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MATHWISE ACADEMY
GRADE 11 MATHEMATICS
June Practice Examination (Paper 2 - Test 1)
Examiner: Vashi S Y
QUESTION 1: ANALYTICAL GEOMETRY
[27 Marks]In the Cartesian plane below, the points $A(-4; 2)$, $B(2; 6)$, and $C(6; -2)$ form the vertices of $\triangle ABC$.
$M$ is the midpoint of $AC$. The straight line $BD$ is perpendicular to $AC$ at point $M$.
Step-by-Step Solution:
Apply the midpoint formula using coordinates $A(-4; 2)$ and $C(6; -2)$:
$$M = \left( \frac{x_A + x_C}{2}; \frac{y_A + y_C}{2} \right)$$
$$M = \left( \frac{-4 + 6}{2}; \frac{2 + (-2)}{2} \right) = \left( \frac{2}{2}; \frac{0}{2} \right)$$
$$M = (1; 0)$$
- Correct formula structure & substitution for $x$ (1 Mark)
- Correct substitution for $y$ (1 Mark)
- Final midpoint coordinate output $M(1; 0)$ (1 Mark)
Step-by-Step Solution:
Apply the distance formula to $A(-4; 2)$ and $C(6; -2)$:
$$AC = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2}$$
$$AC = \sqrt{(6 - (-4))^2 + (-2 - 2)^2} = \sqrt{(10)^2 + (-4)^2}$$
$$AC = \sqrt{100 + 16} = \sqrt{116}$$
$$AC = 2\sqrt{29} \text{ units}$$
- Correct substitution into distance formula (1 Mark)
- Simplification to $\sqrt{116}$ (1 Mark)
- Final simplified surd form $2\sqrt{29}$ (1 Mark)
Step-by-Step Solution:
Apply the gradient formula for points $A(-4; 2)$ and $C(6; -2)$:
$$m_{AC} = \frac{y_2 - y_1}{x_2 - x_1}$$
$$m_{AC} = \frac{-2 - 2}{6 - (-4)} = \frac{-4}{10}$$
$$m_{AC} = -\frac{2}{5} = -0.4$$
- Correct substitution of coordinates (1 Mark)
- Correct final gradient $m_{AC} = -\frac{2}{5}$ (1 Mark)
Step-by-Step Solution:
Since $BD \perp AC$, the product of their gradients is $-1$:
$$m_{BD} \times m_{AC} = -1 \implies m_{BD} \times \left(-\frac{2}{5}\right) = -1$$
$$m_{BD} = \frac{5}{2} = 2.5$$
Substitute $m_{BD} = 2.5$ and midpoint $M(1; 0)$ (where the line intersects $AC$) or $B(2;6)$ into the linear equation:
$$y - y_1 = m(x - x_1) \implies y - 0 = \frac{5}{2}(x - 1)$$
$$y = \frac{5}{2}x - \frac{5}{2} \quad \text{or} \quad y = 2.5x - 2.5$$
- Identifying perpendicular gradient condition & solving $m_{BD} = 2.5$ (2 Marks)
- Substituting point $M(1; 0)$ or $B(2;6)$ correctly into equation (1 Mark)
- Final standard linear equation (1 Mark)
Step-by-Step Solution:
Let $\theta$ be the angle of inclination of line $AC$:
$$\tan \theta = m_{AC} \implies \tan \theta = -0.4$$
Calculate reference angle:
$$\theta_{\text{ref}} = \tan^{-1}(0.4) \approx 21.8^\circ$$
Since the gradient is negative, the angle of inclination is obtuse:
$$\theta = 180^\circ - 21.8^\circ$$
$$\theta \approx 158.2^\circ$$
- Setting up $\tan \theta = -0.4$ (1 Mark)
- Calculating reference angle $21.8^\circ$ (1 Mark)
- Correct final angle of inclination $158.2^\circ$ (1 Mark)
Step-by-Step Solution:
Since $M(1; 0)$ is the midpoint of $BD$ with $B(2; 6)$ and $D(x; y)$:
$$x_M = \frac{x_B + x_D}{2} \implies 1 = \frac{2 + x_D}{2} \implies 2 = 2 + x_D \implies x_D = 0$$
$$y_M = \frac{y_B + y_D}{2} \implies 0 = \frac{6 + y_D}{2} \implies 0 = 6 + y_D \implies y_D = -6$$
$$D = (0; -6)$$
- Setting up $x$-midpoint equation and solving $x_D = 0$ (2 Marks)
- Setting up $y$-midpoint equation and solving $y_D = -6$ (2 Marks)
Step-by-Step Solution:
First, note that the diagonals $AC$ and $BD$ intersect perpendicular at point $M$.
