 5}{2} + 3\right) \\[5ex] = 5\left(\dfrac{7x - 5 + 6}{2}\right) \\[5ex] = 5\left(\dfrac{7x + 1}{2}\right) \\[5ex] (ii.) \\[3ex] f^{-1} \circ g(-2) \\[3ex] = 5\left(\dfrac{7(-2) + 1}{2}\right) \\[5ex] = -\dfrac{65}{2} $

$ a = \text{first term} \\[3ex] d = \text{common difference} \\[3ex] r = \text{common ratio} \\[5ex] \underline{\text{Linear Sequence}} \hspace{7em} \underline{\text{Exponential Sequence}} \\[3ex] \text{1st term} = a \hspace{8.5em} \text{1st term} = a \\[3ex] \text{4th term} = a + 3d \hspace{6em} \text{2nd term} = ar = a + 3d \\[3ex] \text{6th term} = a + 5d \hspace{6em} \text{3rd term} = ar^2 = a + 5d \\[3ex] \text{12th term} = a + 11d = -12 \\[5ex] \text{For Linear Sequence: 12th term} \\[3ex] a = -12 - 11d ...eqn.(1) \\[5ex] \text{Exponential Sequence: r} \\[3ex] r = \dfrac{a + 3d}{a} = \dfrac{a + 5d}{a + 3d} \\[5ex] (a + 3d)(a + 3d) = a(a + 5d) \\[3ex] a^2 + 3ad + 3ad + 9d^2 = a^2 + 5ad \\[3ex] 6ad - 5ad = -9d^2 \\[3ex] ad = -9d^2 \\[3ex] a = -\dfrac{9d^2}{d} \\[5ex] a = -9d ...eqn.(2) \\[5ex] (a.) \\[3ex] eqn.(1) = eqn.(2) \quad\dots\text{Because } a = a \\[3ex] -12 - 11d = -9d \\[3ex] -12 = -9d + 11d \\[3ex] 2d = -12 \\[3ex] d = -\dfrac{12}{2} \\[5ex] d = -6 \\[5ex] (b.) \\[3ex] \text{Substitute for } d \text{ in } eqn.(2) \\[3ex] a = -9(-6) \\[3ex] a = 54 \\[5ex] (c.)\; \underline{\text{Sum of the first n terms of a Linear Sequence}, S_{n}} \\[3ex] S_n = \dfrac{n}{2}[2a + d(n - 1)] \\[5ex] \text{For } n = 14 \\[3ex] S_{14} = \dfrac{14}{2}[2(54) + -6(14 - 1)] \\[5ex] = 7[108 - 6(13)] \\[3ex] = 7(108 - 78) \\[3ex] = 210 $

$ (a.) \\[3ex] \cos 2x - 5\sin^2 x = \dfrac{1}{9} \\[5ex] ................................................... \\[3ex] \sin^2 x + \cos^2 x = 1 \quad\dots\text{Pythagorean Identity} \\[3ex] \sin^2 x = 1 - \cos^2 x \\[5ex] \cos 2x \\[3ex] = \cos(x + x) \\[3ex] = \cos x \cos x - \sin x \sin x \quad\dots\text{Double-Angle Formula} \\[3ex] = \cos^2 x - \sin^2 x \\[3ex] = \cos^2 x - (1 - \cos^2 x)\quad\dots\text{based on Pythagorean Identity} \\[3ex] = \cos^2 x - 1 + \cos^2 x \\[3ex] = 2\cos^2 x - 1 \\[3ex] ................................................... \\[3ex] \implies \\[3ex] 2\cos^2 x - 1 - 5(1 - \cos^2 x) = \dfrac{1}{9} \\[5ex] 2\cos^2 x - 1 - 5 + 5\cos^2 x = \dfrac{1}{9} \\[5ex] 7\cos^2 x = \dfrac{1}{9} + 1 + 5 \\[5ex] 7\cos^2 x = \dfrac{1}{9} + 6 \\[5ex] 7\cos^2 x = \dfrac{1 + 54}{9} \\[5ex] 7\cos^2 x = \dfrac{55}{9} \\[5ex] \cos^2 x = \dfrac{55}{9} * \dfrac{1}{7} \\[5ex] \cos^2 x = \dfrac{55}{63} \\[5ex] \cos x = \pm\sqrt{\dfrac{55}{63}} \\[5ex] \cos x = \pm 0.9343531843 \\[5ex] x = \cos^{-1} (-0.9343531843) \\[3ex] x = 159.1238974 \\[3ex] x \approx 159^\circ\quad\dots\text{2nd Quadrant} \\[5ex] x = \cos^{-1} (0.9343531843) \\[3ex] x = 20.8761026 \\[3ex] x \approx 21^\circ\quad\dots\text{1st Quadrant} \\[5ex] \text{cosine is also positive or negative in the rest of the quadrants} \\[3ex] \underline{\text{3rd Quadrant}} \\[3ex] 180^\circ + 159.1238974^\circ \\[3ex] = 339.1238974 \\[3ex] \approx 339^\circ \\[5ex] 180^\circ + 20.8761026^\circ \\[3ex] = 200.8761026 \\[3ex] \approx 201^\circ \\[5ex] \underline{\text{4th Quadrant}} \\[3ex] 360^\circ - 159.1238974^\circ \\[3ex] = 200.8761026 \\[3ex] \approx 201^\circ \\[5ex] 360^\circ - 20.8761026^\circ \\[3ex] = 339.1238974 \\[3ex] \approx 339^\circ \\[5ex] x \approx 21^\circ, 159^\circ, 201^\circ, 339^\circ \\[3ex] $ Check

