Grade 10 Mathematics P1 Common Assessment Task Study Notes - June 2025

General Examination Guidelines and Paper Structure

This assessment is the Grade 10 Mathematics P1 Common Assessment Task for June 2025, administered by the KwaZulu-Natal Province Department of Education. The examination is designed for a duration of 1 hour and carries a total of 50 marks. This paper consists of 5 pages of questions followed by a 5-page marking guideline.

Students must adhere to the following instructions:

• Answer all 5 questions provided in the paper.
• Clearly show all calculations, diagrams, and graphs used to determine answers; providing answers only will not necessarily result in full marks.
• Use a non-programmable, non-graphical approved scientific calculator unless otherwise specified.
• Round off answers to two decimal places where necessary, unless otherwise stated.
• Ensure all work is neat and legible and that answers are numbered according to the system used in the question paper.

Question 1: Nature of Rational Expressions

This question focuses on analyzing the constraints and properties of the algebraic expression $K = \frac{3n}{\sqrt{n+5}}$.

1.1 Finding values for which the expression equals zero: An algebraic fraction equals zero when its numerator is zero and its denominator is non-zero. For $K = 0$, we set the numerator $3n = 0$. Therefore, $n = 0$. Note that at $n = 0$, the denominator is $\sqrt{5}$, which is valid.

1.2 Finding values for which the expression is undefined: A fraction is undefined when the denominator is equal to zero. Solving $\sqrt{n+5} = 0$ leads to $n+5 = 0$, which gives the solution $n = -5$.

1.3 Finding values for which the expression is non-real: In the real number system, the square root of a negative value results in a non-real number. Therefore, the expression is non-real when the radicand (the part under the square root) is less than zero. Setting $n+5 < 0$, we find the value for non-real results to be $n < -5$.

Question 2: Algebraic Simplification and Factorization

This question assesses the ability to expand brackets and factorize expressions completely.

2.1 Simplification of Expressions:

2.1.1 Expanding a single bracket: The expression $4p(2-5x)$ is expanded by distributing the $4p$ term to both terms inside the parentheses. $4p \times 2 = 8p$$4p \times (-5x) = -20px$ Result: $8p - 20px$

2.1.2 Expanding a binomial product with a coefficient: To simplify $-2(2x+3)(3x-2)$, first expand the binomials using the FOIL method: $(2x \times 3x) + (2x \times -2) + (3 \times 3x) + (3 \times -2) = 6x^2 - 4x + 9x - 6 = 6x^2 + 5x - 6$ Then, multiply the result by the coefficient $-2$: $-2(6x^2 + 5x - 6) = -12x^2 - 10x + 12$

2.1.3 Combining algebraic expansion and subtraction: The expression is $(4x+1)^2 - (3x+1)(3x-1)$.

• Expand the perfect square: $(4x+1)^2 = 16x^2 + 8x + 1$
• Expand the product (difference of squares): $(3x+1)(3x-1) = 9x^2 - 1$
• Subtract the second part from the first: $(16x^2 + 8x + 1) - (9x^2 - 1) = 16x^2 + 8x + 1 - 9x^2 + 1 = 7x^2 + 8x + 2$

2.2 Factorization: The expression provided is $x^2(x-1) + 25(1-x)$. To factorize by common factor, we must recognize that $(1-x)$ is the additive inverse of $(x-1)$. By changing the sign of the constant coefficient from $+25$ to $-25$, we can rewrite the expression as: $x^2(x-1) - 25(x-1)$ Now, take out the common factor $(x-1)$: $(x-1)(x^2 - 25)$ Finally, factorize the difference of squares $(x^2 - 25)$: $(x-1)(x-5)(x+5)$

Question 3: Equations, Inequalities, and Word Problems

3.1 Solving for x:

3.1.1 Quadratic Equation involving a fraction: Solve $x - 3 = \frac{18}{x}$. First, multiply the entire equation by $x$ (noting that $x \neq 0$) to clear the fraction: $x^2 - 3x = 18$ Rearrange into standard quadratic form: $x^2 - 3x - 18 = 0$. Factorize the trinomial: $(x-6)(x+3) = 0$. The solutions are $x = 6$ or $x = -3$.

3.1.2 Linear Inequality: Solve $-3 \leq 2 - x < 0$ for $x \in \mathbb{R}$ and represent the answer on a number line. Subtract 2 from all parts of the inequality: $-5 \leq -x < -2$. Multiply by $-1$ and reverse the inequality signs: $5 \geq x > 2$. This can be rewritten as $2 < x \leq 5$. On a number line, this is represented by an open circle at 2 and a solid (filled) circle at 5, with a connecting line between them.

