MATHWISE ACADEMY
GRADE 11 MATHEMATICS
June Practice Examination (Paper 1 - Test 3)
Examiner: Vashi S Y
Instructions & Guidelines:
- This simulated workspace contains Questions 1 to 3 totaling 100 marks.
- Ensure all algebraic solutions are fully expanded and simplified before checking solutions.
- 💡 Tip: Use your browser's print function (Ctrl/Cmd + P or click the Download / Print PDF button) to save this out as a clean, structured worksheet!
QUESTION 1: EXPONENTS, SURDS & NATURE OF ROOTS
[20 Marks]Step-by-Step Solution:
Express all bases as powers of prime base 5:
$$\frac{5^{2x+1} \cdot (5^2)^{x-2}}{(5^3)^{x-1}}$$
Apply the index rule $(a^m)^n = a^{mn}$ to expand indices:
$$= \frac{5^{2x+1} \cdot 5^{2x-4}}{5^{3x-3}}$$
Apply prime base multiplication (add exponents) and division (subtract exponents) rules:
$$= \frac{5^{(2x+1) + (2x-4)}}{5^{3x-3}} = \frac{5^{4x-3}}{5^{3x-3}}$$
$$= 5^{(4x-3) - (3x-3)} = 5^{4x - 3 - 3x + 3}$$
$$= 5^x$$
- Expressing bases 25 and 125 as base 5 powers (1 Mark)
- Simplifying exponential factors to $5^{2x-4}$ and $5^{3x-3}$ (1 Mark)
- Applying multiplication sum laws to numerator exponents (1 Mark)
- Correctly executing division subtraction law to yield $5^x$ (1 Mark)
Step-by-Step Solution:
Simplify each radical by factorising out the highest perfect square factor:
$$2\sqrt{18} = 2\sqrt{9 \cdot 2} = 2(3\sqrt{2}) = 6\sqrt{2}$$
$$\sqrt{98} = \sqrt{49 \cdot 2} = 7\sqrt{2}$$
$$\sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}$$
Group and combine like terms:
$$6\sqrt{2} - 7\sqrt{2} + 4\sqrt{2} = 3\sqrt{2}$$
- Simplifying at least two of the surd terms correctly (2 Marks)
- Final combined calculation $3\sqrt{2}$ (1 Mark)
Prove algebraically that the roots of the equation are real and unequal for all real values of $p$ where $p \neq 6$ and $p \neq -2$.
Step-by-Step Solution:
To investigate nature of roots, solve for the discriminant ($\Delta = b^2 - 4ac$):
Substitute $a = 1$, $b = -p$, and $c = p + 3$:
$$\Delta = (-p)^2 - 4(1)(p + 3)$$
$$\Delta = p^2 - 4p - 12$$
Factorise the quadratic trinomial expression:
$$\Delta = (p - 6)(p + 2)$$
For roots to be real and unequal, $\Delta > 0$:
$$(p - 6)(p + 2) > 0$$
Applying inequality interval properties, the roots are real and unequal ($\Delta > 0$) for:
$$p < -2 \quad \text{or} \quad p > 6$$
- Identifying correct coefficients $a$, $b$, and $c$ (1 Mark)
- Correctly simplifying discriminant equation $\Delta = p^2 - 4p - 12$ (1 Mark)
- Correct binomial factorisation $(p - 6)(p + 2)$ (1 Mark)
- Stating the positive discriminant inequality requirement $\Delta > 0$ (1 Mark)
- Stating the final inequality solution intervals (1 Mark)
Step-by-Step Solution:
For equal roots, the discriminant must equal zero ($\Delta = 0$):
$$\Delta = b^2 - 4ac = 0$$
Substitute $a = 1$, $b = -4$, and $c = k + 1$:
$$\Delta = (-4)^2 - 4(1)(k + 1) = 0$$
$$16 - 4k - 4 = 0$$
$$12 - 4k = 0 \implies 4k = 12$$
$$k = 3$$
- Stating the equal roots condition requirement $\Delta = 0$ (1 Mark)
- Substituting values and solving equation correctly $12 - 4k = 0$ (1 Mark)
- Final answer $k = 3$ (1 Mark)
Step-by-Step Solution:
Rewrite the area expression in standard quadratic trinomial format:
$$A(x) = -2x^2 + 160x$$
Factorise the coefficient of $x^2$ (which is $-2$):
$$A(x) = -2(x^2 - 80x)$$
Complete the square inside the brackets by adding and subtracting $(\frac{-80}{2})^2 = 1600$:
$$A(x) = -2(x^2 - 80x + 1600 - 1600)$$
$$A(x) = -2\left[(x - 40)^2 - 1600\right]$$
Distribute $-2$ back across the bracket terms:
$$A(x) = -2(x - 40)^2 + 3200$$
The maximum area is $3200\text{ m}^2$ (which occurs when width $x = 40\text{ m}$).