$x = 20.8761026^\circ, 159.1238974^\circ, 200.8761026^\circ, 339.1238974^\circ$

LHS	RHS
$ \cos 2x - 5\sin^2 x \\[3ex] \cos 2(20.8761026) - 5 * \sin^2(20.8761026) \\[3ex] 0.1111111112 $ $ \cos 2x - 5\sin^2 x \\[3ex] \cos 2(159.1238974) - 5 * \sin^2(159.1238974) \\[3ex] 0.1111111112 $ $ \cos 2x - 5\sin^2 x \\[3ex] \cos 2(200.8761026) - 5 * \sin^2(200.8761026) \\[3ex] 0.1111111112 $ $ \cos 2x - 5\sin^2 x \\[3ex] \cos 2(339.1238974) - 5 * \sin^2(339.1238974) \\[3ex] 0.1111111112 $	$ \dfrac{1}{9} \\[5ex] 0.1111111111 $

$ (b.) \\[3ex] 3(\sqrt{4x - 3}) - \sqrt{5x + 1} = 5 \\[3ex] 3\sqrt{4x - 3} = 5 + \sqrt{5x + 1} \\[3ex] \text{Square both sides} \\[3ex] (3\sqrt{4x - 3})^2 = (5 + \sqrt{5x + 1})^2 \\[3ex] 3^2 \cdot (\sqrt{4x - 3})^2 = (5 + \sqrt{5x + 1})(5 + \sqrt{5x + 1}) \\[3ex] 9(4x - 3) = 25 + 5\sqrt{5x + 1} + 5\sqrt{5x + 1} + (\sqrt{5x + 1})^2 \\[3ex] 36x - 27 = 25 + 10\sqrt{5x + 1} + 5x + 1 \\[3ex] 36x - 5x - 27 - 25 - 1 = 10\sqrt{5x + 1} \\[3ex] 31x - 53 = 10\sqrt{5x + 1} \\[3ex] \text{Square both sides again} \\[3ex] (31x - 53)^2 = (10\sqrt{5x + 1})^2 \\[3ex] (31x - 53)(31x - 53) = 10^2 \cdot (\sqrt{5x + 1})^2 \\[3ex] 961x^2 - 1643x - 1643x + 2809 = 100(5x + 1) \\[3ex] 961x^2 - 3286x + 2809 = 500x + 100 \\[3ex] 961x^2 - 3286x - 500x + 2809 - 100 = 0 \\[3ex] 961x^2 - 3786x + 2709 = 0 \\[3ex] \text{Compare to the standard form of a Quadratic Equation: } ax^2 + bx + c = 0 \\[3ex] a = 961 \\[3ex] b = -3786 \\[3ex] c = 2709 \\[3ex] x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[5ex] x = \dfrac{-(-3786) \pm \sqrt{(-3786)^2 - 4(961)(2709)}}{2(961)} \\[5ex] x = \dfrac{3786 \pm \sqrt{14333796 - 10413396}}{1922} \\[5ex] x = \dfrac{3786 \pm \sqrt{3920400}}{1922} \\[5ex] x = \dfrac{3786 \pm 1980}{1922} \\[5ex] x = \dfrac{3786 + 1980}{1922} \hspace{2em}\text{OR}\hspace{2em} x = \dfrac{3786 - 1980}{1922} \\[5ex] x = 3 \hspace{2em}\text{OR}\hspace{2em} x = 0.9396462019 \\[3ex] \text{Because } x \text{ is an integer, } x = 3 \\[3ex] $ Check

