MATHWISE ACADEMY
GRADE 11 MATHEMATICS
June Practice Examination (Paper 1 - Test 2)
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 prime bases of 2:
$$\frac{2^{2x+1} \cdot (2^2)^{x-1}}{(2^3)^{2x-1}}$$
Apply the power law of exponents $ (a^m)^n = a^{mn} $:
$$= \frac{2^{2x+1} \cdot 2^{2x-2}}{2^{6x-3}}$$
Simplify numerator index sums and divide exponents:
$$= \frac{2^{(2x+1) + (2x-2)}}{2^{6x-3}} = \frac{2^{4x-1}}{2^{6x-3}}$$
$$= 2^{(4x-1) - (6x-3)} = 2^{-2x+2}$$
$$= 2^{2(1-x)} \quad \text{or} \quad \frac{4}{2^{2x}}$$
- Expressing bases 4 and 8 as base 2 power values (1 Mark)
- Correctly expanding exponents of powers (1 Mark)
- Multiplying indices ($4x-1$) in numerator (1 Mark)
- Final simplified division index result $2^{2-2x}$ (1 Mark)
Step-by-Step Solution:
Simplify each surd by factorising with a perfect square:
$$\sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3}$$
$$\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}$$
$$2\sqrt{12} = 2\sqrt{4 \cdot 3} = 2(2\sqrt{3}) = 4\sqrt{3}$$
Combine the like terms:
$$3\sqrt{3} - 5\sqrt{3} + 4\sqrt{3} = 2\sqrt{3}$$
- Simplifying at least two of the surd terms correctly (2 Marks)
- Final simplified calculation of $2\sqrt{3}$ (1 Mark)
Step-by-Step Solution:
Multiply both numerator and denominator by the conjugate of the denominator, which is $\sqrt{3} + 1$:
$$= \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}$$
Expand binomials in both numerator and denominator:
$$= \frac{3 + 2\sqrt{3} + 1}{3 - 1} = \frac{4 + 2\sqrt{3}}{2}$$
Divide all terms of numerator by 2:
$$= 2 + \sqrt{3}$$
- Multiplying by correct conjugate fraction $\frac{\sqrt{3}+1}{\sqrt{3}+1}$ (1 Mark)
- Correct numerator expansion ($4 + 2\sqrt{3}$) (1 Mark)
- Correct denominator expansion ($2$) (1 Mark)
- Final simplified answer $2 + \sqrt{3}$ (1 Mark)
Step-by-Step Solution:
To investigate nature of roots, calculate the discriminant ($\Delta = b^2 - 4ac$):
$$\Delta = [-(2k-1)]^2 - 4(1)(k^2 - k)$$
$$\Delta = (4k^2 - 4k + 1) - (4k^2 - 4k)$$
$$\Delta = 4k^2 - 4k + 1 - 4k^2 + 4k$$
$$\Delta = 1$$
Since $\Delta = 1$ (which is strictly positive and a perfect square, $1 > 0$), the roots are guaranteed to be real, unequal, and rational for all real values of $k$.
