Grade 11 Maths Assignment (May 2026)
MEMORANDUM
QUESTION 1
1.1.1
2x^2 - 7x = 0
x(2x - 7) = 0
x = 0 \quad \text{or} \quad x = \frac{7}{2}
1.1.2
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-8)}}{2(2)}
x = \frac{5 \pm \sqrt{25 + 64}}{4}
x = \frac{5 \pm \sqrt{89}}{4}
x \approx 3,61 \quad \text{or} \quad x \approx -1,11
1.1.3
x^2 + 3x - 10 < 0
(x + 5)(x - 2) < 0
Critical values: x = -5 \quad ; \quad x = 2
-5 < x < 2
1.1.4
x - 4 = \sqrt{x + 2}
(x - 4)^2 = (\sqrt{x + 2})^2
x^2 - 8x + 16 = x + 2
x^2 - 9x + 14 = 0
(x - 7)(x - 2) = 0
x = 7 \quad \text{or} \quad x = 2
Check: x \neq 2
\therefore x = 7
1.2.1
2^x + 2^x \cdot 2^2 = -5y + 10
2^x(1 + 4) = -5y + 10
5 \cdot 2^x = -5y + 10
2^x = \frac{-5y + 10}{5}
2^x = -y + 2
1.2.2
2^x = -(\frac{15}{8}) + 2
2^x = -\frac{15}{8} + \frac{16}{8}
2^x = \frac{1}{8}
2^x = 2^{-3}
x = -3
1.3
2y - x = 1 \implies x = 2y - 1 \quad \text{--- (1)}
x^2 - 6y = 37 \quad \text{--- (2)}
Substitute (1) into (2):
(2y - 1)^2 - 6y = 37
4y^2 - 4y + 1 - 6y - 37 = 0
4y^2 - 10y - 36 = 0
2y^2 - 5y - 18 = 0
(2y - 9)(y + 2) = 0
y = \frac{9}{2} \quad \text{or} \quad y = -2
Substitute y back into (1):
If y = \frac{9}{2} : x = 2(\frac{9}{2}) - 1 \implies x = 8
If y = -2 : x = 2(-2) - 1 \implies x = -5
QUESTION 2
2.1.1
For two equal roots: \Delta = 0
-7m + 3 = 0
-7m = -3
m = \frac{3}{7}
2.1.2
For non-real roots: \Delta < 0
-7m + 3 < 0
-7m < -3
m > \frac{3}{7}
2.2.1
m + n = 3(m - n)
m + n = 3m - 3n
4n = 2m
m = 2n
2.2.2
\frac{5mn}{m^2 + n^2}
Substitute m = 2n:
= \frac{5(2n)(n)}{(2n)^2 + n^2}
= \frac{10n^2}{4n^2 + n^2}
= \frac{10n^2}{5n^2}
= 2
QUESTION 4
4.1
3 \sin\theta = 2 \implies \sin\theta = \frac{2}{3}
y = 2 \quad ; \quad r = 3
x^2 + y^2 = r^2 \implies x^2 + (2)^2 = (3)^2
x^2 = 5 \implies x = -\sqrt{5} (Since \cos\theta < 0, \theta is in Quadrant 2)
4.1.1
3 \cos^2\theta - 1
= 3\left(\frac{-\sqrt{5}}{3}\right)^2 - 1
= 3\left(\frac{5}{9}\right) - 1
= \frac{5}{3} - 1
= \frac{2}{3}
4.1.2
\tan(-\theta - 180^\circ)
= \tan(-(\theta + 180^\circ))
= -\tan(\theta + 180^\circ)
= -\tan\theta
= -\left(\frac{2}{-\sqrt{5}}\right)
= \frac{2}{\sqrt{5}} \quad (\text{or } \frac{2\sqrt{5}}{5})
4.2
t \cos 15^\circ = 4 \implies \cos 15^\circ = \frac{4}{t}
x = 4 \quad ; \quad r = t
y^2 = t^2 - 16 \implies y = \sqrt{t^2 - 16} (Quadrant 1)
4.2.1
\sin 15^\circ = \frac{y}{r} = \frac{\sqrt{t^2 - 16}}{t}
4.2.2
\sin 75^\circ = \sin(90^\circ - 15^\circ)
= \cos 15^\circ
= \frac{4}{t}
4.3
\frac{\cos(90^\circ - \alpha) \sin(-\alpha - 540^\circ)}{\tan 225^\circ + \sin\alpha \cdot \sin(180^\circ + \alpha)}