$x = 3$

LHS	RHS
$ 3(\sqrt{4x - 3}) - \sqrt{5x + 1} \\[3ex] 3(\sqrt{4(3) - 3}) - \sqrt{5(3) + 1} \\[3ex] 3(\sqrt{9}) - \sqrt{16} \\[3ex] 3(3) - 4 \\[3ex] 9 - 4 \\[3ex] 5 $	$5$

Let us represent the table as:

Frequency Table

Marks	Mid-Mark, $X$	Number of Students (Frequency), $F$	$FX$
11 — 15	$ \dfrac{11 + 15}{2} = 13 $	p	13p
16 — 20	$ \dfrac{16 + 20}{2} = 18 $	4	72
21 — 25	$ \dfrac{21 + 25}{2} = 23 $	11	253
26 — 30	$ \dfrac{26 + 40}{2} = 28 $	k	28k
31 — 35	$ \dfrac{31 + 35}{2} = 33 $	6	198
36 — 40	$ \dfrac{36 + 40}{2} = 38 $	1	38
		$\Sigma F = p + k + 22 = 40$	$\Sigma FX = 13p + 28k + 561$

$ \underline{\text{Frequency}} \\[3ex] \Sigma F = p + k + 22 = 40 \\[3ex] p + k = 40 - 22 \\[3ex] p + k = 18 ...eqn.(1) \\[5ex] \underline{\text{Mean}} \\[3ex] \text{Mean} = \dfrac{\Sigma FX}{\Sigma F} \\[5ex] 24.75 = \dfrac{13p + 28k + 561}{40} \\[5ex] 13p + 28k + 561 = 24.75(40) \\[3ex] 13p + 28k = 990 - 561 \\[3ex] 13p + 28k = 429 ...eqn.(2) \\[5ex] \underline{\text{Substitution Method}} \\[3ex] \text{From } eqn.(1) \\[3ex] p = 18 - k ...eqn.(3) \\[3ex] \text{Substitute for } p \text{ in } eqn.(2) \\[3ex] 13(18 - k) + 28k = 429 \\[3ex] 234 - 13k + 28k = 429 \\[3ex] 15k = 429 - 234 \\[3ex] k = \dfrac{195}{15} \\[5ex] k = 13 \\[3ex] \text{Substitute for } k \text{ in } eqn.(3) \\[3ex] p = 18 - 13 \\[3ex] p = 5 $

Let the probability that:
John secures a job = P(J) = 70% = 0.7
Ngozi secures a job = P(N) = 60% = 0.6
Ade secures a job = P(A) = a
Assuming the three events (John, Ngozi, Ade securing jobs) are independent.

$ (a.) \\[3ex] \underline{\text{Complementary Rule}} \\[3ex] P(J) + P(J') = 1 \\[3ex] 0.7 + P(J') = 1 \\[3ex] P(J') = 1 - 0.7 = 0.3 \\[5ex] P(N) + P(N') = 1 \\[3ex] 0.6 + P(N') = 1 \\[3ex] P(N') = 1 - 0.6 = 0.4 \\[5ex] P(A) + P(A') = 1 \\[3ex] a + P(A') = 1 \\[3ex] P(A') = 1 - a \\[5ex] $ Only one secures a job implies:
JN'A': John secures a job, Ngozi does not, Ade does not
J'NA': John does not, Ngozi does, Ade does not
J'N'A: John does not, Ngozi does not, Ade does

