Algebra and Functions — Diagnostic Tests

Unit Tests

UT-1: Algebraic Manipulation

Question:

(a) Simplify the following expressions fully: (i) $3x(2x - 5) - 4x(x + 3)$ (ii) $\frac{2x^2 - 8}{x^2 - 4x + 4}$

(b) Solve the equation $\frac{3x + 1}{4} - \frac{x - 2}{3} = 2$.

(c) Factorise fully: $x^3 - x^2 - 12x$.

(d) Express $(2x - 1)^2 - (x + 3)^2$ as a product of linear factors.

Solution:

(a)

(i) $3x(2x - 5) - 4x(x + 3) = 6x^2 - 15x - 4x^2 - 12x = 2x^2 - 27x$

(ii) $\frac{2x^2 - 8}{x^2 - 4x + 4} = \frac{2(x^2 - 4)}{(x - 2)^2} = \frac{2(x - 2)(x + 2)}{(x - 2)^2} = \frac{2(x + 2)}{x - 2}$ (for $x \neq 2$)

(b) Multiply through by 12 (the LCM of 4 and 3):

$12 \times \frac{3x + 1}{4} - 12 \times \frac{x - 2}{3} = 12 \times 2$

$3(3x + 1) - 4(x - 2) = 24$

$9x + 3 - 4x + 8 = 24$

$5x + 11 = 24$

$5x = 13$

$x = \frac{13}{5}$

(c) $x^3 - x^2 - 12x = x(x^2 - x - 12) = x(x - 4)(x + 3)$

(d) Using the difference of two squares: $a^2 - b^2 = (a - b)(a + b)$ where $a = 2x - 1$ and $b = x + 3$:

$(2x - 1)^2 - (x + 3)^2 = [(2x - 1) - (x + 3)][(2x - 1) + (x + 3)]$

$= (2x - 1 - x - 3)(2x - 1 + x + 3)$

$= (x - 4)(3x + 2)$

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UT-2: Functions and Composite Functions

Question:

(a) Given that $f(x) = 2x + 3$ and $g(x) = x^2 - 1$, find: (i) $f(g(2))$ (ii) $gf(x)$ (the composite function $g$ of $f$) (iii) $f^{-1}(x)$

(b) The function $h(x) = \frac{x + 2}{x - 1}$ for $x \neq 1$. Find $h^{-1}(x)$.

(c) State the domain and range of $f(x) = \sqrt{2x - 6}$.

(d) Two functions are defined as $p(x) = 3x - 1$ and $q(x) = \frac{x}{2} + 1$. Show that $p(q(x)) = x$ and hence explain the relationship between $p$ and $q$.

Solution:

(a)

(i) $g(2) = 2^2 - 1 = 3$. Then $f(3) = 2(3) + 3 = 9$. So $f(g(2)) = 9$.

(ii) $gf(x) = g(f(x)) = g(2x + 3) = (2x + 3)^2 - 1 = 4x^2 + 12x + 9 - 1 = 4x^2 + 12x + 8$

(iii) Let $y = 2x + 3$. Then $x = \frac{y - 3}{2}$, so $f^{-1}(x) = \frac{x - 3}{2}$.

(b) Let $y = \frac{x + 2}{x - 1}$. Then:

$y(x - 1) = x + 2$ $xy - y = x + 2$ $xy - x = y + 2$ $x(y - 1) = y + 2$ $x = \frac{y + 2}{y - 1}$

So $h^{-1}(x) = \frac{x + 2}{x - 1}$, which equals $h(x)$ itself (this function is self-inverse).

(c) The expression under the square root must be non-negative: $2x - 6 \geq 0$, so $x \geq 3$.

Domain: $x \geq 3$, or $[3, \infty)$.

Since $\sqrt{2x - 6}$ is always non-negative, the range is $y \geq 0$, or $[0, \infty)$.

(d) $q(x) = \frac{x}{2} + 1$. Then:

$p(q(x)) = 3\left(\frac{x}{2} + 1\right) - 1 = \frac{3x}{2} + 3 - 1 = \frac{3x}{2} + 2$

This does not equal $x$. Let me verify: $p(q(x)) = 3(\frac{x}{2} + 1) - 1 = \frac{3x}{2} + 3 - 1 = \frac{3x}{2} + 2$.

This is not equal to $x$ for all $x$. However, let us check $q(p(x))$:

$q(p(x)) = \frac{3x - 1}{2} + 1 = \frac{3x - 1 + 2}{2} = \frac{3x + 1}{2}$

This is also not $x$. Let me re-examine the question. For $p$ and $q$ to be inverses, we need $p(q(x)) = x$. The given functions do not satisfy this, so the question may contain an error. If $q(x)$ were instead $\frac{x + 1}{3}$, then $p(q(x)) = 3(\frac{x+1}{3}) - 1 = x + 1 - 1 = x$, confirming $p$ and $q$ are inverse functions.

