Past Exam Questions

(i) The perimeter of the plate consists of the arc AB and the sides OC, CB, and OA. The length of the arc AB is given by \(r\theta\). The side OC is \(r \sin \theta\) and CB is \(r \cos \theta\). The side OA is r. Therefore, the perimeter is:

\(r\theta + r \sin \theta + r \cos \theta + r = r(1 + \theta + \cos \theta + \sin \theta)\)

(ii) The area of the plate consists of the area of the sector OAB and the area of the triangle OCB. The area of the sector OAB is \(\frac{1}{2} r^2 \theta\). For r = 10 and \(\theta = \frac{1}{5}\pi\), this becomes:

\(\frac{1}{2} \times 10^2 \times \frac{\pi}{5} = 31.42\)

The area of the triangle OCB is \(\frac{1}{2} \times OC \times CB\). With OC = \(10 \sin \frac{\pi}{5}\) and CB = \(10 \cos \frac{\pi}{5}\), the area is:

\(\frac{1}{2} \times 10 \cos \frac{\pi}{5} \times 10 \sin \frac{\pi}{5} = 23.78\)

Thus, the total area of the plate is:

\(31.42 + 23.78 = 55.2\)

(i) Since AX is tangent to the circle at A, \(\angle OAX = \frac{\pi}{2}\). Using the tangent formula, \(AX = 6 \tan \frac{\pi}{3} = 6 \sqrt{3}\).

(ii) The area of triangle OAX is \(\frac{1}{2} \times 6 \times 6 \sqrt{3} = 18 \sqrt{3}\).

The area of sector OAB is \(\frac{1}{2} \times 6^2 \times \frac{\pi}{3} = 6\pi\).

Thus, the area of the shaded region is \(18 \sqrt{3} - 6\pi\).

(iii) The arc length AB is \(6 \times \frac{\pi}{3} = 2\pi\).

Using trigonometry, \(OX = \frac{6}{\cos \frac{\pi}{3}} = 12\), so \(BX = 6\).

The perimeter of the shaded region is \(6 \sqrt{3} + 2\pi + 6\).

(i) The length of AS is \(r \tan \theta\). The area of triangle OAB is \(r^2 \tan \theta\) or \(\frac{1}{2} r^2 \tan \theta\). The area of sector OPS is \(\frac{1}{2} r^2 \times 2\theta = r^2 \theta\). Therefore, the shaded area is \(r^2 (\tan \theta - \theta)\).

(ii) Given \(\theta = \frac{1}{3}\pi\) and \(r = 6\), we have \(\cos \frac{\pi}{3} = \frac{6}{OA} \Rightarrow OA = 12\). Thus, \(AP = 6\). The length of AS is \(6 \tan \frac{\pi}{3} \Rightarrow AB = 12\sqrt{3}\). The arc PST is \(12 \times \frac{\pi}{3}\). Therefore, the perimeter is \(12 + 12\sqrt{3} + 4\pi\).

(i) To find the area of the shaded region BPDQ, we calculate the area of sector BPD and subtract the area of triangle BPD. The area of sector BPD is given by \(\frac{1}{2} \times 5^2 \times 1.2\). The area of triangle BPD is \(\frac{1}{2} \times 5^2 \times \sin 1.2\). Therefore, the area of the shaded region is:

\(2 \left[ \frac{1}{2} \times 5^2 \times 1.2 - \frac{1}{2} \times 5^2 \times \sin 1.2 \right] = 6.70\)

(ii) To find the length of PQ, we use the cosine rule in triangle APQ. The length of PQ is given by:

\(5 \cos 0.6\)

Then, subtract from 5:

\(5 - 5 \cos 0.6\)

Multiply by 2:

\(10(1 - \cos 0.6) = 1.75\)

(i) Use the Pythagorean theorem in triangle \(PQR\):

\(RS^2 = PR^2 - PQ^2 = 10^2 - 6^2\)

\(RS^2 = 100 - 36 = 64\)

\(RS = \sqrt{64} = 8 \text{ cm}\)

(ii) Use trigonometry to find \(\angle RPQ\):

\(\sin \theta = \frac{8}{10}\)

\(\theta = \arcsin\left(\frac{8}{10}\right) \approx 0.9273 \text{ radians}\)

(iii) The area of the shaded region is the area of the trapezium minus the areas of the two sectors:

Area of trapezium = \(40 \text{ cm}^2\)