$ P(\text{only one secures a job}) = 0.29 \quad\dots\text{Given} \\[3ex] But: \\[3ex] P(\text{only one secures a job}) \\[3ex] = P(JN'A') + P(J'NA') + P(J'N'A) \quad\dots\text{Addition Rule for Mutually Exclusive Events} \\[3ex] $

$ \underline{\text{Multiplication Rule for Independent Events}} \\[3ex] P(JN'A') = 0.7(0.4)(1 - a) = 0.28(1 - a) = 0.28 - 0.28a \\[3ex] P(J'NA') = 0.3(0.6)(1 - a) = 0.18(1 - a) = 0.18 - 0.18a \\[3ex] P(J'N'A) = 0.3(0.4)(a) = 0.12a \\[3ex] \implies \\[3ex] (0.28 - 0.28a) + (0.18 - 0.18a) + 0.12a = 0.29 \\[3ex] 0.28 - 0.28a + 0.18 - 0.18a + 0.12a = 0.29 \\[3ex] -0.34a = 0.29 - 0.28 - 0.18 \\[3ex] a = \dfrac{-0.17}{-0.34} \\[5ex] a = 0.5 \\[3ex] P(A) = a = 0.5 \\[3ex] P(A') = 1 - 0.5 = 0.5 \\[3ex] $ Two of them secures a job implies:
JNA': John and Ngozi secure jobs, Ade does not
J'NA: Ngozi and Ade secure jobs, John does not
JN'A: John and Ade does secure jobs, Ngozi does not

$ (b.) \\[3ex] P(\text{two of them secure jobs}) \\[3ex] = P(JNA') + P(J'NA) + P(JN'A) \quad\dots\text{Addition Rule for Mutually Exclusive Events} \\[3ex] ............................................................... \\[3ex] \underline{\text{Multiplication Rule for Independent Events}} \\[3ex] P(JNA') = 0.7(0.6)(0.5) = 0.21 \\[3ex] P(J'NA) = 0.3(0.6)(0.5) = 0.09 \\[3ex] P(JN'A) = 0.7(0.4)(0.5) = 0.14 \\[3ex] ............................................................... \\[3ex] = 0.21 + 0.09 + 0.14 \\[3ex] = 0.44 \\[5ex] (c.) \\[3ex] P(\text{all of them secure jobs}) \\[3ex] = P(J) * P(N) * P(A) \quad\dots\text{Multiplication Rule for Independent Events} \\[3ex] = 0.7(0.6)(0.5) \\[3ex] = 0.21 $

$ S = \text{distance} \\[3ex] t = \text{time} \\[3ex] v = \text{velocity} \\[3ex] a = \text{acceleration} \\[5ex] S(t) = 8t^3 - 7t^2 - 13t + 11 \\[3ex] (a.) \\[3ex] \text{At the start of the journey, } t = 0 \\[3ex] S(0) = 8(0)^3 - 7(0)^2 - 13(0) + 11 \\[3ex] = 0 - 0 - 0 + 11 \\[3ex] = 11\;m \\[5ex] (b.) \\[3ex] \text{When the car is momentarily at rest, } v = 0 \\[3ex] v = \dfrac{dS}{dt} \\[5ex] v = 24t^2 - 14t - 13 \quad\dots\text{Power Rule} \\[3ex] ................................................ \\[3ex] 24t^2 - 14t - 13 = 0 \\[3ex] 24t^2(-13) = -312t^2 \\[3ex] \text{Factors are: } -26t \text{ and } 12t \\[3ex] 24t^2 - 26t + 12t - 13 = 0 \\[3ex] 2t(12t - 13) + 1(12t - 13) = 0 \\[3ex] (12t - 13)(2t + 1) = 0 \\[3ex] 12t - 13 = 0 \hspace{2em}OR\hspace{2em} 2t + 1 = 0 \\[3ex] 12t = 13 \hspace{2em}OR\hspace{2em} 2t = -1 \\[3ex] t = \dfrac{13}{12} \hspace{2em}OR\hspace{2em} t = -\dfrac{1}{2} \\[5ex] ................................................ \\[3ex] \text{Because the time cannot be negative, } t = \dfrac{13}{12}\;s \\[5ex] (c.) \\[3ex] a = \dfrac{dv}{dt} \\[5ex] a(t) = 48t - 14 \\[3ex] \text{At the start of the journey, } t = 0 \\[3ex] a(0) = 48(0) - 14 \\[3ex] a = -14\;m/s^2 \\[5ex] $ (d.) Distance covered in the 5th second
= Distance covered in 5 seconds — Distance covered in 4 seconds