Assuming the intended functions are $p(x) = 3x - 1$ and $q(x) = \frac{x + 1}{3}$: $p(q(x)) = x$ confirms that $p$ and $q$ are inverse functions of each other, meaning $q = p^{-1}$.

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UT-3: Graphs of Functions

Question:

(a) Sketch the graph of $y = x^2 - 4x + 3$, labelling the y-intercept, the x-intercepts, and the coordinates of the turning point.

(b) The function $y = f(x)$ is transformed to $y = -f(x + 2) + 1$. Describe the sequence of transformations applied to the graph of $y = f(x)$.

(c) A quadratic function has a minimum value of $-5$ at $x = 3$ and passes through the point $(1, 3)$. Find the equation of the quadratic.

(d) Determine whether the function $f(x) = x^3 - 3x + 1$ has any stationary points. If so, find their coordinates and determine their nature.

Solution:

(a) $y = x^2 - 4x + 3 = (x - 1)(x - 3)$

x-intercepts: $x = 1$ and $x = 3$ (when $y = 0$).

y-intercept: When $x = 0$, $y = 3$.

Turning point: $x = -\frac{b}{2a} = -\frac{-4}{2} = 2$. When $x = 2$, $y = 4 - 8 + 3 = -1$. Turning point at $(2, -1)$.

The graph is a parabola opening upwards (since $a = 1 > 0$), with vertex at $(2, -1)$, crossing the x-axis at $(1, 0)$ and $(3, 0)$, and the y-axis at $(0, 3)$.

(b) The sequence of transformations (applied in the correct order) is:

1. Translate the graph 2 units to the left (replace $x$ with $x + 2$).
2. Reflect the graph in the x-axis (multiply by $-1$).
3. Translate the graph 1 unit up (add 1).

Note: the order of operations for combined transformations is: horizontal translation first, then vertical reflection, then vertical translation.

(c) The quadratic has the form $y = a(x - h)^2 + k$ where the minimum is at $(h, k) = (3, -5)$.

So $y = a(x - 3)^2 - 5$. Using the point $(1, 3)$:

$3 = a(1 - 3)^2 - 5$ $3 = a(4) - 5$ $8 = 4a$ $a = 2$

The equation is $y = 2(x - 3)^2 - 5$, or expanded: $y = 2x^2 - 12x + 13$.

(d) To find stationary points, set $f'(x) = 0$:

$f'(x) = 3x^2 - 3$ $3x^2 - 3 = 0$ $x^2 = 1$ $x = 1 \text{ or } x = -1$

At $x = 1$: $y = 1 - 3 + 1 = -1$. Point: $(1, -1)$. At $x = -1$: $y = -1 + 3 + 1 = 3$. Point: $(-1, 3)$.

Using the second derivative $f''(x) = 6x$:

At $x = 1$: $f''(1) = 6 > 0$, so $(1, -1)$ is a local minimum. At $x = -1$: $f''(-1) = -6 < 0$, so $(-1, 3)$ is a local maximum.

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Integration Tests

IT-1: Functions and Quadratic Applications

Question:

(a) The revenue $R$ (in pounds) from selling $x$ items is given by $R(x) = -2x^2 + 80x$, and the cost is given by $C(x) = 15x + 200$. Find the number of items that maximises profit and calculate the maximum profit.

(b) A ball is thrown upwards and its height $h$ metres after $t$ seconds is given by $h(t) = -5t^2 + 20t + 1$. Find the maximum height reached and the time at which the ball hits the ground.

(c) Given $f(x) = 2x + 1$ and $g(x) = \frac{1}{x}$ for $x \neq 0$, solve the equation $fg(x) = g(x) + 3$.

(d) The function $y = x^2 + px + q$ has a turning point at $(2, -3)$. Find the values of $p$ and $q$.

Solution:

(a) Profit $P(x) = R(x) - C(x) = -2x^2 + 80x - 15x - 200 = -2x^2 + 65x - 200$.

Maximum profit occurs at the vertex: $x = -\frac{b}{2a} = -\frac{65}{-4} = 16.25$.

Since we cannot sell a fraction of an item, check $x = 16$ and $x = 17$:

$P(16) = -2(256) + 65(16) - 200 = -512 + 1040 - 200 = 328$ $P(17) = -2(289) + 65(17) - 200 = -578 + 1105 - 200 = 327$

Maximum profit is $\pounds 328$ when 16 items are sold.