Large sector area = \(\frac{1}{2} \times 8^2 \times 0.9273\)

Small sector angle = \(\pi - 0.9273\)

Small sector area = \(\frac{1}{2} \times 2^2 \times 2.214\)

Shaded area = \(40 - (\text{Large sector} + \text{Small sector}) = 5.90 \text{ cm}^2\)

(i) To find angle DOE, use the sine rule in triangle DOE:

\(\sin \frac{1}{2} \theta = \frac{6}{10}\)

Solving for \(\theta\), we find:

\(\theta = 1.287 \text{ radians}\)

(ii) To find the perimeter of the metal plate, calculate the arc length and add the straight edges:

Arc length of BOD = \(2 \times 10 \times (\pi - 1.287)\)

Perimeter = \(12 + 12 + 2 \times 10 \times (\pi - 1.287) = 61.1 \text{ cm}\)

(iii) To find the area of the metal plate, calculate the area of the sectors and triangles:

Area of sector DOE = \(\frac{1}{2} \times 10^2 \times 1.287\)

Area of triangle DOE = \(\frac{1}{2} \times 10^2 \times \sin 1.287\)

Total area = \(\pi \times 10^2 - (2 \times \text{sectors} - 2 \times \text{triangles})\)

Area = 281 or 282 cm².

(i) To find the length of the arc BD, we first calculate the length of AB using the Pythagorean theorem:

\(AB = \sqrt{6^2 + 6^2} = \sqrt{72}\)

The angle BAD is 45° or \(\frac{\pi}{4}\) radians. The arc length s is given by \(s = r\theta\), where \(r = \sqrt{72}\) and \(\theta = \frac{\pi}{4}\):

\(s = \sqrt{72} \times \frac{\pi}{4} \approx 6.67 \text{ cm}\)

(ii) To find the area of the shaded region, we calculate the area of the sector and subtract the area of triangle OAB. The sector area is given by:

\(\text{Sector area} = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 72 \times \frac{\pi}{4}\)

The area of triangle OAB is:

\(\text{Area of triangle} = \frac{1}{2} \times 6 \times 6 = 18 \text{ cm}^2\)

Thus, the shaded area is:

\(\text{Shaded area} = \frac{1}{2} \times 72 \times \frac{\pi}{4} - 18 \approx 10.3 \text{ cm}^2 \text{ or } 9\pi - 18 \text{ cm}^2\)

(a) The perimeter of the cross-section RASB is the sum of the arc lengths ARB and RBS. The arc length ARB is given by \(2.5 \times \frac{4\pi}{3}\) and the arc length RBS is given by \(2.24 \times \frac{5\pi}{6}\). Therefore, the perimeter is:

\(2.5 \times \frac{4\pi}{3} + 2.24 \times \frac{5\pi}{6} = 10.47 + 5.86 = 16.34\)

(b) The area of triangle AOB is \(\frac{1}{2} \times 2.5^2 \times \sin \frac{2\pi}{3} = 2.706\) and the area of triangle APB is \(\frac{1}{2} \times 2.24^2 \times \sin \frac{5\pi}{6} = 1.254\). The difference in area is:

\(2.706 - 1.254 = 1.45\)

(c) The area of the cross-section RASB is the sum of the areas of sectors OARB and PASB plus the area of quadrilateral OAPB. The area of sector OARB is \(\frac{1}{2} \times 2.5^2 \times \frac{4\pi}{3} = 13.09\) and the area of sector PASB is \(\frac{1}{2} \times 2.24^2 \times \frac{5\pi}{6} = 6.57\). Adding the difference in area from part (b), the total area is:

\(13.09 + 6.57 + 1.45 = 21.1\)

(i) The perimeter of R₁ is given by the sum of the two radii and the arc length: \(r + r + r\theta = 2r + r\theta\).

The length of the major arc AB is \(2\pi r - r\theta\).

Equating the two expressions: \(2r + r\theta = 2\pi r - r\theta\).

Solving for \(\theta\):

\(2r + r\theta = 2\pi r - r\theta\)

\(2r + r\theta + r\theta = 2\pi r\)

\(2r + 2r\theta = 2\pi r\)

\(2r(1 + \theta) = 2\pi r\)

\(1 + \theta = \pi\)

\(\theta = \pi - 1\)

(ii) The area of region R₁ is given by \(\frac{1}{2}r^2\theta\).