$ S(5) = 8(5)^3 - 7(5)^2 - 13(5) + 11 \\[3ex] = 8(125) - 7(25) - 65 + 11 \\[3ex] = 1000 - 175 - 65 + 11 \\[3ex] = 771\;m \\[5ex] S(4) = 8(4)^3 - 7(4)^2 - 13(4) + 11 \\[3ex] = 8(64) - 7(16) - 52 + 11 \\[3ex] = 512 - 112 - 52 + 11 \\[3ex] = 359\;m \\[5ex] \text{Distance covered in the 5th second} \\[3ex] = S(5) - S(4) \\[3ex] = 771 - 359 \\[3ex] = 412\;m $

$ \underline{\text{Beam } PQ} \\[3ex] |PQ| = 80\;cm \\[5ex] \underline{\text{Weight}} \\[3ex] \text{The weight acts downwards at the center of the beam} \\[3ex] \text{Weight, }W = \text{Mass, }m * \text{Acceleration due to gravity, } g \\[3ex] W = m * g \\[3ex] W = 30 * 10 \\[3ex] W = 300\;N \\[3ex] \text{Centre of Beam} = \dfrac{|PQ|}{2} = \dfrac{80}{2} = 40\;cm \\[5ex] \text{Other Lengths } (cm) \\[3ex] 15 - 10 = 5 \\[3ex] 40 - 15 = 25 \\[3ex] 40 - 32 = 8 \\[3ex] 32 - 18 = 14 \\[5ex] \text{Other Weights } (N = kg * m/s^2) \\[3ex] x * 10 = 10x \\[3ex] y * 10 = 10y \\[3ex] $ Let us represent the information diagrammatically as shown:



$ \underline{\text{Vertical Forces Equilibrium}} \\[3ex] \text{Upward Forces } (N) \\[3ex] 320 + 430 = 750 \\[5ex] \text{Downward Forces } (N) \\[3ex] 10x + 10y + 300 \\[5ex] \text{Downward Forces} = \text{Upward Forces} \\[3ex] 10x + 10y + 300 = 750 \\[3ex] 10x + 10y = 750 - 300 \\[3ex] 10(x + y) = 450 \\[3ex] x + y = \dfrac{450}{10} \\[5ex] x + y = 45 ...eqn.(1) \\[5ex] \underline{\text{Moments Equilibrium about Point E}} \\[3ex] \text{Clockwise Moments } (Ncm) \\[3ex] 320(5 + 25 + 8) + 10y(14) \\[3ex] 320(38) + 140y \\[3ex] 12160 + 140y \\[5ex] \text{AntiClockwise Moments } (Ncm) \\[3ex] 300(8) + 10x(25 + 8) \\[3ex] 2400 + 10x(33) \\[3ex] 2400 + 330x \\[5ex] \text{AntiClockwise Moments} = \text{Clockwise Moments} \\[3ex] 2400 + 330x = 12160 + 140y \\[3ex] 330x - 140y = 12160 - 2400 \\[3ex] 10(33x - 14y) = 9760 \\[3ex] 33x - 14y = \dfrac{9760}{10} \\[5ex] 33x - 14y = 976 ...eqn.(2) \\[5ex] \underline{\text{Substitution Method}} \\[3ex] \text{From } eqn.(1) \\[3ex] x = 45 - y ...eqn.(3) \\[3ex] \text{Substitute for } x \text{ in } eqn.(2) \\[3ex] 33(45 - y) - 14y = 976 \\[3ex] 1485 - 33y - 14y = 976 \\[3ex] -47y = 976 - 1485 \\[3ex] y = \dfrac{-509}{-47} \\[5ex] y = 10.82978723 \\[3ex] y \approx 10.83\;kg \quad\dots\text{to 2 decimal places} \\[5ex] \text{Substitute for } y \text{ in } eqn.(3) \\[3ex] x = 45 - 10.82978723 \\[3ex] x = 34.17021277 \\[3ex] x \approx 34.17\;kg \quad\dots\text{to 2 decimal places} $