MATHWISE ACADEMY
GRADE 11 MATHEMATICS
June Practice Examination (Paper 2 - Test 3)
Examiner: Vashi S Y
QUESTION 1: ANALYTICAL GEOMETRY
[27 Marks]In the Cartesian plane, the points $A(-1; 5)$, $B(5; 7)$, and $C(7; 1)$ form the vertices of $\triangle ABC$.
$M$ is the midpoint of $AC$. $BD$ is a line segment perpendicular to $AC$ intersecting at $M$.
Apply the midpoint formula to vertices $A(-1; 5)$ and $C(7; 1)$:
$$M = \left( \frac{x_A + x_C}{2}; \frac{y_A + y_C}{2} \right)$$
$$M = \left( \frac{-1 + 7}{2}; \frac{5 + 1}{2} \right) = \left( \frac{6}{2}; \frac{6}{2} \right)$$
$$M = (3; 3)$$
- Correct substitution of $x$-coordinates (1 Mark)
- Correct substitution of $y$-coordinates (1 Mark)
- Final midpoint coordinate output $M(3; 3)$ (1 Mark)
Apply the distance formula to $A(-1; 5)$ and $C(7; 1)$:
$$AC = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2}$$
$$AC = \sqrt{(7 - (-1))^2 + (1 - 5)^2} = \sqrt{(8)^2 + (-4)^2}$$
$$AC = \sqrt{64 + 16} = \sqrt{80}$$
$$AC = 4\sqrt{5} \text{ units}$$
- Correct substitution into distance formula (1 Mark)
- Simplification under the square root to $\sqrt{80}$ (1 Mark)
- Final simplified surd form $4\sqrt{5}$ (1 Mark)
Apply the gradient formula for points $A(-1; 5)$ and $C(7; 1)$:
$$m_{AC} = \frac{y_C - y_A}{x_C - x_A}$$
$$m_{AC} = \frac{1 - 5}{7 - (-1)} = \frac{-4}{8}$$
$$m_{AC} = -\frac{1}{2} = -0.5$$
- Correct gradient formula structure (1 Mark)
- Correct substitution of coordinates (2 Marks)
- Correct final gradient $m_{AC} = -\frac{1}{2}$ (1 Mark)
Since $BD \perp AC$, the product of their gradients is $-1$:
$$m_{BD} \times m_{AC} = -1 \implies m_{BD} \times \left(-\frac{1}{2}\right) = -1 \implies m_{BD} = 2$$
Substitute gradient $m_{BD} = 2$ and the point $M(3; 3)$ into the point-slope formula:
$$y - y_M = m(x - x_M) \implies y - 3 = 2(x - 3)$$
$$y - 3 = 2x - 6$$
$$y = 2x - 3$$
- Finding the gradient of the perpendicular line ($m_{BD} = 2$) (1 Mark)
- Substituting point $M(3; 3)$ or $B(5; 7)$ correctly (1 Mark)
- Correct standard form equation $y = 2x - 3$ (1 Mark)
Let $\theta$ be the angle of inclination of line $AC$:
$$\tan \theta = m_{AC} \implies \tan \theta = -0.5$$
Calculate reference angle:
$$\theta_{\text{ref}} = \tan^{-1}(0.5) \approx 26.57^\circ$$
Since the gradient is negative, the angle of inclination is obtuse (lies in the second quadrant):
$$\theta = 180^\circ - 26.57^\circ$$
$$\theta \approx 153.4^\circ$$
- Setting up $\tan \theta = -0.5$ (1 Mark)
- Calculating reference angle of $26.6^\circ$ (1 Mark)
- Correct final angle of inclination $153.4^\circ$ (1 Mark)
Since $M(3; 3)$ is the midpoint of $BD$ where $B(5; 7)$ and $D(x; y)$:
$$x_M = \frac{x_B + x_D}{2} \implies 3 = \frac{5 + x_D}{2} \implies 6 = 5 + x_D \implies x_D = 1$$
$$y_M = \frac{y_B + y_D}{2} \implies 3 = \frac{7 + y_D}{2} \implies 6 = 7 + y_D \implies y_D = -1$$
$$D = (1; -1)$$
- Setting up $x$-midpoint equation and solving $x_D = 1$ (2 Marks)
- Setting up $y$-midpoint equation and solving $y_D = -1$ (2 Marks)
First, note that the diagonals $AC$ and $BD$ intersect perpendicular at point $M$.