- Factoring out $-2$ coefficient and completing square correctly (1 Mark)
- Expanding to standard vertex form $-2(x - 40)^2 + 3200$ (1 Mark)
- Stating the maximum possible area of $3200\text{ m}^2$ (1 Mark)
QUESTION 2: EQUATIONS & INEQUALITIES
[40 Marks]Step-by-Step Solution:
Set each linear factor of the binomial brackets equal to zero:
$$x - 3 = 0 \quad \text{or} \quad 2x + 1 = 0$$
$$x = 3 \quad \text{or} \quad x = -\frac{1}{2}$$
- First correct root $x = 3$ (1 Mark)
- Second correct root $x = -\frac{1}{2}$ (1 Mark)
Step-by-Step Solution:
Apply the quadratic formula with parameters $a = 2$, $b = 5$, and $c = -4$:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
$$x = \frac{-(5) \pm \sqrt{(5)^2 - 4(2)(-4)}}{2(2)}$$
$$x = \frac{-5 \pm \sqrt{25 + 32}}{4} = \frac{-5 \pm \sqrt{57}}{4}$$
Calculate decimal values using a scientific calculator:
$$x \approx 0.64 \quad \text{or} \quad x \approx -3.14$$
- Writing down or correctly implying standard formula use (1 Mark)
- Correct substitution of values (1 Mark)
- Correct decimal evaluation for first root $0.64$ (1 Mark)
- Correct decimal evaluation for second root $-3.14$ (1 Mark)
Step-by-Step Solution:
Square both sides of the equation to eliminate the radical (restrictions require $x \ge \frac{2}{3}$ and $x \ge 2$):
$$3x - 2 = (x - 2)^2$$
$$3x - 2 = x^2 - 4x + 4$$
Rearrange into standard quadratic form:
$$x^2 - 7x + 6 = 0$$
Factorise the quadratic expression:
$$(x - 6)(x - 1) = 0 \implies x = 6 \quad \text{or} \quad x = 1$$
Check Solutions:
- Check $x = 6$: LHS $= \sqrt{3(6) - 2} = \sqrt{16} = 4$; RHS $= 6 - 2 = 4$. (Valid)
- Check $x = 1$: LHS $= \sqrt{3(1) - 2} = \sqrt{1} = 1$; RHS $= 1 - 2 = -1$. (Invalid, LHS $\neq$ RHS)
$$x = 6 \text{ only}$$
- Squaring both sides of the equation (1 Mark)
- Correct expansion and standard form $x^2 - 7x + 6 = 0$ (1 Mark)
- Factorisation to find both roots $x = 6$ and $x = 1$ (1 Mark)
- Showing valid check substitution step (1 Mark)
- Concluding $x = 6$ as only valid solution (1 Mark)
Step-by-Step Solution:
Factorise the quadratic expression to determine the critical values:
$$(x - 5)(x + 4) > 0$$
The critical values are $x = 5$ and $x = -4$.
Analyze the intervals (the parabola opens upwards since $a > 0$):
- For $x < -4$ or $x > 5$, the expression is positive.
- For $-4 < x < 5$, the expression is negative.
Solution Set: $$x < -4 \quad \text{or} \quad x > 5$$
Number Line Representation:
Draw a straight horizontal line. Plot points at $-4$ and $5$. Place an open circle above both $-4$ and $5$, drawing arrows extending outwards to the left of $-4$ and to the right of $5$.