(b) Maximum height at the vertex: $t = -\frac{20}{2(-5)} = 2$ seconds.

$h(2) = -5(4) + 20(2) + 1 = -20 + 40 + 1 = 21$ metres.

The ball hits the ground when $h = 0$: $-5t^2 + 20t + 1 = 0$, so $5t^2 - 20t - 1 = 0$.

$t = \frac{20 \pm \sqrt{400 + 20}}{10} = \frac{20 \pm \sqrt{420}}{10} = \frac{20 \pm 20.49}{10}$

Taking the positive root: $t = \frac{40.49}{10} \approx 4.05$ seconds.

(c) $fg(x) = f(g(x)) = f\left(\frac{1}{x}\right) = 2\left(\frac{1}{x}\right) + 1 = \frac{2}{x} + 1$.

$\frac{2}{x} + 1 = \frac{1}{x} + 3$ $\frac{1}{x} = 2$ $x = \frac{1}{2}$

(d) At the turning point, $f'(x) = 2x + p = 0$ when $x = 2$: $2(2) + p = 0$, so $p = -4$.

Substituting $x = 2, y = -3$: $-3 = 4 - 8 + q$, so $q = 1$.

The equation is $y = x^2 - 4x + 1$.

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IT-2: Advanced Function Analysis

Question:

(a) The function $f(x) = x^2 - 6x + 5$ is defined on the domain $x \geq 3$. Find $f^{-1}(x)$ and state its domain.

(b) Solve the inequality $x^2 - 5x + 6 > 0$.

(c) The graph of $y = f(x)$ passes through the points $(0, 2)$, $(2, 6)$, and $(4, 2)$. Sketch a possible graph and determine whether $f$ could be a quadratic function. Explain your reasoning.

(d) A function is defined by $f(x) = ax^3 + bx^2 + cx + d$. Given that $f(0) = 4$, $f(1) = 2$, $f(2) = 6$, and $f(3) = 28$, find the values of $a$, $b$, $c$, and $d$.

Solution:

(a) Completing the square: $f(x) = x^2 - 6x + 5 = (x - 3)^2 - 4$.

Since the domain is $x \geq 3$, $f(x) \geq -4$ (range). Let $y = (x - 3)^2 - 4$:

$y + 4 = (x - 3)^2$ $x - 3 = \sqrt{y + 4}$ (taking positive root since $x \geq 3$) $x = 3 + \sqrt{y + 4}$

$f^{-1}(x) = 3 + \sqrt{x + 4}$.

Domain of $f^{-1}$: $x \geq -4$ (which is the range of $f$).

(b) $x^2 - 5x + 6 > 0$. Factorising: $(x - 2)(x - 3) > 0$.

The quadratic is positive when $x < 2$ or $x > 3$. So the solution is $x \in (-\infty, 2) \cup (3, \infty)$.

(c) If $f$ is quadratic, its graph is a parabola. A parabola is symmetric about its axis of symmetry. The points $(0, 2)$ and $(4, 2)$ are symmetric about $x = 2$, which is consistent. The turning point would be at $x = 2$ with $y = 6$ (maximum, since the parabola opens downward). The quadratic $f(x) = -a(x - 2)^2 + 6$: using $(0, 2)$: $2 = -4a + 6$, so $a = 1$. Thus $f(x) = -(x - 2)^2 + 6 = -x^2 + 4x + 2$. Yes, this could be a quadratic function.

(d) From $f(0) = 4$: $d = 4$.

From $f(1) = 2$: $a + b + c + 4 = 2$, so $a + b + c = -2$ … (1) From $f(2) = 6$: $8a + 4b + 2c + 4 = 6$, so $8a + 4b + 2c = 2$, i.e., $4a + 2b + c = 1$ … (2) From $f(3) = 28$: $27a + 9b + 3c + 4 = 28$, so $27a + 9b + 3c = 24$, i.e., $9a + 3b + c = 8$ … (3)

Subtract (1) from (2): $3a + b = 3$ … (4) Subtract (2) from (3): $5a + b = 7$ … (5) Subtract (4) from (5): $2a = 4$, so $a = 2$.

From (4): $b = 3 - 6 = -3$. From (1): $c = -2 - 2 + 3 = -1$.

$f(x) = 2x^3 - 3x^2 - x + 4$.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Common Pitfalls

• Forgetting to consider the domain when finding inverse functions, especially for restricted quadratics and square root functions.
• Applying transformations in the wrong order: horizontal transformations must be applied before vertical transformations.
• Sign errors when completing the square or when solving quadratic inequalities (forgetting to reverse the inequality when multiplying by a negative value).

- Confusing composite function notation: $fg(x)$ means $f(g(x))$, applying $g$ first, then $f$.