3.2 Age Word Problem: A father is twice as old as his son. Twelve years ago, the father was three times the son's age. Let the son's current age be $x$ years. The father's current age is $2x$. Twelve years ago:

• Son's age: $x - 12$
• Father's age: $2x - 12$ The problem states the father was three times the son's age then: $2x - 12 = 3(x - 12)$. Expand: $2x - 12 = 3x - 36$ Solve for $x$: $-x = -24$, so $x = 24$. The son is currently 24 years old.

3.3 Algebraic Identity Application: Given $x^3 + \frac{1}{x^3} = 12$ and $x^2 + \frac{1}{x^2} = 7$, determine the value of $x + \frac{1}{x} + 3$. Using the sum of cubes factorization: $x^3 + \frac{1}{x^3} = (x + \frac{1}{x})(x^2 - x(\frac{1}{x}) + \frac{1}{x^2}) = (x + \frac{1}{x})(x^2 - 1 + \frac{1}{x^2})$. Substitute the known values: $12 = (x + \frac{1}{x})(7 - 1)$$12 = (x + \frac{1}{x})(6)$, leading to $x + \frac{1}{x} = 2$. Therefore, the final value is $2 + 3 = 5$.

Question 4: Functions & Transformations

4.1 Parabolic Function: Given $f(x) = -x^2 + 4$.

4.1.1 Sketching the graph: The graph of $f$ is a downward-turning parabola (concave down) because the coefficient of $x^2$ is negative. The y-intercept is at $(0,4)$. To find x-intercepts, set $-x^2 + 4 = 0$, giving $x^2 = 4$, so $x = 2$ or $x = -2$. The intercepts are $(2,0)$ and $(-2,0)$.

4.1.2 Range: Since the maximum value (turning point) of the parabola is at $y = 4$, the range is $y \leq 4$ or $y \in (-\infty, 4]$.

4.1.3 Interval and Inequality: Determine $x$ for which $f(x) > 0$. This occurs when the graph is above the x-axis, which is between the two x-intercepts: $-2 < x < 2$ or $x \in (-2, 2)$.

4.1.4 Transformation: If $h(x) = -(x-1)(x+1)$, expanding this gives $h(x) = -(x^2 - 1) = -x^2 + 1$. Comparing $f(x) = -x^2 + 4$ and $h(x) = -x^2 + 1$, the transformation is a vertical shift (translation) downward by 3 units.

4.2 Hyperbolic Function: Determine the equation of $K(x) = \frac{a}{x} + q$ with horizontal asymptote $y = 3$ and passing through point $(2,4)$.

• The horizontal asymptote is given by $q$, so $q = 3$.
• Substitute point $(2,4)$ into $K(x) = \frac{a}{x} + 3$: $4 = \frac{a}{2} + 3$$1 = \frac{a}{2}$$a = 2$
• The equation is $K(x) = \frac{2}{x} + 3$.

Question 5: Intersection of Functions

Given the linear function $f(x) = -x+1$ and the hyperbolic function $g(x) = \frac{-2}{x} + 2$, where $R$ and $S$ are points of intersection.

5.1 Line of Symmetry: Determine the equation of the line of symmetry of $g$ with a negative gradient. For a hyperbola shifted only vertically ($q=2$), the lines of symmetry are $y = x + q$ and $y = -x + q$. The line with the negative gradient is $y = -x + 2$.

5.2 Coordinates of R and S: Equate the two functions to find intersection points: $-x + 1 = \frac{-2}{x} + 2$$-x - 1 = \frac{-2}{x}$ Multiply by $x$: $-x^2 - x = -2$ Rearrange: $x^2 + x - 2 = 0$ Factorize: $(x+2)(x-1) = 0$

• If $x = 1$, $y = -1 + 1 = 0$, so $S(1, 0)$.
• If $x = -2$, $y = -(-2) + 1 = 3$, so $R(-2, 3)$.

5.3 Length of DB: Line $DBQ$ is perpendicular to the x-axis with $Q$ at $(-1, 0)$. This means the x-coordinate for both $D$ and $B$ is $-1$.

• Point $D$ is on $g(x)$: $g(-1) = \frac{-2}{-1} + 2 = 2 + 2 = 4$. Coordinates of $D$ are $(-1, 4)$.
• Point $B$ is on $f(x)$: $f(-1) = -(-1) + 1 = 2$. Coordinates of $B$ are $(-1, 2)$.
• The length of vertical line segment $DB$ is the difference in y-values: $y_D - y_B = 4 - 2 = 2$ units.