- Factorising to find critical values $-4$ and $5$ (2 Marks)
- Correct inequality interval solution set (2 Marks)
- Accurate number line representation with open boundary circles (1 Mark)
Step-by-Step Solution:
Deconstruct indices using exponential properties:
$$5^x \cdot 5^1 - 5^x \cdot 5^{-1} = 120$$
Factorise out the common base term $5^x$:
$$5^x\left(5 - \frac{1}{5}\right) = 120$$
$$5^x\left(\frac{24}{5}\right) = 120$$
Multiply both sides by reciprocal $\frac{5}{24}$:
$$5^x = 120 \times \frac{5}{24} = 25$$
Express 25 in base 5 and equate the exponents:
$$5^x = 5^2$$
$$x = 2$$
- Splitting power components using index properties (1 Mark)
- Factorising out common $5^x$ factor (1 Mark)
- Simplifying bracket expression to $\frac{24}{5}$ (1 Mark)
- Isolating to $5^x = 25$ (1 Mark)
- Final solution $x = 2$ (1 Mark)
Step-by-Step Solution:
Isolate $y$ in the linear equation:
$$y = 3x + 1 \quad \text{--- (Equation 1)}$$
Substitute Equation 1 into the quadratic equation:
$$x^2 + x(3x + 1) - (3x + 1)^2 = -11$$
$$x^2 + 3x^2 + x - (9x^2 + 6x + 1) = -11$$
$$4x^2 + x - 9x^2 - 6x - 1 = -11$$
$$-5x^2 - 5x + 10 = 0$$
Divide the entire equation by $-5$ to simplify:
$$x^2 + x - 2 = 0$$
Factorise the quadratic expression:
$$(x + 2)(x - 1) = 0 \implies x = -2 \quad \text{or} \quad x = 1$$
Substitute these $x$ values back into Equation 1 to find the corresponding $y$ values:
- If $x = 1 \implies y = 3(1) + 1 = 4 \implies (1;\ 4)$
- If $x = -2 \implies y = 3(-2) + 1 = -5 \implies (-2;\ -5)$
Solution Coordinates: $$(1;\ 4) \quad \text{and} \quad (-2;\ -5)$$
- Isolating $y = 3x + 1$ correctly (1 Mark)
- Substituting linear equation into quadratic expression (1 Mark)
- Correctly expanding binomial $(3x+1)^2$ (1 Mark)
- Simplification into standard form $x^2 + x - 2 = 0$ (2 Marks)
- Correct binomial factors $(x + 2)(x - 1) = 0$ (1 Mark)
- Both correct $x$ values: $x = 1$ and $x = -2$ (1 Mark)
- Both correct matching $y$ values: $y = 4$ and $y = -5$ (1 Mark)
Step-by-Step Solution:
First, factorise out the coefficient of $x^2$, which is $-2$, from the variable terms:
$$R(x) = -2(x^2 - 8x) + 20$$
Complete the square inside the brackets by adding and subtracting $(\frac{-8}{2})^2 = 16$:
$$R(x) = -2(x^2 - 8x + 16 - 16) + 20$$
$$R(x) = -2[(x - 4)^2 - 16] + 20$$
Distribute the $-2$ back across the bracket terms:
$$R(x) = -2(x - 4)^2 + 32 + 20$$
$$R(x) = -2(x - 4)^2 + 52$$
The maximum revenue is R52 000 (which occurs when $x = 4$ units are produced).
- Factoring out $-2$ coefficient correctly (1 Mark)
- Completing square inside bracket ($+ 16 - 16$) (1 Mark)
- Grouping square terms to $(x - 4)^2$ (1 Mark)
- Distributing back correctly to get $+32$ (1 Mark)
- Final maximum revenue value of R52 000 (1 Mark)
Step-by-Step Solution:
For non-real roots, the discriminant must be strictly less than zero ($\Delta < 0$):
$$\Delta = b^2 - 4ac < 0$$
Substitute $a = 1$, $b = -p$, and $c = 9$:
$$(-p)^2 - 4(1)(9) < 0 \implies p^2 - 36 < 0$$
Factorise the quadratic inequality:
$$(p - 6)(p + 6) < 0$$
$$-6 < p < 6$$
- Stating the non-real roots condition requirement $\Delta < 0$ (1 Mark)
- Substituting coefficients to get $p^2 - 36 < 0$ (1 Mark)
- Identifying critical values $-6$ and $6$ (1 Mark)
- Correct final inequality range $-6 < p < 6$ (1 Mark)
QUESTION 3: FUNCTIONS & GRAPHS
[40 Marks]Given the functions:
$f(x) = -(x+1)^2 + 9$ (parabola) and $g(x) = \frac{4}{x-2} + 2$ (hyperbola).
