MATHWISE ACADEMY
GRADE 11 MATHEMATICS
June Practice Examination (Paper 1 - Test 5)
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 2:
$$\frac{2^{2x+2} \cdot (2^4)^{x-1}}{(2^2)^{3x-1}}$$
Apply the index rule $(a^m)^n = a^{mn}$ to expand indices:
$$= \frac{2^{2x+2} \cdot 2^{4x-4}}{2^{6x-2}}$$
Apply prime base multiplication (add exponents) and division (subtract exponents) laws:
$$= \frac{2^{(2x+2) + (4x-4)}}{2^{6x-2}} = \frac{2^{6x-2}}{2^{6x-2}}$$
$$= 2^{(6x-2) - (6x-2)} = 2^0$$
$$= 1$$
- Expressing bases 16 and 4 as base 2 power values (1 Mark)
- Simplifying exponential factors to $2^{4x-4}$ and $2^{6x-2}$ (1 Mark)
- Applying multiplication sum laws to numerator exponents to yield $6x-2$ (1 Mark)
- Correctly executing division subtraction law to yield $1$ (1 Mark)
Step-by-Step Solution:
Simplify each radical by factorising out the highest perfect square factor:
$$3\sqrt{20} = 3\sqrt{4 \cdot 5} = 3(2\sqrt{5}) = 6\sqrt{5}$$
$$\sqrt{80} = \sqrt{16 \cdot 5} = 4\sqrt{5}$$
$$2\sqrt{45} = 2\sqrt{9 \cdot 5} = 2(3\sqrt{5}) = 6\sqrt{5}$$
Group and combine like terms:
$$6\sqrt{5} - 4\sqrt{5} + 6\sqrt{5} = 8\sqrt{5}$$
- Simplifying at least two of the surd terms correctly (2 Marks)
- Final combined calculation $8\sqrt{5}$ (1 Mark)
Prove algebraically that the roots of the equation are real and unequal for all real values of $k$ where $k \neq 8$ and $k \neq 0$.
Step-by-Step Solution:
To investigate nature of roots, solve for the discriminant ($\Delta = b^2 - 4ac$):
Substitute $a = 1$, $b = -(k-2)$, and $c = k + 1$:
$$\Delta = [-(k-2)]^2 - 4(1)(k + 1)$$
$$\Delta = (k^2 - 4k + 4) - (4k + 4)$$
$$\Delta = k^2 - 8k$$
Factorise the quadratic expression:
$$\Delta = k(k - 8)$$
For roots to be real and unequal, the discriminant must be strictly positive ($\Delta > 0$):
$$k(k - 8) > 0$$
Applying inequality interval properties, the roots are real and unequal for:
$$k < 0 \quad \text{or} \quad k > 8$$
- Identifying correct coefficients $a$, $b$, and $c$ (1 Mark)
- Correct substitution and expansion into discriminant formula (1 Mark)
- Simplifying discriminant to $\Delta = k^2 - 8k$ (1 Mark)
- Stating the positive discriminant inequality requirement $\Delta > 0$ (1 Mark)
- Correct final inequality range solution set (1 Mark)
Step-by-Step Solution:
For equal roots, the discriminant must equal zero ($\Delta = 0$):
$$\Delta = b^2 - 4ac = 0$$
Substitute $a = 1$, $b = -6$, and $c = p - 2$:
$$(-6)^2 - 4(1)(p - 2) = 0$$
$$36 - 4p + 8 = 0$$
$$44 - 4p = 0 \implies 4p = 44$$
$$p = 11$$
- Stating equal roots condition $\Delta = 0$ (1 Mark)
- Substituting values correctly to get $44 - 4p = 0$ (1 Mark)
- Final algebraic result $p = 11$ (1 Mark)
Step-by-Step Solution:
To find the minimum vertex value, complete the square for $C(x) = 2x^2 - 24x + 100$:
1. Factorise out the coefficient of $x^2$ ($2$) from the variable terms:
$$C(x) = 2(x^2 - 12x) + 100$$
2. Complete the square inside the bracket by adding and subtracting $(\frac{-12}{2})^2 = 36$:
$$C(x) = 2(x^2 - 12x + 36 - 36) + 100$$
$$C(x) = 2\left[(x - 6)^2 - 36\right] + 100$$
3. Distribute the 2 coefficient and group constant values:
$$C(x) = 2(x - 6)^2 - 72 + 100 = 2(x - 6)^2 + 28$$
The minimum daily cost is R28 000 (represented by $28$ thousands of rands) which is achieved when producing exactly $6$ components daily.
