Algebra — Introduction & Intermediate

Question 1

(a) $h_0 + ut - \tfrac12 g t^2 = H \Rightarrow \tfrac12 g t^2 - ut + (H - h_0) = 0$. So $a = \tfrac{g}{2}$, $b = -u$, $c = H - h_0$. (3) (1 for rearranging, 2 for correct signs of a,b,c.)

(b) Equal roots ⟺ discriminant $= 0$: $b^2 - 4ac = u^2 - 4\cdot\tfrac{g}{2}(H - h_0) = u^2 - 2g(H - h_0) = 0.$ Thus $H = h_0 + \dfrac{u^2}{2g}$. (3) Physically this is the maximum height: at exactly one instant the ball reaches it (apex), beyond which the trajectory never gets. (2) (consistency with energy $\tfrac12 u^2 = g\Delta h$).

(c) $g=10, u=20, h_0=5, H=20$: $5t^2 - 20t + 15 = 0 \Rightarrow t^2 - 4t + 3 = 0$. Discriminant $= 16 - 12 = 4$. $t = \dfrac{4 \pm 2}{2}$. So $t = 3$ or $t = 1$ (seconds). (6) (2 setup, 2 discriminant, 2 both roots.)

(d) From $t^2 - 4t + 3 = 0$: sum $= 4$, product $= 3$. (2) Meaning of sum: $t_1 + t_2 = 4$ s; the times are symmetric about the apex time $t_{\text{apex}} = \tfrac{t_1+t_2}{2} = 2$ s $= u/g$. (1)

(e) (3)

def reaches(H, g=10, u=20, h0=5):
    a, b, c = g/2, -u, H - h0
    D = b*b - 4*a*c
    return 0 if D < 0 else (1 if D == 0 else 2)

Question 2

(a) Remainder Theorem: dividing $P(x)$ by $(x-r)$ gives $P(x) = (x-r)Q(x) + R$ where $R$ is constant (degree $< 1$). Setting $x = r$: $P(r) = R$. (2) Factor Theorem: $(x-r)$ divides $P(x)$ ⟺ remainder $R = 0$ ⟺ $P(r) = 0$ (by the above). Both directions:

• If $P(r)=0$ then $R=0$ so $P(x)=(x-r)Q(x)$, i.e. $(x-r)\mid P(x)$.
• If $(x-r)\mid P(x)$ then $R=0$ so $P(r)=R=0$. (3)

(b) Candidates $\pm\frac{p}{q}$, $p \mid 6$, $q \mid 2$: $\pm1,\pm2,\pm3,\pm6,\pm\tfrac12,\pm\tfrac32$. (2) Test $x=3$: $2(27)-3(9)-33+6 = 54-27-33+6 = 0$. So $x=3$ is a root. (2)

(c) Synthetic division by $3$ on coefficients $2, -3, -11, 6$: $\begin{array}{c|cccc} 3 & 2 & -3 & -11 & 6\\ & & 6 & 9 & -6\\\hline & 2 & 3 & -2 & 0\end{array}$ Quotient $2x^2 + 3x - 2 = (2x - 1)(x + 2)$. (4) So $P(x) = (x - 3)(2x - 1)(x + 2)$. Roots $3, \tfrac12, -2$. (2)

(d) Expand: $(2x-1)(x+2) = 2x^2 + 3x - 2$; times $(x-3)$: $2x^3 -6x^2 +3x^2 -9x -2x +6 = 2x^3 -3x^2 -11x +6$ ✓. (3) Vieta check: sum of roots $3 + \tfrac12 - 2 = \tfrac32 = -\tfrac{-3}{2} = -b/a$ ✓. Product $3\cdot\tfrac12\cdot(-2) = -3 = -\tfrac{6}{2} = -d/a$ ✓. (2)

Question 3

(a) Square: $2x+7 = (x-4)^2 = x^2 -8x +16 \Rightarrow x^2 -10x +9 = 0 \Rightarrow (x-1)(x-9)=0$. (3) Candidates $x=1, x=9$. Domain requires RHS $x-4 \ge 0 \Rightarrow x \ge 4$. $x=1$: RHS $=-3<0$, extraneous — reject. $x=9$: $\sqrt{25}=5 = 9-4$ ✓. Solution: $x = 9$. (3) (reject reasoning essential.)

(b) Multiply by conjugate: $\frac{3}{\sqrt5-\sqrt2}\cdot\frac{\sqrt5+\sqrt2}{\sqrt5+\sqrt2} = \frac{3(\sqrt5+\sqrt2)}{5-2} = \frac{3(\sqrt5+\sqrt2)}{3} = \sqrt5+\sqrt2.$ (4)

(c) $|2x-3|\le 5 \Rightarrow -5 \le 2x-3 \le 5 \Rightarrow -2 \le 2x \le 8 \Rightarrow -1 \le x \le 4$. (3) Number line: closed dots at $-1$ and $4$, shaded segment between. (1)

(d) $-1 < \frac{3x-2}{4} \le 2 \Rightarrow -4 < 3x-2 \le 8 \Rightarrow -2 < 3x \le 10 \Rightarrow -\tfrac23 < x \le \tfrac{10}{3}$. (3)

(e) (3)

def is_solution(x, tol=1e-9):
    return x - 4 >= -tol and 2*x + 7 >= -tol and \
           abs((2*x + 7)**0.5 - (x - 4)) < tol

[
  {"claim":"Q1c roots of t^2-4t+3 are 1 and 3","code":"t=symbols('t'); r=solve(Eq(t**2-4*t+3,0),t); result=set(r)=={1,3}"},
  {"claim":"Q1b max height H=h0+u^2/(2g) for g=10,u=20,h0=5 equals 25","code":"result=(5+20**2/(2*10))==25"},
  {"claim":"Q2 P(x)=2x^3-3x^2-11x+6 factors as (x-3)(2x-1)(x+2)","code":"x=symbols('x'); result=expand((x-3)*(2*x-1)*(x+2))==2*x**3-3*x**2-11*x+6"},
  {"claim":"Q2 roots sum=3/2 and product=-3","code":"x=symbols('x'); rs=solve(2*x**3-3*x**2-11*x+6,x); result=(sum(rs)==Rational(3,2)) and (prod(rs)==-3)"},
  {"claim":"Q3a only x=9 solves sqrt(2x+7)=x-4","code":"x=symbols('x'); sols=solve(Eq(sqrt(2*x+7),x-4),x); result=sols==[9]"},
  {"claim":"Q3b rationalisation equals sqrt5+sqrt2","code":"result=simplify(3/(sqrt(5)-sqrt(2))-(sqrt(5)+sqrt(2)))==0"}
]