Step-by-Step Solution:
The parabola is given in standard vertex form: $f(x) = a(x - p)^2 + q$ where turning point is $(p;\ q)$:
For $f(x) = -(x + 1)^2 + 9$:
Turning Point $= (-1;\ 9)$
- Correct $x$-coordinate value ($-1$) (1 Mark)
- Correct $y$-coordinate value ($9$) (1 Mark)
Step-by-Step Solution:
To find the $y$-intercept, substitute $x = 0$ into $f(x)$:
$$f(0) = -(0 + 1)^2 + 9$$
$$f(0) = -(1)^2 + 9 = -1 + 9 = 8$$
$$y\text{-intercept} = (0;\ 8)$$
- Substituting $x = 0$ (1 Mark)
- Correct coordinate pair $(0;\ 8)$ (1 Mark)
Step-by-Step Solution:
To find $x$-intercepts, substitute $f(x) = 0$:
$$0 = -(x + 1)^2 + 9$$
$$(x + 1)^2 = 9$$
$$x + 1 = \pm 3$$
Separate into two linear equations:
$$x + 1 = 3 \implies x = 2$$
$$x + 1 = -3 \implies x = -4$$
$$x\text{-intercepts} = (2;\ 0) \quad \text{and} \quad (-4;\ 0)$$
- Substituting $f(x) = 0$ (1 Mark)
- Isolating square term $(x + 1)^2 = 9$ (1 Mark)
- Solving for both $x$ values: $x = 2$ and $x = -4$ (1 Mark)
- Writing as complete coordinate pairs $(2;\ 0)$ and $(-4;\ 0)$ (1 Mark)
Step-by-Step Solution:
Analyze the boundaries of the function $f(x) = -(x + 1)^2 + 9$:
Domain: Represents all possible $x$ input values. There are no restrictions (such as division by zero) on a standard parabola:
$$x \in \mathbb{R}$$
Range: Represents all possible $y$ output values. Since the parabola opens downwards ($a = -1 < 0$) and has a maximum turning point value of $9$:
$$y \in (-\infty;\ 9] \quad \text{or} \quad y \le 9,\ y \in \mathbb{R}$$
- Correct Domain expression ($x \in \mathbb{R}$) (1 Mark)
- Correct Range expression ($y \le 9$) (2 Marks)
Step-by-Step Solution:
For the hyperbola $g(x) = \frac{4}{x-2} + 2$:
1. Vertical asymptote occurs where denominator is zero:
$$x = 2$$
2. Horizontal asymptote represents the standalone constant term:
$$y = 2$$
- Correct equation of vertical asymptote ($x = 2$) (1 Mark)
- Correct equation of horizontal asymptote ($y = 2$) (1 Mark)
Step-by-Step Solution:
1. Calculate the $y$-intercept (substitute $x = 0$):
$$g(0) = \frac{4}{0 - 2} + 2 = -2 + 2 = 0$$
Coordinate: $$(0;\ 0)$$
2. Calculate the $x$-intercept (substitute $g(x) = 0$):
$$0 = \frac{4}{x - 2} + 2 \implies -2 = \frac{4}{x - 2}$$
$$-2(x - 2) = 4 \implies -2x + 4 = 4 \implies -2x = 0 \implies x = 0$$
Coordinate: $$(0;\ 0)$$
- Substituting $x = 0$ for $y$-intercept (1 Mark)
- Correct $y$-intercept coordinate $(0;\ 0)$ (1 Mark)
- Substituting $g(x) = 0$ to solve for $x$ (1 Mark)
- Correct $x$-intercept coordinate $(0;\ 0)$ (1 Mark)
Step-by-Step Solution:
The axis of symmetry of the hyperbola with negative gradient is modeled by:
$$y = -(x - p) + q$$
Substitute the asymptote intersection points $p = 2$ and $q = 2$:
$$y = -(x - 2) + 2$$
$$y = -x + 2 + 2$$
$$y = -x + 4$$
- Selecting correct negative gradient model formula $y = -x + c$ (1 Mark)
- Substituting horizontal/vertical shift coordinates $(2;\ 2)$ (1 Mark)
- Correct final equation $y = -x + 4$ (1 Mark)
Step-by-Step Solution:
First, calculate coordinate outputs for $x_1 = -2$ and $x_2 = 1$ using parabola function $f(x) = -(x + 1)^2 + 9$:
$$f(-2) = -(-2 + 1)^2 + 9 = -(-1)^2 + 9 = -1 + 9 = 8 \implies (-2;\ 8)$$
$$f(1) = -(1 + 1)^2 + 9 = -(2)^2 + 9 = -4 + 9 = 5 \implies (1;\ 5)$$
Now apply average gradient formula:
$$\text{Avg Gradient} = \frac{f(1) - f(-2)}{1 - (-2)}$$
$$= \frac{5 - 8}{1 + 2} = \frac{-3}{3}$$
$$= -1$$
- Calculating $f(-2) = 8$ and $f(1) = 5$ (1 Mark)
- Substituting values into average gradient formula (1 Mark)
- Correct final average gradient of $-1$ (1 Mark)
Step-by-Step Solution:
Find where the parabola graph lies on or above the x-axis:
We calculated the horizontal x-intercepts of $f(x)$ to be $x = 2$ and $x = -4$ in Question 3.1.3.