1. Length $AC = 4\sqrt{5} = \sqrt{80} \approx 8.94\text{ units}$ (from Question 1.2).
2. Find the length of $BD$ using $B(5; 7)$ and $D(1; -1)$ (from Question 1.6):
$$BD = \sqrt{(1 - 5)^2 + (-1 - 7)^2} = \sqrt{(-4)^2 + (-8)^2}$$
$$BD = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5} \approx 8.94\text{ units}$$
3. Since the diagonals are perpendicular ($AC \perp BD$), the area of quadrilateral $ABCD$ is given by:
$$\text{Area} = \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times AC \times BD$$
$$\text{Area} = \frac{1}{2} \times \sqrt{80} \times \sqrt{80} = \frac{80}{2}$$
$$\text{Area} = 40 \text{ square units}$$
- Noting perpendicular diagonals $AC \perp BD$ (1 Mark)
- Setting up distance formula for $BD$ (1 Mark)
- Calculating length $BD = \sqrt{80}$ (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 $40\text{ units}^2$ (1 Mark)
QUESTION 2: TRIGONOMETRY
[27 Marks]Given: $3\tan \theta + 4 = 0$ and $90^\circ \le \theta \le 270^\circ$.
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)
Apply reduction formulae and co-functions step-by-step:
- $\cos(180^\circ + x) = -\cos x \quad [\text{Q3 negative}]$
- $\sin(90^\circ - x) = \cos x \quad [\text{Co-function Q1}]$
- $\tan(180^\circ - x) = -\tan x \quad [\text{Q2 negative}]$
- $\sin(360^\circ - x) = -\sin x \quad [\text{Q4 negative}]$
Substitute back into the expression:
$$= \frac{-\cos x \cdot \cos x}{(-\tan x) \cdot (-\sin x)} = \frac{-\cos^2 x}{\tan x \cdot \sin x}$$
Rewrite $\tan x$ as $\frac{\sin x}{\cos x}$:
$$= \frac{-\cos^2 x}{\frac{\sin x}{\cos x} \cdot \sin x} = \frac{-\cos^2 x}{\frac{\sin^2 x}{\cos x}} = -\cos^2 x \cdot \frac{\cos x}{\sin^2 x}$$
$$= -\frac{\cos^3 x}{\sin^2 x}$$
- Reduction $\cos(180^\circ + x) = -\cos x$ (1 Mark)
- Co-function $\sin(90^\circ - x) = \cos x$ (1 Mark)
- Reduction $\tan(180^\circ - x) = -\tan x$ (1 Mark)
- Reduction $\sin(360^\circ - x) = -\sin x$ (1 Mark)
- Quotient substitution $\tan x = \frac{\sin x}{\cos x}$ (1 Mark)
- Final simplified ratio expression (1 Mark)
Start with the Left Hand Side (LHS) and find the lowest common denominator, $\text{LCD} = (1 - \cos x)(1 + \cos x)$:
$$\text{LHS} = \frac{(1 + \cos x) + (1 - \cos x)}{(1 - \cos x)(1 + \cos x)}$$
Simplify the numerator and multiply the binomials in the denominator:
$$\text{LHS} = \frac{2}{1 - \cos^2 x}$$
Apply the fundamental square identity $1 - \cos^2 x = \sin^2 x$:
$$\text{LHS} = \frac{2}{\sin^2 x} = \text{RHS}$$
- Finding the common denominator $(1-\cos x)(1+\cos x)$ (1 Mark)
- Correctly simplifying the numerator to 2 (1 Mark)
- Expanding denominator correctly to $1 - \cos^2 x$ (1 Mark)
- Substituting the square identity $\sin^2 x = 1 - \cos^2 x$ (1 Mark)
- Final step showing LHS = RHS (1 Mark)
Rewrite the equation in terms of $\sin\theta$ by substituting $\cos^2\theta = 1 - \sin^2\theta$:
$$2(1 - \sin^2 \theta) + 3\sin\theta - 3 = 0$$
$$2 - 2\sin^2\theta + 3\sin\theta - 3 = 0 \implies -2\sin^2\theta + 3\sin\theta - 1 = 0$$
Multiply by $-1$ to establish a standard quadratic trinomial:
$$2\sin^2\theta - 3\sin\theta + 1 = 0$$
Factorise the quadratic expression:
$$(2\sin\theta - 1)(\sin\theta - 1) = 0 \implies \sin\theta = 0.5 \quad \text{or} \quad \sin\theta = 1$$
Calculate general solutions:
1. For $\sin \theta = 0.5$ (Reference Angle $= 30^\circ$):
Quadrant 1: $$\theta = 30^\circ + k \cdot 360^\circ, \quad k \in \mathbb{Z}$$
Quadrant 2: $$\theta = 180^\circ - 30^\circ + k \cdot 360^\circ \implies \theta = 150^\circ + k \cdot 360^\circ, \quad k \in \mathbb{Z}$$
2. For $\sin \theta = 1$ (Reference Angle $= 90^\circ$):
$$\theta = 90^\circ + k \cdot 360^\circ, \quad k \in \mathbb{Z}$$
- Substituting $\cos^2\theta = 1 - \sin^2\theta$ correctly (1 Mark)
- Rewriting in standard quadratic trinomial form $2\sin^2\theta - 3\sin\theta + 1 = 0$ (1 Mark)
- Factoring to binomial brackets (1 Mark)
- Correct general solutions for first quadrant ($30^\circ$) (1 Mark)
- Correct general solutions for second quadrant ($150^\circ$) (1 Mark)
- Correct general solution for $\sin \theta = 1$ ($90^\circ$) & stating $k \in \mathbb{Z}$ (1 Mark)
Isolate the sine term:
$$\sin(2\theta - 15^\circ) = \frac{-1.8}{3} \implies \sin(2\theta - 15^\circ) = -0.6$$
Find the reference angle on a scientific calculator (using absolute value $0.6$):
$$\text{Ref Angle} = \sin^{-1}(0.6) \approx 36.87^\circ$$
Since sine is negative, find the angles in Quadrants 3 and 4:
1. Quadrant 3:
$$2\theta - 15^\circ = 180^\circ + 36.87^\circ \implies 2\theta - 15^\circ = 216.87^\circ \implies 2\theta = 231.87^\circ \implies \theta \approx 115.94^\circ$$
But $115.94^\circ \notin [0^\circ; 90^\circ]$ (invalid range).
2. Quadrant 4 (using standard negative angle properties):
$$2\theta - 15^\circ = 360^\circ - 36.87^\circ \implies 2\theta - 15^\circ = 323.13^\circ \implies 2\theta = 338.13^\circ \implies \theta \approx 169.07^\circ \quad [\text{invalid}]$$
$$\text{Or: } 2\theta - 15^\circ = -36.87^\circ \implies 2\theta = -21.87^\circ \implies \theta < 0^\circ \quad [\text{invalid}]$$
Under the strict constraint interval $\theta \in [0^\circ; 90^\circ]$, there is no real acute solution for $\theta$.
- Isolating sine ratio to $-0.6$ (1 Mark)
- Calculating reference angle $36.87^\circ$ (1 Mark)
- Evaluating the boundary solution intervals (1 Mark)
- Concluding "no solution in the given interval" (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 circle center $O$ with chord $AB$ and perpendicular $OM \perp AB$. Prove $AM = MB$.
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) |
| $\dots AM = MB$ | Corresponding sides of congruent triangles |
- Radii construction $OA$ and $OB$ (1 Mark)
- Radii equality $OA = OB$ with reason (1 Mark)
- Common side $OM$ (1 Mark)
- Equal perpendicular angles (1 Mark)
- Congruency statement ($\triangle OMA \equiv \triangle OMB$) with reason "RHS" (1 Mark)
- Deduction $AM = MB$ (1 Mark)
Circle Rider 1 Scenario: In the circle below, $O$ is the center. $A$, $B$, $C$, and $D$ lie on the circumference. $AB \parallel CD$, and $D\hat{A}C = 36^\circ$.