- Completing square correctly to vertex form $2(x - 6)^2 + 28$ (1 Mark)
- Stating number of components is $6$ (1 Mark)
- Stating minimum cost value of R28 000 (1 Mark)
QUESTION 2: EQUATIONS & INEQUALITIES
[40 Marks]Step-by-Step Solution:
Factorise the quadratic trinomial:
$$(x + 7)(x - 2) = 0$$
Set each linear factor equal to zero:
$$x + 7 = 0 \quad \text{or} \quad x - 2 = 0$$
$$x = -7 \quad \text{or} \quad x = 2$$
- Correct binomial factors $(x + 7)(x - 2) = 0$ (1 Mark)
- First correct root value $x = -7$ (1 Mark)
- Second correct root value $x = 2$ (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:
$$x \approx 0.64 \quad \text{or} \quad x \approx -3.14$$
- Correct substitution of values into standard quadratic formula (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 $x \ge -\frac{1}{3}$ and $x \ge 3$):
$$3x + 1 = (x - 3)^2$$
$$3x + 1 = x^2 - 6x + 9$$
Rearrange into standard quadratic form:
$$x^2 - 9x + 8 = 0$$
Factorise the quadratic expression:
$$(x - 8)(x - 1) = 0 \implies x = 8 \quad \text{or} \quad x = 1$$
Check Solutions:
- Check $x = 8$: LHS $= \sqrt{3(8) + 1} = \sqrt{25} = 5$; RHS $= 8 - 3 = 5$. (Valid)
- Check $x = 1$: LHS $= \sqrt{3(1) + 1} = \sqrt{4} = 2$; RHS $= 1 - 3 = -2$. (Invalid, LHS $\neq$ RHS)
$$x = 8 \text{ only}$$
- Squaring both sides of the equation to form a standard quadratic equation $x^2 - 9x + 8 = 0$ (1 Mark)
- Factorisation to find both roots $x = 8$ and $x = 1$ (1 Mark)
- Showing valid check substitution step (1 Mark)
- Concluding $x = 8$ as only valid solution (1 Mark)
Step-by-Step Solution:
Factorise the quadratic expression to determine critical values:
$$(2x + 1)(x - 3) \le 0$$
The critical values are $x = -\frac{1}{2}$ (or $-0.5$) and $x = 3$.
Analyze intervals (parabola opens upwards since $a > 0$):
- For $x < -0.5$ or $x > 3$, the expression is positive.
- For $-0.5 \le x \le 3$, the expression is negative/equal to zero.
Solution Set: $$-\frac{1}{2} \le x \le 3 \quad \text{or} \quad x \in [-0.5;\ 3]$$
Number Line Representation:
Draw a straight horizontal line. Plot points at $-0.5$ and $3$. Place a solid dot (closed circle) on both $-0.5$ and $3$, and connect them with a bold segment line.