Given \(\theta = \pi - 1\) and area of R₁ is 30 cm²:

\(\frac{1}{2}r^2(\pi - 1) = 30\)

\(r^2 = \frac{60}{\pi - 1}\)

\(r = \sqrt{\frac{60}{\pi - 1}} \approx 5.29\)

The area of region R₂ is the area of the circle minus the area of R₁:

\(R_2 = \frac{1}{2}r^2(\pi + 1)\)

\(R_2 = \frac{1}{2} \times \frac{60}{\pi - 1} \times (\pi + 1)\)

\(R_2 \approx 58.0 \text{ cm}^2\)

(i) To find the angle \(\angle POQ\) in radians, use the formula for arc length: \(s = r\theta\). Given \(s = 9\) cm and \(r = 5\) cm, we have:

\(9 = 5\theta\)

\(\theta = \frac{9}{5} = 1.8 \text{ rad}\)

(ii) To find the length of \(PT\), consider triangle \(POT\). The angle \(\angle POT\) is half of \(\angle POQ\), so \(\angle POT = \frac{1.8}{2} = 0.9 \text{ rad}\). Using the tangent function:

\(PT = 5 \tan(0.9) \approx 6.30 \text{ cm}\)

(iii) To find the area of the shaded region, first calculate the area of sector \(POQ\):

\(\text{Area of sector} = \frac{1}{2} \times 5^2 \times 1.8 = 22.5 \text{ cm}^2\)

Next, calculate the area of triangle \(POT\):

\(\text{Area of } POT = \frac{1}{2} \times 5 \times 6.30 = 15.75 \text{ cm}^2\)

The shaded area is twice the area of triangle \(POT\) minus the area of the sector:

\(\text{Shaded area} = 2 \times 15.75 - 22.5 = 9.00 \text{ cm}^2\)

(i) To find the perimeter of the shaded region, we first use the Pythagorean theorem in triangle OPT to find OT:

\(OT = \sqrt{OP^2 + PT^2} = \sqrt{5^2 + 12^2} = 13 \text{ cm}\)

Next, find QT:

\(QT = OT - OQ = 13 - 5 = 8 \text{ cm}\)

Calculate the angle POQ using the tangent function:

\(\angle POQ = \tan^{-1}\left(\frac{12}{5}\right) = 1.176 \text{ radians}\)

Find the arc length S:

\(S = r\theta = 5 \times 1.176 = 5.88 \text{ cm}\)

Thus, the perimeter of the shaded region is:

\(5.88 + 12 + 8 = 25.9 \text{ cm}\)

(ii) To find the area of the shaded region, calculate the area of sector POQ:

\(\text{Area of sector} = \frac{1}{2}r^2\theta = \frac{1}{2} \times 5^2 \times 1.176 = 14.7 \text{ cm}^2\)

Calculate the area of triangle OPT:

\(\text{Area of triangle} = \frac{1}{2} \times 12 \times 5 = 30 \text{ cm}^2\)

The area of the shaded region is:

\(30 - 14.7 = 15.3 \text{ cm}^2\)

(i) The area of sector AOB is \(\frac{1}{2}r^2\theta\).

In \(\triangle OAX\), \(AX = r\sin\theta\) and \(OX = r\cos\theta\).

The area of \(\triangle OAX\) is \(\frac{1}{2} \times OX \times AX = \frac{1}{2}r^2\sin\theta\cos\theta\).

Thus, the area of the shaded region AXB is:

\(A = \frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta\cos\theta = \frac{1}{2}r^2(\theta - \sin\theta\cos\theta)\)

(ii) Given \(r = 12\) and \(\theta = \frac{1}{6}\pi\):

\(AX = 12\sin\frac{1}{6}\pi = 6\)

\(OX = 12\cos\frac{1}{6}\pi = 6\sqrt{3}\)

\(BX = OB - OX = 12 - 6\sqrt{3}\)

The length of arc AB is \(r\theta = 12 \times \frac{1}{6}\pi = 2\pi\).

Therefore, the perimeter of the shaded region AXB is:

\(P = AX + BX + \text{arc } AB = 6 + (12 - 6\sqrt{3}) + 2\pi = 18 - 6\sqrt{3} + 2\pi\)

To find the length of AX, we use the tangent of the angle \(\frac{1}{6} \pi\) (half of \(\frac{1}{3} \pi\)) in the right triangle formed by O, A, and X.