= \frac{(\sin\alpha) \cdot \sin(-(\alpha + 540^\circ))}{\tan(180^\circ + 45^\circ) + \sin\alpha(-\sin\alpha)}
= \frac{(\sin\alpha) \cdot (-\sin(\alpha + 180^\circ))}{1 - \sin^2\alpha}
= \frac{(\sin\alpha) \cdot (\sin\alpha)}{\cos^2\alpha}
= \frac{\sin^2\alpha}{\cos^2\alpha}
= \tan^2\alpha
4.4
1 - \cos\theta = 2 \sin^2\theta
1 - \cos\theta = 2(1 - \cos^2\theta)
1 - \cos\theta = 2 - 2\cos^2\theta
2\cos^2\theta - \cos\theta - 1 = 0
(2\cos\theta + 1)(\cos\theta - 1) = 0
\cos\theta = -\frac{1}{2} \quad \text{or} \quad \cos\theta = 1
For \cos\theta = -\frac{1}{2}:
Ref angle = 60^\circ
\theta = 180^\circ - 60^\circ + k \cdot 360^\circ = 120^\circ + k \cdot 360^\circ
\theta = 180^\circ + 60^\circ + k \cdot 360^\circ = 240^\circ + k \cdot 360^\circ
For \cos\theta = 1:
\theta = 0^\circ + k \cdot 360^\circ
(k \in \mathbb{Z})
QUESTION 5
5.1.1
\hat{D} = 55^\circ (\angles in same segment)
5.1.2
\hat{O}_2 = 110^\circ (\angle at centre = 2 \times \angle at circumference)
5.1.3
\hat{O}_1 = 180^\circ - 110^\circ = 70^\circ (\angles on a straight line)
In right-angled \triangle OHE:
\hat{E}_2 = 180^\circ - 90^\circ - 70^\circ (Sum of \angles in \triangle)
\hat{E}_2 = 20^\circ
5.1.4
In \triangle OEF, OE = OF (radii)
\angle OEF (\hat{E}_3) = \frac{180^\circ - 110^\circ}{2} (\angles opp equal sides)
\hat{E}_3 = 35^\circ
5.2
Diameter = 10 \implies Radius OE = 5 units
DF \perp EG \implies EH = \frac{9,4}{2} = 4,7 units (line from centre \perp to chord bisects chord)
In \triangle OHE:
OH^2 + EH^2 = OE^2 (Pythagoras)
OH^2 + (4,7)^2 = (5)^2
OH^2 = 25 - 22,09
OH^2 = 2,91
OH \approx 1,71 units
QUESTION 6
6.1
1. \hat{T}_2 = x (Tan-chord theorem)
2. In \triangle BCV, \hat{B}_1 = \hat{V}_1 (\angles opp equal sides BC = CV). Sum of \angles \implies \hat{C} = 180^\circ - 2\hat{V}_1.
AC \parallel TV \implies \hat{C} + \angle CVT = 180^\circ (Co-interior \angles).
(180^\circ - 2\hat{V}_1) + (\hat{V}_1 + x) = 180^\circ \implies -\hat{V}_1 + x = 0 \implies \hat{V}_1 = x
3. \hat{B}_1 = x (\angles opp equal sides BC = CV, since \hat{V}_1 = x).
6.2
- Exterior \angle A\hat{B}T (\hat{B}_3) of cyclic quad CBTV = interior opposite \angle C\hat{V}T.
- \hat{B}_3 = \hat{V}_1 + \hat{V}_2 = x + x = 2x.
- If ATVC is a parallelogram, opposite angles are equal \implies \angle A = \angle C\hat{V}T = 2x.
- In \triangle ABT, \angle A = 2x and \hat{B}_3 = 2x.
- \therefore \angle A = \hat{B}_3 \implies AT = BT (Sides opposite equal \angles).
6.3
- In \triangle ABT, the sum of angles is 180^\circ.