- Factorising to find critical values $-0.5$ and $3$ (1 Mark)
- Correct inequality interval solution set (2 Marks)
- Accurate number line representation with closed boundary circles (1 Mark)
Step-by-Step Solution:
Separate exponents using exponential multiplication properties:
$$3^x \cdot 3^1 - 2 \cdot 3^x \cdot 3^{-1} = 21$$
Factorise out the common base term $3^x$:
$$3^x\left(3 - \frac{2}{3}\right) = 21$$
$$3^x\left(\frac{7}{3}\right) = 21$$
Multiply both sides by reciprocal $\frac{3}{7}$ to isolate $3^x$:
$$3^x = 21 \times \frac{3}{7} = 9$$
Express 9 in base 3 and equate the exponents:
$$3^x = 3^2 \implies x = 2$$
$$x = 2$$
- Splitting power components using index properties (1 Mark)
- Factorising out common $3^x$ factor (1 Mark)
- Simplifying bracket expression to $\frac{7}{3}$ (1 Mark)
- Isolating to $3^x = 9$ (1 Mark)
- Final solution $x = 2$ (1 Mark)
Step-by-Step Solution:
Isolate $y$ in the linear equation:
$$y = x + 2 \quad \text{--- (Equation 1)}$$
Substitute Equation 1 into the quadratic equation:
$$2x^2 - x(x + 2) + (x + 2)^2 = 8$$
$$2x^2 - (x^2 + 2x) + (x^2 + 4x + 4) = 8$$
$$2x^2 - x^2 - 2x + x^2 + 4x + 4 - 8 = 0$$
$$2x^2 + 2x - 4 = 0$$
Divide by 2 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 $x = 1 \implies y = 1 + 2 = 3 \implies (1;\ 3)$
- If $x = -2 \implies y = -2 + 2 = 0 \implies (-2;\ 0)$
Solution Coordinates: $$(1;\ 3) \quad \text{and} \quad (-2;\ 0)$$
- Isolating $y = x + 2$ correctly (1 Mark)
- Substituting linear equation into quadratic expression (1 Mark)
- Correctly expanding binomial $(x+2)^2$ (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: $y = 3$ and $y = 0$ (2 Marks)
Step-by-Step Solution:
For the expression to be defined in real numbers, two conditions must be met:
1. The value under the square root must be non-negative:
$$x - 2 \ge 0 \implies x \ge 2$$
2. The denominator cannot be zero:
$$x - 5 \neq 0 \implies x \neq 5$$
Combine both conditions:
$$x \ge 2 \quad \text{and} \quad x \neq 5 \quad \text{or} \quad x \in [2;\ 5) \cup (5;\ \infty)$$
- Identifying numerator root restriction condition $x \ge 2$ (2 Marks)
- Identifying denominator restriction condition $x \neq 5$ (1 Mark)
- Writing final interval correctly in either inequality or interval form (1 Mark)
QUESTION 3: FUNCTIONS & GRAPHS
[40 Marks]Given the functions:
$f(x) = -(x - 1)^2 + 9$ (parabola) and $g(x) = \frac{6}{x + 2} + 1$ (hyperbola).
Step-by-Step Solution:
The parabola is given in standard vertex form: $f(x) = a(x - p)^2 + q$ where vertex 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:
1. Calculate the $y$-intercept (substitute $x = 0$):
$$f(0) = -(0 - 1)^2 + 9 = -(-1)^2 + 9 = -1 + 9 = 8 \implies (0;\ 8)$$
2. Calculate the $x$-intercepts (substitute $f(x) = 0$):
$$0 = -(x - 1)^2 + 9 \implies (x - 1)^2 = 9$$
$$x - 1 = \pm 3 \implies x = 1 + 3 = 4 \quad \text{or} \quad x = 1 - 3 = -2$$
$$x\text{-intercepts}: \ (4;\ 0) \quad \text{and} \quad (-2;\ 0)$$
- Correct $y$-intercept coordinate $(0;\ 8)$ (1 Mark)
- Setting $f(x) = 0$ and solving square root equation (1 Mark)
- Solving both correct roots $x = -2$ and $x = 4$ (1 Mark)
- Writing horizontal intercepts as coordinate pairs $(4;\ 0)$ and $(-2;\ 0)$ (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 $g(x) = \frac{6}{x + 2} + 1$:
1. Vertical asymptote occurs where denominator is zero:
$$x = -2$$
2. Horizontal asymptote represents the standalone constant term:
$$y = 1$$
- Correct equation of vertical asymptote ($x = -2$) (1 Mark)
- Correct equation of horizontal asymptote ($y = 1$) (1 Mark)
