Secondary 3 Additional Mathematics Algebra Functions Quiz

Secondary 3 Additional Mathematics Quiz - Algebra Functions (Answer Key)

1.
$2x^2 - 8x + 5 = 2(x^2 - 4x) + 5$
$= 2[(x-2)^2 - 4] + 5$
$= 2(x-2)^2 - 8 + 5$
$= 2(x-2)^2 - 3$
Answer: $a=2, h=2, k=-3$
[M1 for completing square inside bracket, M1 for expanding and simplifying, A1]

2.
(a) Minimum value is $k = -3$.
(b) Axis of symmetry is $x = 2$.
[A1 for each]

3.
For no real roots, discriminant $\Delta < 0$.
$\Delta = b^2 - 4ac = (k-2)^2 - 4(1)(4) < 0$
$(k-2)^2 - 16 < 0$
$(k-2)^2 < 16$
$-4 < k-2 < 4$
$-2 < k < 6$
Answer: $-2 < k < 6$
[M1 for setting up $\Delta < 0$, M1 for solving inequality, A1]

4.
Intersection: $x^2 - 4x + 7 = 2x + c$
$x^2 - 6x + (7-c) = 0$
For tangent, $\Delta = 0$.
$(-6)^2 - 4(1)(7-c) = 0$
$36 - 28 + 4c = 0$
$8 + 4c = 0 \Rightarrow 4c = -8 \Rightarrow c = -2$
Answer: $c = -2$
[M1 for forming quadratic, M1 for $\Delta=0$, M1 for solving, A1]

5.
$x^2 - 5x + 6 \le 0$
$(x-2)(x-3) \le 0$
Critical values: $x=2, x=3$.
Since coefficient of $x^2$ is positive, the curve is below the axis between the roots.
Answer: $2 \le x \le 3$
[M1 for factors, M1 for critical values/inequality logic, A1 for final range. Number line should show solid dots at 2 and 3 and shading between them.]

6.
$P(1) = -4 \Rightarrow 2(1)^3 - 5(1)^2 + a(1) + b = -4$
$2 - 5 + a + b = -4 \Rightarrow a + b = -1$ --- (1)
$P(-2) = -20 \Rightarrow 2(-2)^3 - 5(-2)^2 + a(-2) + b = -20$
$-16 - 20 - 2a + b = -20 \Rightarrow -36 - 2a + b = -20 \Rightarrow -2a + b = 16$ --- (2)
(1) - (2): $(a+b) - (-2a+b) = -1 - 16$
$3a = -17 \Rightarrow a = -\frac{17}{3}$
Substitute $a$ into (1): $-\frac{17}{3} + b = -1 \Rightarrow b = \frac{14}{3}$
Answer: $a = -\frac{17}{3}, b = \frac{14}{3}$
[M1 for $P(1)$, M1 for $P(-2)$, M1 for solving simultaneous, A1]

7.
$\frac{3}{\sqrt{5}-\sqrt{2}} = \frac{3(\sqrt{5}+\sqrt{2})}{5-2} = \frac{3\sqrt{5}+3\sqrt{2}}{3} = \sqrt{5}+\sqrt{2}$
$\frac{2}{\sqrt{5}+\sqrt{2}} = \frac{2(\sqrt{5}-\sqrt{2})}{5-2} = \frac{2\sqrt{5}-2\sqrt{2}}{3}$
Sum $= \sqrt{5}+\sqrt{2} + \frac{2\sqrt{5}-2\sqrt{2}}{3}$
$= \frac{3\sqrt{5}+3\sqrt{2} + 2\sqrt{5}-2\sqrt{2}}{3}$
$= \frac{5\sqrt{5}+\sqrt{2}}{3}$
Answer: $\frac{5\sqrt{5}+\sqrt{2}}{3}$
[M1 for rationalizing first term, M1 for rationalizing second term, M1 for combining, A1]

8.
$\sqrt{2x+3} = x$
Square both sides: $2x + 3 = x^2$
$x^2 - 2x - 3 = 0$
$(x-3)(x+1) = 0$
$x = 3$ or $x = -1$.
Check:
If $x=3$, LHS $=\sqrt{9}=3$, RHS $=3$. Valid.
If $x=-1$, LHS $=\sqrt{1}=1$, RHS $=-1$. Invalid.
Answer: $x = 3$
[M1 for squaring, M1 for solving quadratic, M1 for checking, A1]