- Identifying correct variables $a = 1$, $b = -(2k-1)$, and $c = k^2-k$ (1 Mark)
- Correct substitution of values into discriminant formula (1 Mark)
- Expanding $-(2k-1)^2$ to $4k^2 - 4k + 1$ (1 Mark)
- Simplifying to find $\Delta = 1$ (1 Mark)
- Deducing final real, unequal nature of roots from positive discriminant (1 Mark)
For equal roots, set the discriminant equal to zero ($\Delta = 0$):
$$\Delta = b^2 - 4ac = 0$$
$$(m + 2)^2 - 4(1)(2m + 1) = 0$$
$$(m^2 + 4m + 4) - (8m + 4) = 0$$
$$m^2 - 4m = 0$$
Factorise the quadratic expression:
$$m(m - 4) = 0$$
$$m = 0 \quad \text{or} \quad m = 4$$
- Stating the equal roots condition requirement $\Delta = 0$ (1 Mark)
- Substituting values and expanding correctly to $m^2 - 4m = 0$ (1 Mark)
- Common factor grouping $m(m-4) = 0$ (1 Mark)
- Correct roots $m = 0$ or $m = 4$ (1 Mark)
QUESTION 2: EQUATIONS & INEQUALITIES
[40 Marks]Step-by-Step Solution:
Factorise the quadratic expression (find factors of $3 \times -8 = -24$ that sum to $-10$):
The factors are $-12$ and $+2$. Split the middle term:
$$3x^2 - 12x + 2x - 8 = 0$$
$$3x(x - 4) + 2(x - 4) = 0$$
$$(3x + 2)(x - 4) = 0$$
Set each factor equal to zero:
$$x = -\frac{2}{3} \quad \text{or} \quad x = 4$$
- Correct binomial factorisation factors $(3x + 2)(x - 4)$ (1 Mark)
- First correct root value $x = -\frac{2}{3}$ (1 Mark)
- Second correct root value $x = 4$ (1 Mark)
Step-by-Step Solution:
Distribute and rearrange into standard quadratic form:
$$2x^2 - 5x - 1 = 0$$
Apply the quadratic formula with parameters $a = 2$, $b = -5$, and $c = -1$:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
$$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-1)}}{2(2)}$$
$$x = \frac{5 \pm \sqrt{25 + 8}}{4} = \frac{5 \pm \sqrt{33}}{4}$$
Calculate decimal values using a scientific calculator:
$$x \approx 2.69 \quad \text{or} \quad x \approx -0.19$$
- Writing down equation standard form $2x^2 - 5x - 1 = 0$ (1 Mark)
- Correct substitution into quadratic formula (1 Mark)
- Correct decimal evaluation for first root $2.69$ (1 Mark)
- Correct decimal evaluation for second root $-0.19$ (1 Mark)
Step-by-Step Solution:
Square both sides of the equation to eliminate the radical sign (restrictions require $x \ge -3.5$ and $x \ge 4$):
$$2x + 7 = (x - 4)^2$$
$$2x + 7 = x^2 - 8x + 16$$
Rearrange into standard quadratic form:
$$x^2 - 10x + 9 = 0$$
Factorise the quadratic expression:
$$(x - 9)(x - 1) = 0 \implies x = 9 \quad \text{or} \quad x = 1$$
Check Solutions:
- Check $x = 9$: LHS $= \sqrt{2(9) + 7} = \sqrt{25} = 5$; RHS $= 9 - 4 = 5$. (Valid)
- Check $x = 1$: LHS $= \sqrt{2(1) + 7} = \sqrt{9} = 3$; RHS $= 1 - 4 = -3$. (Invalid, LHS $\neq$ RHS)
$$x = 9 \text{ only}$$
- Squaring both sides of the equation (1 Mark)
- Expanding correctly to standard form $x^2 - 10x + 9 = 0$ (1 Mark)
- Factorisation to find both roots $x = 9$ and $x = 1$ (1 Mark)
- Showing valid check substitution step (1 Mark)
- Concluding $x = 9$ as only valid solution (1 Mark)
Step-by-Step Solution:
Deconstruct coefficients of powers using exponential properties:
$$3^x \cdot 3^1 - 3^x \cdot 3^{-1} = 24$$
Factorise out the common base term $3^x$:
$$3^x\left(3 - \frac{1}{3}\right) = 24$$
$$3^x\left(\frac{8}{3}\right) = 24$$
Isolate $3^x$ by multiplying by the reciprocal $\frac{3}{8}$:
$$3^x = 24 \times \frac{3}{8} = 9$$
Write 9 in base 3 and equate exponents:
$$3^x = 3^2 \implies x = 2$$
- Splitting exponents correctly (1 Mark)
- Factorising out $3^x$ correctly (1 Mark)
- Isolating base equation to $3^x = 9$ (1 Mark)
- Final answer $x = 2$ (1 Mark)
Step-by-Step Solution:
Factorise the quadratic expression to determine the critical values:
$$(x - 5)(x + 2) \le 0$$
The critical values are $x = 5$ and $x = -2$.
Analyze intervals (parabola opens upwards since $a > 0$):
- For $x < -2$ or $x > 5$, the expression is positive.
- For $-2 \le x \le 5$, the expression is negative/equal to zero.