Since the parabola is concave down ($a = -1$), the graph is above the x-axis ($f(x) \ge 0$) between the intercepts:
$$-4 \le x \le 2 \quad \text{or} \quad x \in [-4;\ 2]$$
- Identifying critical values $-4$ and $2$ from intercepts (1 Mark)
- Correct inequality interval representation (1 Mark)
- Correct inclusive boundary bracket formatting ($\le$) (1 Mark)
Step-by-Step Solution:
For the hyperbola $g(x) = \frac{4}{x - 2} + 2$, the numerator is positive ($a = 4 > 0$).
This means the curves are strictly decreasing throughout their domains.
The only point of exclusion is the vertical asymptote $x \neq 2$.
Decreasing for: $$x \in \mathbb{R}, \ x \neq 2 \quad \text{or} \quad x < 2 \ \cup \ x > 2$$
- Identifying that the function is decreasing everywhere except at the vertical asymptote point (1 Mark)
- Correct interval notation $x \neq 2$ (1 Mark)
Step-by-Step Solution:
Identify the two coordinates:
- Turning point of $f(x) = (-1;\ 9)$ (from Question 3.1.1)
- Asymptote intersection center of $g(x) = (2;\ 2)$ (from Question 3.2.1)
Calculate gradient $m$ of $h(x)$:
$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 9}{2 - (-1)} = \frac{-7}{3}$$
Solve for equation using $y - y_1 = m(x - x_1)$ and coordinate point $(2; 2)$:
$$y - 2 = -\frac{7}{3}(x - 2) \implies y = -\frac{7}{3}x + \frac{14}{3} + 2$$
$$h(x) = -\frac{7}{3}x + \frac{20}{3} \quad \text{or} \quad h(x) \approx -2.33x + 6.67$$
- Identifying both points $(-1; 9)$ and $(2; 2)$ correctly (1 Mark)
- Calculating correct gradient $m = -\frac{7}{3}$ (1 Mark)
- Correct substitution of coordinate points (1 Mark)
- Final standard straight line equation $y = -\frac{7}{3}x + \frac{20}{3}$ (1 Mark)
Step-by-Step Solution:
Analyze the equation $f(x) = k$:
$$-(x + 1)^2 + 9 = k \implies -(x + 1)^2 = k - 9 \implies (x + 1)^2 = 9 - k$$
For the equation to have no real roots, the term on the right must be strictly negative:
$$9 - k < 0 \implies 9 < k \implies k > 9$$
$$k > 9$$
Alternative Graphical Method: Since the maximum range of the parabola $f(x)$ is $y \le 9$, the horizontal line $y = k$ will not intersect the parabola if $k > 9$. Thus, no real roots occur.
- Identifying maximum turning point value of $y = 9$ (1 Mark)
- Setting up algebraic inequality condition $9 - k < 0$ or stating graphical non-intersection (1 Mark)
- Correct final inequality range $k > 9$ (1 Mark)
Sketch Properties Solution:
A correct sketch of the parabola $f(x) = -(x + 1)^2 + 9$ must feature:
- Symmetrical concave-down shape.
- Turning point vertex labeled clearly at point $(-1;\ 9)$.
- Vertical axis y-intercept labeled at $(0;\ 8)$.
- Horizontal axis x-intercepts labeled at $(-4;\ 0)$ and $(2;\ 0)$.
- Correct turning point plotted at $(-1;\ 9)$ (1 Mark)
- Correct y-intercept plotted at $(0;\ 8)$ (1 Mark)
- Both x-intercepts plotted at $(-4;\ 0)$ and $(2;\ 0)$ (1 Mark)
- Symmetric concave-down parabola shape curve (1 Mark)