1. Length $AC = 2\sqrt{29} = \sqrt{116} \approx 10.77\text{ units}$ (from Question 1.2).
2. Find the length of $BD$ using $B(2; 6)$ and $D(0; -6)$ (from Question 1.6):
$$BD = \sqrt{(0 - 2)^2 + (-6 - 6)^2} = \sqrt{(-2)^2 + (-12)^2} = \sqrt{4 + 144} = \sqrt{148} = 2\sqrt{37} \approx 12.17\text{ units}$$
3. Since the diagonals are perpendicular ($AC \perp BD$), the area of quadrilateral $ABCD$ is given by:
$$\text{Area} = \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times AC \times BD$$
$$\text{Area} = \frac{1}{2} \times \sqrt{116} \times \sqrt{148} = \frac{1}{2} \times 2\sqrt{29} \times 2\sqrt{37} = 2\sqrt{1073}$$
$$\text{Area} \approx 65.51 \text{ square units}$$
- Noting perpendicular diagonals $AC \perp BD$ (1 Mark)
- Setting up distance formula for $BD$ (1 Mark)
- Calculating length $BD = \sqrt{148}$ (2 Marks)
- Stating the area formula $\text{Area} = \frac{1}{2} \cdot AC \cdot BD$ (1 Mark)
- Substituting lengths into formula (2 Marks)
- Correct final area $65.51\text{ units}^2$ (1 Mark)
QUESTION 2: TRIGONOMETRY
[27 Marks]Given: $3\tan \theta + 4 = 0$ and $90^\circ \le \theta \le 270^\circ$.
Step-by-Step Solution:
Isolate the trig ratio: $3\tan\theta = -4 \implies \tan \theta = -\frac{4}{3} = \frac{y}{x}$
Since $\tan \theta < 0$ (negative in Q2 & Q4) and the domain is $90^\circ \le \theta \le 270^\circ$ (encompasses Q2 & Q3), the angle $\theta$ lies in Quadrant 2.
Thus, $x$ is negative and $y$ is positive: $x = -3$ and $y = 4$.
Apply Pythagoras to solve for hypotenuse $r$ (always positive):
$$r^2 = x^2 + y^2 = (-3)^2 + 4^2 = 9 + 16 = 25 \implies r = 5$$
Write down trig ratios: $\sin \theta = \frac{y}{r} = \frac{4}{5}$ and $\cos \theta = \frac{x}{r} = -\frac{3}{5}$.
Evaluate the expression:
$$\sin\theta - \cos\theta = \frac{4}{5} - \left(-\frac{3}{5}\right) = \frac{7}{5}$$
- Identifying Quadrant 2 with a sketch (2 Marks)
- Using Pythagoras to find $r = 5$ (1 Mark)
- Stating correct $\sin \theta = \frac{4}{5}$ ratio (1 Mark)
- Stating correct $\cos \theta = -\frac{3}{5}$ ratio (1 Mark)
- Final calculation $\frac{7}{5}$ (1 Mark)
Step-by-Step Solution:
Apply reduction formulae and co-functions step-by-step:
- $\sin(180^\circ - x) = \sin x \quad [\text{Q2 positive}]$
- $\cos(90^\circ - x) = \sin x \quad [\text{Co-function Q1}]$
- $\tan(180^\circ + x) = \tan x \quad [\text{Q3 positive}]$
- $\cos(360^\circ - x) = \cos x \quad [\text{Q4 positive}]$
Substitute back into the expression:
$$= \frac{\sin x \cdot \sin x}{\tan x \cdot \cos x}$$
Rewrite $\tan x$ as $\frac{\sin x}{\cos x}$:
$$= \frac{\sin^2 x}{\frac{\sin x}{\cos x} \cdot \cos x} = \frac{\sin^2 x}{\sin x}$$
$$= \sin x$$
- Reduction $\sin(180^\circ - x) = \sin x$ (1 Mark)
- Co-function $\cos(90^\circ - x) = \sin x$ (1 Mark)
- Reduction $\tan(180^\circ + x) = \tan x$ (1 Mark)
- Reduction $\cos(360^\circ - x) = \cos x$ (1 Mark)
- Quotient substitution $\tan x = \frac{\sin x}{\cos x}$ (1 Mark)
- Final simplified ratio $\sin x$ (1 Mark)
Step-by-Step Proof Solution:
Start with the Left Hand Side (LHS) and find the lowest common denominator, $\text{LCD} = \cos x(1 + \sin x)$:
$$\text{LHS} = \frac{(1 + \sin x) - \cos^2 x}{\cos x(1 + \sin x)}$$
Substitute the square identity $\cos^2 x = 1 - \sin^2 x$ into numerator:
$$\text{LHS} = \frac{1 + \sin x - (1 - \sin^2 x)}{\cos x(1 + \sin x)}$$
$$\text{LHS} = \frac{\sin x + \sin^2 x}{\cos x(1 + \sin x)}$$
Factorise the numerator by taking out the common factor $\sin x$:
$$\text{LHS} = \frac{\sin x(1 + \sin x)}{\cos x(1 + \sin x)}$$
Cancel the matching binomial factor $(1 + \sin x)$:
$$\text{LHS} = \frac{\sin x}{\cos x} = \tan x = \text{RHS}$$
- Finding common denominator LHS $\cos x(1+\sin x)$ (1 Mark)
- Writing numerator correct algebra $(1+\sin x) - \cos^2 x$ (1 Mark)
- Substituting $\cos^2 x = 1 - \sin^2 x$ identity (1 Mark)
- Simplifying numerator to $\sin x(1+\sin x)$ by common factoring (2 Marks)
- Canceling common bracket to show quotient identity $\frac{\sin x}{\cos x} = \tan x$ (1 Mark)
Step-by-Step Solution:
Let $\sin\theta = k$, this yields a standard quadratic trinomial:
$$4k^2 + 5k - 6 = 0$$
Factorise the quadratic:
$$(4k - 3)(k + 2) = 0 \implies k = \frac{3}{4} \quad \text{or} \quad k = -2$$
Substitute back $\sin \theta$:
$$\sin \theta = \frac{3}{4} = 0.75 \quad \text{or} \quad \sin \theta = -2 \quad [\text{Invalid, no solution since } -1 \le \sin\theta \le 1]$$
Calculate general solutions for $\sin \theta = 0.75$ (Reference Angle $\approx 48.59^\circ$):
Quadrant 1 Solution: $$\theta \approx 48.59^\circ + k \cdot 360^\circ, \quad k \in \mathbb{Z}$$
Quadrant 2 Solution: $$\theta \approx 180^\circ - 48.59^\circ + k \cdot 360^\circ \implies \theta \approx 131.41^\circ + k \cdot 360^\circ, \quad k \in \mathbb{Z}$$
- Correct factorisation factors $(4\sin\theta - 3)(\sin\theta + 2) = 0$ (1 Mark)
- Identifying $\sin \theta = \frac{3}{4}$ and rejecting $\sin \theta = -2$ (1 Mark)
- Calculating correct reference angle $48.59^\circ$ (1 Mark)
- First quadrant solution: $\theta = 48.59^\circ + k \cdot 360^\circ$ (1 Mark)
- Second quadrant solution: $\theta = 131.41^\circ + k \cdot 360^\circ$ (1 Mark)
- Stating parameter constant $k \in \mathbb{Z}$ (1 Mark)
QUESTION 3: EUCLIDEAN GEOMETRY
[46 Marks]In Circle geometry, you are required to reproduce the proofs of formal core theorems. Ensure your reasons match the exact, standard examination conditions.
Hint: Draw a circle with centre $O$, chord $AB$, and line $OM \perp AB$. Prove $AM = MB$.
Step-by-Step Proof:
Construction: Draw circle center $O$ with chord $AB$. Draw perpendicular line $OM \perp AB$ where $M$ is on $AB$. Join radii $OA$ and $OB$.
| Statement | Reason |
|---|---|
| In $\triangle OMA$ and $\triangle OMB$: | |
| $1. \quad OA = OB$ | Radii of circle |
| $2. \quad OM = OM$ | Common side |
| $3. \quad \hat{M}_1 = \hat{M}_2 = 90^\circ$ | Given ($OM \perp AB$) |
| $\therefore \triangle OMA \equiv \triangle OMB$ | Right-angle, Hypotenuse, Side (RHS) |
| $\therefore AM = MB$ | Corresponding sides of congruent triangles |
- Construction of radii $OA$ and $OB$ (1 Mark)
- Stating $OA = OB$ with reason "radii" (1 Mark)
- Stating $OM$ is common (1 Mark)
- Stating perpendicular angle equality $90^\circ$ (1 Mark)
- Congruence conclusion ($\triangle OMA \equiv \triangle OMB$) with reason "RHS" (1 Mark)
- Deduction $AM = MB$ (1 Mark)
Hint: Draw circle center $O$ with arc $AB$ subtending angle $A\hat{O}B$ at center and angle $A\hat{C}B$ at circumference. Show $A\hat{O}B = 2A\hat{C}B$.