Step-by-Step Solution:
1. Calculate the $y$-intercept (substitute $x = 0$):
$$g(0) = \frac{6}{0 + 2} + 1 = \frac{6}{2} + 1 = 3 + 1 = 4$$
Coordinate: $$(0;\ 4)$$
2. Calculate the $x$-intercept (substitute $g(x) = 0$):
$$0 = \frac{6}{x + 2} + 1 \implies -1 = \frac{6}{x + 2}$$
$$-1(x + 2) = 6 \implies -x - 2 = 6 \implies -x = 8 \implies x = -8$$
Coordinate: $$(-8;\ 0)$$
- Substituting $x = 0$ for $y$-intercept (1 Mark)
- Correct $y$-intercept coordinate $(0;\ 4)$ (1 Mark)
- Substituting $g(x) = 0$ to solve for $x$ (1 Mark)
- Correct $x$-intercept coordinate $(-8;\ 0)$ (1 Mark)
Step-by-Step Solution:
An axis of symmetry of the hyperbola with positive gradient is modeled by:
$$y = (x - p) + q$$
Substitute the asymptote intersection points $p = -2$ and $q = 1$:
$$y = (x - (-2)) + 1$$
$$y = x + 2 + 1$$
$$y = x + 3$$
- Selecting correct positive gradient model formula $y = x + c$ (1 Mark)
- Substituting horizontal/vertical shift coordinates $(-2;\ 1)$ (1 Mark)
- Correct final equation $y = x + 3$ (1 Mark)
Step-by-Step Solution:
First, calculate coordinate outputs for $x_1 = -1$ and $x_2 = 2$ using parabola function $f(x) = -(x - 1)^2 + 9$:
$$f(-1) = -(-1 - 1)^2 + 9 = -(-2)^2 + 9 = -4 + 9 = 5 \implies (-1;\ 5)$$
$$f(2) = -(2 - 1)^2 + 9 = -(1)^2 + 9 = -1 + 9 = 8 \implies (2;\ 8)$$
Now apply average gradient formula:
$$\text{Avg Gradient} = \frac{f(2) - f(-1)}{2 - (-1)}$$
$$= \frac{8 - 5}{2 + 1} = \frac{3}{3}$$
$$= 1$$
- Calculating $f(-1) = 5$ and $f(2) = 8$ (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 3.1.2.
Since the parabola is concave down ($a = -1$), the graph is above the x-axis ($f(x) \ge 0$) between the intercepts:
$$-2 \le x \le 4 \quad \text{or} \quad x \in [-2;\ 4]$$
- Identifying critical values $-2$ and $4$ 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 $g(x) = \frac{6}{x + 2} + 1$, the numerator is positive ($a = 6 > 0$).
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 $g(x) = (-2;\ 1)$ (from Question 3.2.1)
Calculate gradient $m$ of $h(x)$:
$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 9}{-2 - 1} = \frac{-8}{-3} = \frac{8}{3}$$
Solve for equation using $y - y_1 = m(x - x_1)$ and coordinate point $(1; 9)$:
$$y - 9 = \frac{8}{3}(x - 1) \implies y = \frac{8}{3}x - \frac{8}{3} + 9$$
$$h(x) = \frac{8}{3}x + \frac{19}{3} \quad \text{or} \quad h(x) \approx 2.67x + 6.33$$
- Identifying both points $(1;\ 9)$ and $(-2;\ 1)$ correctly (1 Mark)
- Calculating correct gradient $m = \frac{8}{3}$ (1 Mark)
- Correct substitution of coordinate points (1 Mark)
- Final standard straight line equation $y = \frac{8}{3}x + \frac{19}{3}$ (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 real and unequal roots, the term on the right must be strictly positive:
$$9 - k > 0 \implies 9 > k \implies k < 9$$
$$k < 9$$
Alternative Graphical Method: Since the maximum range value of the parabola $f(x)$ is $y = 9$, the horizontal line $y = k$ will intersect the parabola at two distinct points if $k < 9$. Therefore, roots are real and unequal when $k < 9$.
- Identifying maximum turning point value of $y = 9$ (1 Mark)
- Setting up algebraic inequality condition $9 - k > 0$ or stating graphical intersections (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 $(-2;\ 0)$ and $(4;\ 0)$.
- Correct turning point plotted at $(1;\ 9)$ (1 Mark)
- Correct y-intercept plotted at $(0;\ 8)$ (1 Mark)
- Both x-intercepts plotted at $(-2;\ 0)$ and $(4;\ 0)$ (1 Mark)
- Symmetric concave-down parabola shape curve (1 Mark)