Solution Set: $$-2 \le x \le 5 \quad \text{or} \quad x \in [-2;\ 5]$$
Number Line Representation:
Draw a straight horizontal line. Plot points at $-2$ and $5$. Place a solid dot (closed circle) on both $-2$ and $5$, and connect them with a bold segment line.
- Factorising to find critical values $-2$ and $5$ (2 Marks)
- Correct inequality interval solution set (2 Marks)
- Accurate number line representation with closed boundary circles (1 Mark)
Step-by-Step Solution:
Determine the LCD of the equation: $\text{LCD} = (x + 1)(x - 2)$ (restrictions: $x \neq -1$ and $x \neq 2$).
Multiply all terms by LCD to eliminate denominators:
$$(3x - 2)(x - 2) - 2(x + 1) = 1(x + 1)(x - 2)$$
$$(3x^2 - 8x + 4) - (2x + 2) = x^2 - x - 2$$
$$3x^2 - 10x + 2 = x^2 - x - 2$$
Rearrange into standard quadratic form:
$$2x^2 - 9x + 4 = 0$$
Factorise the quadratic expression:
$$(2x - 1)(x - 4) = 0$$
$$x = \frac{1}{2} \quad \text{or} \quad x = 4$$
Both roots satisfy the restriction parameters, so both are valid.
- Identifying correct LCD of $(x + 1)(x - 2)$ (1 Mark)
- Correct multiplication expansion step (1 Mark)
- Simplification into standard trinomial form $2x^2 - 9x + 4 = 0$ (1 Mark)
- Factorisation of quadratic expression (1 Mark)
- Both final correct roots $x = \frac{1}{2}$ or $x = 4$ (1 Mark)
Step-by-Step Solution:
Isolate $y$ in the linear equation:
$$y = 2x + 1 \quad \text{--- (Equation 1)}$$
Substitute Equation 1 into the quadratic equation:
$$x^2 + x(2x + 1) - (2x + 1)^2 = -5$$
$$x^2 + 2x^2 + x - (4x^2 + 4x + 1) = -5$$
$$3x^2 + x - 4x^2 - 4x - 1 = -5$$
$$-x^2 - 3x - 1 = -5 \implies -x^2 - 3x + 4 = 0$$
Multiply entire equation by $-1$:
$$x^2 + 3x - 4 = 0$$
Factorise the trinomial equation:
$$(x + 4)(x - 1) = 0 \implies x = -4 \quad \text{or} \quad x = 1$$
Substitute these $x$ values back into Equation 1 to find the corresponding $y$ values:
- If $x = 1 \implies y = 2(1) + 1 = 3 \implies (1;\ 3)$
- If $x = -4 \implies y = 2(-4) + 1 = -7 \implies (-4;\ -7)$
Solution Coordinates: $$(1;\ 3) \quad \text{and} \quad (-4;\ -7)$$
- Isolating $y = 2x + 1$ correctly (1 Mark)
- Substituting linear equation into quadratic expression (1 Mark)
- Expanding $(2x+1)^2$ correctly (1 Mark)
- Simplification into standard form $x^2 + 3x - 4 = 0$ (2 Marks)
- Correct factors $(x + 4)(x - 1) = 0$ (1 Mark)
- Both correct $x$ values: $x = 1$ and $x = -4$ (1 Mark)
- Both correct matching $y$ values: $y = 3$ and $y = -7$ (1 Mark)
First, factorise out the coefficient of $x^2$, which is $-3$, from the variable terms:
$$P(x) = -3(x^2 - 8x) + 15$$
Complete the square inside the brackets by adding and subtracting $(\frac{-8}{2})^2 = 16$:
$$P(x) = -3(x^2 - 8x + 16 - 16) + 15$$
$$P(x) = -3[(x - 4)^2 - 16] + 15$$
Distribute the $-3$ back across the bracket terms:
$$P(x) = -3(x - 4)^2 + 48 + 15$$
$$P(x) = -3(x - 4)^2 + 63$$
The maximum profit is R63 000 (which occurs when $x = 4$ units are produced).
- Factoring out $-3$ 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 $+48$ (1 Mark)
- Simplifying to standard vertex form $+ 63$ (1 Mark)
- Stating the correct maximum profit value of R63 000 (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: Parabolic functions have no domain restrictions, so:
$$x \in \mathbb{R}$$
Range: 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 asymptote (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)