Step-by-Step Proof:
Construction: Draw line $CO$ and extend it to point $D$. Mark angles $\hat{C}_1, \hat{C}_2, \hat{O}_1, \hat{O}_2$.
Consider $\triangle AOC$:
$$OA = OC \quad [\text{radii}]$$
$$\therefore \hat{A} = \hat{C}_1 \quad [\text{angles opp equal sides}]$$
$$\hat{O}_1 = \hat{A} + \hat{C}_1 \quad [\text{ext } \angle \text{ of } \triangle AOC]$$
$$\therefore \hat{O}_1 = 2\hat{C}_1 \quad [1]$$
Similarly, in $\triangle BOC$:
$$\hat{O}_2 = 2\hat{C}_2 \quad [2]$$
Add equation [1] and [2] to get full subtended angles:
$$\hat{O}_1 + \hat{O}_2 = 2\hat{C}_1 + 2\hat{C}_2 \implies A\hat{O}B = 2(\hat{C}_1 + \hat{C}_2)$$
$$\therefore A\hat{O}B = 2A\hat{C}B$$
- Construction of line $COD$ (1 Mark)
- Identifying $\hat{A} = \hat{C}_1$ with radii reason (1 Mark)
- Stating exterior angle equation $\hat{O}_1 = \hat{A} + \hat{C}_1$ (1 Mark)
- Deducing $\hat{O}_1 = 2\hat{C}_1$ (1 Mark)
- Stating similar step $\hat{O}_2 = 2\hat{C}_2$ (1 Mark)
- Final summation step proving $A\hat{O}B = 2A\hat{C}B$ (1 Mark)
Circle Rider Scenario: In the circle below, $O$ is the center. $AB$ is a diameter. Point $C$ and $D$ lie on the circumference. $AB$ is parallel to $CD$ ($AB \parallel CD$). Let $D\hat{A}C = 36^\circ$.
Step-by-Step Solution:
Angle $A\hat{C}B$ is subtended by diameter $AB$:
$$A\hat{C}B = 90^\circ \quad [\text{angle in a semi-circle}]$$
- Identifying correct size $90^\circ$ (2 Marks)
- Providing correct CAPS reason: "angle in semi-circle" (1 Mark)
Step-by-Step Solution:
We know that $ABCD$ is a cyclic quadrilateral because all four vertices lie on the circumference.
Apply the alternate angles theorem since $AB \parallel CD$:
$$\hat{C}_1 = B\hat{A}C \quad [\text{alt } \angle \text{s; } AB \parallel CD]$$
But we also know that $A\hat{C}D = A\hat{B}D$ (not needed here). Instead, use cyclic quad theorem:
$$\hat{B} + A\hat{D}C = 180^\circ \quad [\text{opp } \angle\text{s of cyclic quad}]$$
Alternatively, since $AB \parallel CD$, the interior angles sum to $180^\circ$ and $\hat{C}_{\text{ext}} = \hat{A}$:
Let's find the correct angle size:
$$A\hat{D}C = B\hat{C}D = 90^\circ + 36^\circ \quad [\text{since } AB \parallel CD, \text{ diagonals/angles are symmetric}]$$
$$A\hat{D}C = 144^\circ$$
- Identifying cyclic quad opposite angle condition (2 Marks)
- Calculating correct size $144^\circ$ (1 Mark)
- Reason "opp angles of cyclic quad" (1 Mark)
Cyclic Quad & Tangent Scenario: In the diagram below, $TAD$ is a tangent to the circle at point $A$. $B$, $C$, and $E$ lie on the circle. Chord $BC$ is produced to $D$. Join chord $AE$. Let $T\hat{A}B = 58^\circ$ and $A\hat{E}C = 62^\circ$.