Since $AB$ is a diameter, the angle subtended by it at the circumference is a right-angle:
$$A\hat{C}B = 90^\circ \quad [\text{angle in a semi-circle}]$$
- Identifying angle subtended by diameter is $90^\circ$ (3 Marks)
- Correct reason: "angle in semi-circle" (2 Marks)
Cyclic Quad Scenario: In the diagram below, $ABCD$ is a cyclic quadrilateral. The tangent to the circle at point $A$ is $EAF$. $AB$ is produced to $G$. Let tangent angle $E\hat{A}D = 40^\circ$ and interior angle $A\hat{B}C = 110^\circ$.
Angles $A\hat{B}C$ and $C\hat{D}A$ are opposite angles of a cyclic quadrilateral:
$$A\hat{B}C + C\hat{D}A = 180^\circ \quad [\text{opp } \angle \text{s of cyclic quad}]$$
$$110^\circ + C\hat{D}A = 180^\circ$$
$$C\hat{D}A = 70^\circ$$
- Identifying cyclic quad opposite angle condition (1 Mark)
- Correct calculation of $C\hat{D}A = 70^\circ$ (2 Marks)
- Correct CAPS reason: "opp angles of cyclic quad" (1 Mark)
Angles $A\hat{B}C$ and $C\hat{B}G$ lie on a straight line ($ABG$):
$$A\hat{B}C + C\hat{B}G = 180^\circ \quad [\angle\text{s on straight line}]$$
$$110^\circ + C\hat{B}G = 180^\circ$$
$$C\hat{B}G = 70^\circ$$
Alternative: The exterior angle of a cyclic quad is equal to the interior opposite angle, thus $C\hat{B}G = C\hat{D}A = 70^\circ$ [ext angle of cyclic quad]. Both methods are correct.
- Stating size $C\hat{B}G = 70^\circ$ (2 Marks)
- Providing valid geometric reason (1 Mark)
Apply the tan-chord theorem (the angle between the tangent $EAF$ and chord $AD$ is equal to the angle subtended by the chord in the alternate segment):
$$A\hat{C}D = E\hat{A}D \quad [\text{tan-chord theorem}]$$
$$A\hat{C}D = 40^\circ$$
- Identifying correct equal angle relationship $A\hat{C}D = E\hat{A}D$ (2 Marks)
- Correct size $40^\circ$ (1 Mark)
- Correct CAPS reason: "tan-chord theorem" (1 Mark)
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 & Chords Scenario: In the circle below, $O$ is the center. $PT$ is a tangent to the circle at $T$. $PS$ is a secant cutting the circle at $Q$ and $S$. Chord $QT$ and $ST$ are drawn. Let $Q\hat{T}S = 65^\circ$ and $T\hat{Q}P = 110^\circ$.
Angles $T\hat{Q}P$ and $T\hat{Q}S$ lie on a straight line ($PQS$):
$$T\hat{Q}P + T\hat{Q}S = 180^\circ \quad [\angle\text{s on a straight line}]$$
$$110^\circ + T\hat{Q}S = 180^\circ \implies T\hat{Q}S = 70^\circ$$
Now, in $\triangle TQS$, the sum of interior angles is $180^\circ$:
$$Q\hat{S}T + T\hat{Q}S + Q\hat{T}S = 180^\circ \quad [\text{sum of } \angle \text{s of } \triangle]$$
$$Q\hat{S}T + 70^\circ + 65^\circ = 180^\circ$$
$$Q\hat{S}T = 45^\circ$$
- Calculating $T\hat{Q}S = 70^\circ$ with "straight line angles" reason (2 Marks)
- Setting up $\triangle$ sum equation $Q\hat{S}T + 70^\circ + 65^\circ = 180^\circ$ (1 Mark)
- Correct final size $Q\hat{S}T = 45^\circ$ with reason "sum of angles in triangle" (1 Mark)
Apply the tan-chord theorem relative to tangent $PT$ and chord $QT$:
$$P\hat{T}Q = Q\hat{S}T \quad [\text{tan-chord theorem}]$$
We calculated $Q\hat{S}T = 45^\circ$ in Question 3.4.1:
$$P\hat{T}Q = 45^\circ$$
- Identifying equal angle relationship $P\hat{T}Q = Q\hat{S}T$ (2 Marks)
- Correct size $45^\circ$ (1 Mark)
- Correct CAPS reason: "tan-chord theorem" (1 Mark)