9.
$(1+2x)^5 = 1 + 5(2x) + 10(2x)^2 + \dots = 1 + 10x + 40x^2 + \dots$
Multiply by $(3-x)$:
$(1 + 10x + 40x^2)(3 - x)$
Term in $x^2$:
From $10x \cdot (-x) = -10x^2$
From $40x^2 \cdot 3 = 120x^2$
Total coeff: $120 - 10 = 110$.
Answer: 110
[M1 for expansion of binomial up to $x^2$, M1 for identifying relevant products, A1]

10.
General term of $(2+kx)^6$: $\binom{6}{r} (2)^{6-r} (kx)^r$.
For $x^2$, $r=2$.
Coeff $= \binom{6}{2} (2)^4 (k)^2 = 15 \cdot 16 \cdot k^2 = 240k^2$.
Given coeff is 60:
$240k^2 = 60$
$k^2 = \frac{60}{240} = \frac{1}{4}$
$k = \pm \frac{1}{2}$.
Answer: $k = \frac{1}{2}, -\frac{1}{2}$
[M1 for general term/r=2, M1 for setting up eq, A1 for both values]

11.
$\frac{5x-1}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1}$
$5x - 1 = A(x+1) + B(x-2)$
Let $x=2$: $9 = 3A \Rightarrow A=3$.
Let $x=-1$: $-6 = -3B \Rightarrow B=2$.
Answer: $\frac{3}{x-2} + \frac{2}{x+1}$
[M1 for form, M1 for solving A, A1 for B and final answer]

12.
$\frac{3x^2+10x+4}{(x+2)(x^2+1)} = \frac{A}{x+2} + \frac{Bx+C}{x^2+1}$
$3x^2+10x+4 = A(x^2+1) + (Bx+C)(x+2)$
Let $x=-2$: $3(4)-20+4 = -4$.
$-4 = A(5) \Rightarrow A = -\frac{4}{5}$.
Compare $x^2$: $3 = A + B \Rightarrow B = 3 - (-\frac{4}{5}) = \frac{19}{5}$.
Compare const: $4 = A + 2C \Rightarrow 4 = -\frac{4}{5} + 2C \Rightarrow 2C = \frac{24}{5} \Rightarrow C = \frac{12}{5}$.
Answer: $\frac{-4/5}{x+2} + \frac{19x/5 + 12/5}{x^2+1}$ or $\frac{1}{5} \left[ \frac{-4}{x+2} + \frac{19x+12}{x^2+1} \right]$
[M1 for form, M1 for A, M1 for B/C, A1]

13.
(a) $y = \frac{2x+1}{x-3}$
$y(x-3) = 2x+1$
$xy - 3y = 2x + 1$
$xy - 2x = 3y + 1$
$x(y-2) = 3y + 1$
$x = \frac{3y+1}{y-2}$
$f^{-1}(x) = \frac{3x+1}{x-2}$
[M1 for swapping/rearranging, M1 for isolating x, A1]

(b) Domain of $f^{-1}$ is Range of $f$.
$f(x) = \frac{2x+1}{x-3} = \frac{2(x-3)+7}{x-3} = 2 + \frac{7}{x-3}$.
As $x \to \infty$, $f(x) \to 2$. $f(x) \ne 2$.
Alternatively, denominator of $f^{-1}(x)$ cannot be zero.
$x - 2 \ne 0 \Rightarrow x \ne 2$.
Answer: $x \in \mathbb{R}, x \ne 2$
[A1]

14.
(a) $g(2) = 2^2 - 4 = 0$.
$f(g(2)) = f(0) = \frac{2(0)+1}{0-3} = -\frac{1}{3}$.
Answer: $-\frac{1}{3}$
[M1 for $g(2)$, A1 for $f(0)$]

(b) $gf(x) = g(f(x)) = (f(x))^2 - 4$.
$f(x)$ is defined for $x \ne 3$.
However, the range of $f(x)$ is $\mathbb{R} \setminus \{2\}$.
The domain of $g$ is $x \ge 0$.
For $gf(x)$ to be defined, $f(x)$ must be $\ge 0$.
$\frac{2x+1}{x-3} \ge 0$. This is not true for all real $x$ (e.g., $x=0 \Rightarrow f(0)=-1/3 < 0$).
Thus, $g(f(x))$ is undefined when $f(x) < 0$.
Answer: Because the range of $f$ includes negative values, which are not in the domain of $g$ ($x \ge 0$).
[M1 for identifying domain constraint of g, A1 for explanation]