Step-by-Step Solution:
Apply the tan-chord theorem relative to tangent $TAD$ at point $A$ and chord $AB$:
$$A\hat{C}B = T\hat{A}B = 58^\circ \quad [\text{tan-chord theorem}]$$
- Identifying correct size $58^\circ$ (2 Marks)
- Providing correct CAPS reason: "tan-chord theorem" (1 Mark)
Step-by-Step Solution:
Apply the theorem of angles subtended by the same arc at the circle ($AC$):
$$\hat{B} = A\hat{E}C = 62^\circ \quad [\text{angles in same circle segment}]$$
- Correct size $62^\circ$ (2 Marks)
- Correct reason: "angles in same segment" (1 Mark)
Step-by-Step Solution:
We know that $ABCE$ is a cyclic quadrilateral because all four vertices lie on the circumference.
By cyclic quad exterior angle theorem, the exterior angle at $C$ is equal to the interior opposite angle $\hat{E}$ (which is $A\hat{E}C$):
$$A\hat{C}D = A\hat{E}C \quad [\text{ext } \angle \text{ of cyclic quad}]$$
$$A\hat{C}D = 62^\circ$$
- Identifying cyclic quad vertices and opposite exterior angle $A\hat{C}D$ (2 Marks)
- Stating size $62^\circ$ (2 Marks)
- Stating the reason "ext angle of cyclic quad" (1 Mark)
Step-by-Step Solution:
We know that angles on straight line $BCD$ sum to $180^\circ$:
$$A\hat{C}B + A\hat{C}D + B\hat{C}D = 180^\circ$$
Actually, we can directly find $A\hat{C}B$ and the interior angle $\hat{C}$:
$$A\hat{C}B = 58^\circ \quad [\text{from 3.3.1}]$$
Now, in $\triangle ABC$, the sum of interior angles is $180^\circ$:
$$B\hat{A}C + \hat{B} + A\hat{C}B = 180^\circ \quad [\text{sum of } \angle \text{s in } \triangle]$$
We know $\hat{B} = 62^\circ$ (from Question 3.3.2) and $A\hat{C}B = 58^\circ$ (from Question 3.3.1):
$$B\hat{A}C + 62^\circ + 58^\circ = 180^\circ$$
$$B\hat{A}C + 120^\circ = 180^\circ$$
$$B\hat{A}C = 60^\circ$$
- Noting the sum of angles of $\triangle ABC$ is $180^\circ$ (1 Mark)
- Correct substitution of values $\hat{B}$ and $A\hat{C}B$ (2 Marks)
- Correct final angle $B\hat{A}C = 60^\circ$ (1 Mark)
- Correct theorem reason stated (1 Mark)
Tangents Intersection Scenario: In the diagram below, tangents $PA$ and $PB$ are drawn to a circle from an external point $P$. $O$ is the center of the circle. Join radii $OA$ and $OB$. Let $A\hat{P}B = 40^\circ$.
Step-by-Step Solution:
Apply the tangent-radius theorem (tangent is perpendicular to radius at point of contact):
$$O\hat{A}P = 90^\circ \quad [\text{radius } \perp \text{ tangent}]$$
$$O\hat{B}P = 90^\circ \quad [\text{radius } \perp \text{ tangent}]$$
- Correct size $O\hat{A}P = 90^\circ$ (1 Mark)
- Correct size $O\hat{B}P = 90^\circ$ (1 Mark)
- Providing correct CAPS reason: "radius $\perp$ tangent" (2 Marks)
Step-by-Step Solution:
Consider the quadrilateral $OAPB$. The sum of angles in a quadrilateral is $360^\circ$:
$$A\hat{O}B + O\hat{A}P + A\hat{P}B + O\hat{B}P = 360^\circ \quad [\text{sum of } \angle \text{s of quad}]$$
Substitute $O\hat{A}P = 90^\circ$ and $O\hat{B}P = 90^\circ$ (from Question 3.4.1), and $A\hat{P}B = 40^\circ$:
$$A\hat{O}B + 90^\circ + 40^\circ + 90^\circ = 360^\circ$$
$$A\hat{O}B + 220^\circ = 360^\circ$$
$$A\hat{O}B = 140^\circ$$
Alternative: Since opposite angles $O\hat{A}P + O\hat{B}P = 180^\circ$, quadrilateral $OAPB$ is a cyclic quad. Thus, $A\hat{O}B = 180^\circ - 40^\circ = 140^\circ \quad [\text{opp angles of cyclic quad}]$.
- Identifying sum of angles of quad is $360^\circ$ (1 Mark)
- Correct substitution of values (1 Mark)
- Correct final angle $A\hat{O}B = 140^\circ$ (1 Mark)
- Stating correct geometric reason (1 Mark)