15.
(a) $\frac{1}{x} = x - 2$
$1 = x(x-2)$
$1 = x^2 - 2x$
$x^2 - 2x - 1 = 0$
[Shown]
[M1 for equating, A1 for correct quadratic]

(b) $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2}$
$x = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$.
If $x = 1 + \sqrt{2}$, $y = \frac{1}{1+\sqrt{2}} = \sqrt{2}-1$. (Or $y = x-2 = \sqrt{2}-1$).
If $x = 1 - \sqrt{2}$, $y = \frac{1}{1-\sqrt{2}} = -1-\sqrt{2}$. (Or $y = x-2 = -1-\sqrt{2}$).
Answer: $(1+\sqrt{2}, \sqrt{2}-1)$ and $(1-\sqrt{2}, -1-\sqrt{2})$
[M1 for solving x, M1 for finding y, A1 for both coordinates]

16.
$y = |2x - 4|$.
Vertex at $2x-4=0 \Rightarrow x=2, y=0$. Point $(2,0)$.
y-intercept: $x=0 \Rightarrow y=|-4|=4$. Point $(0,4)$.
Endpoint $x=5 \Rightarrow y=|10-4|=6$. Point $(5,6)$.
Endpoint $x=-1 \Rightarrow y=|-2-4|=6$. Point $(-1,6)$.
Graph is V-shaped with vertex at $(2,0)$, passing through $(0,4), (-1,6), (5,6)$.
[M1 for vertex, M1 for intercepts/endpoints, A1 for correct shape and labels]

17.
Case 1: $2x - 4 \ge 0 \Rightarrow x \ge 2$.
$2x - 4 = 3 - x \Rightarrow 3x = 7 \Rightarrow x = 7/3$.
$7/3 \ge 2$, so valid.
Case 2: $2x - 4 < 0 \Rightarrow x < 2$.
$-(2x - 4) = 3 - x \Rightarrow -2x + 4 = 3 - x \Rightarrow 1 = x$.
$1 < 2$, so valid.
Answer: $x = 1, x = \frac{7}{3}$
[M1 for setting up cases, M1 for solving one case, A1 for both solutions]

18.
(a) Asymptotes: Vertical $x=1$, Horizontal $y=0$.
Graph is in 1st quadrant relative to asymptotes (since $x>1 \Rightarrow y>0$).
Passes through $(2,1), (1.5, 2)$, etc.
[M1 for asymptotes, M1 for shape, A1 for correct quadrant/position]

(b) Since $x > 1$, $x-1 > 0$, so $\frac{1}{x-1} > 0$.
As $x \to 1^+, y \to \infty$. As $x \to \infty, y \to 0$.
Answer: $y > 0$ (or $h(x) \in \mathbb{R}^+$)
[A1]

19.
(a) $gf(x) = g(3x-2) = (3x-2)^2 + 1$
$= 9x^2 - 12x + 4 + 1$
$= 9x^2 - 12x + 5$.
Answer: $9x^2 - 12x + 5$
[M1 for substitution, A1 for expansion]

(b) $9x^2 - 12x + 5 = 10$
$9x^2 - 12x - 5 = 0$
$(3x-5)(3x+1) = 0$
$x = \frac{5}{3}$ or $x = -\frac{1}{3}$.
Answer: $x = \frac{5}{3}, -\frac{1}{3}$
[M1 for setting up eq, M1 for solving, A1]

20.
$y = ax^2 + bx + c$.
Pt $(0,5) \Rightarrow c = 5$.
Pt $(1,2) \Rightarrow a + b + 5 = 2 \Rightarrow a + b = -3$ --- (1)
Pt $(2,3) \Rightarrow 4a + 2b + 5 = 3 \Rightarrow 4a + 2b = -2 \Rightarrow 2a + b = -1$ --- (2)
(2) - (1): $(2a+b) - (a+b) = -1 - (-3)$
$a = 2$.
Sub into (1): $2 + b = -3 \Rightarrow b = -5$.
Answer: $a=2, b=-5, c=5$
[M1 for finding c, M1 for setting up simultaneous eqs, M1 for solving, A1]