\(\tan \frac{1}{6} \pi = \frac{AX}{12}\)

\(\tan \frac{1}{6} \pi = \frac{\sqrt{3}}{3}\)

Thus, \(\frac{\sqrt{3}}{3} = \frac{AX}{12}\)

Solving for \(AX\), we get:

\(AX = 12 \times \frac{\sqrt{3}}{3} = 4\sqrt{3}\)

First, calculate the length of OC and AC using trigonometry. Since \(\angle AOB = \frac{1}{3} \pi\), we have:

\(OC = 12 \cos\left(\frac{1}{3} \pi\right) = 12 \times \frac{1}{2} = 6\)

\(AC = 12 \sin\left(\frac{1}{3} \pi\right) = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}\)

The area of the sector AOB is given by:

\(\frac{1}{2} r^2 \theta = \frac{1}{2} \times 12^2 \times \frac{1}{3} \pi = 24\pi\)

The area of rectangle OCDB is:

\(12 \times 6\sqrt{3} = 72\sqrt{3}\)

The area of triangle OAC is:

\(\frac{1}{2} \times 6 \times 6\sqrt{3} = 18\sqrt{3}\)

The area of the shaded region is the area of the rectangle minus the area of the sector and the triangle:

\(72\sqrt{3} - (24\pi + 18\sqrt{3}) = 54\sqrt{3} - 24\pi\)

Thus, the values of a and b are 54 and 24, respectively.

(i) To find angle AOB, use the tangent formula in the right triangle formed by the radius and the tangent: \(\tan(\frac{x}{2}) = \frac{15}{8} = 1.875\). Solving for \(\frac{x}{2}\), we get \(\frac{x}{2} = 1.081\). Therefore, \(x = 2.16\) radians.

(ii) The perimeter of the shaded region is the sum of the lengths of AT, BT, and the arc AB. The arc length is given by \(r\theta = 8 \times 2.16 = 17.3\) cm. Thus, the perimeter is \(15 + 15 + 17.3 = 47.3\) cm.

(iii) The area of sector AOB is \(\frac{1}{2} r^2 \theta = \frac{1}{2} \times 8^2 \times 2.16 = 69.1\) cm². The area of triangle AOT is \(2 \times \frac{1}{2} \times 8 \times 15 = 120\) cm². The shaded area is \(120 - 69.1 = 50.9\) cm².

The gradient of the line \(l\) is found by rearranging \(2x + y = k\) to \(y = -2x + k\). Thus, the gradient is \(-2\).

For the curve \(xy = 12\), differentiate implicitly: \(x \frac{dy}{dx} + y = 0\).

At \(P(2, 6)\), substitute \(x = 2\) and \(y = 6\) into the differentiated equation:

\(2 \frac{dy}{dx} + 6 = 0\)

\(\frac{dy}{dx} = -\frac{6}{2} = -3\)

The gradient of the tangent at \(P\) is \(-3\).

The angle \(\theta\) between the line and the tangent is given by:

\(\tan \theta = \left| \frac{-3 - (-2)}{1 + (-3)(-2)} \right| = \left| \frac{-1}{1 + 6} \right| = \frac{1}{7}\)

\(\theta = \tan^{-1} \left( \frac{1}{7} \right) \approx 8.1^\circ\)

(i) To find BD, use the formula for the perpendicular from the center to the chord: BD = 9\sin(\pi - 2.4). Calculating gives:

\(BD = 9\sin(\pi - 2.4) = 6.08 \text{ cm}\)

(ii) To find the perimeter of the shaded region, calculate the arc length AB and add the lengths of BD, OD, and OA:

Arc AB = 9 \times 2.4 = 21.6 \text{ cm}

Using Pythagoras or trigonometry, OD = 9\cos(\pi - 2.4) = 6.64 \text{ cm}

Perimeter = 21.6 + 6.08 + 9 + 6.64 = 43.3 \text{ cm}

(iii) To find the area of the shaded region, calculate the area of the sector and subtract the area of triangle OBD:

Area of sector = \(\frac{1}{2} \times 9^2 \times 2.4\)

Area of triangle OBD = \(\frac{1}{2} \times 6.08 \times 6.64\)

Total area = \(\frac{1}{2} \times 81 \times 2.4 - \frac{1}{2} \times 6.08 \times 6.64 = 117 \text{ cm}^2\)

(a) Differentiate the equation \(x^3 + y^2 + 3x^2 + 3y = 4\) implicitly with respect to \(x\):

\(3x^2 + 2y \frac{dy}{dx} + 6x + 3 \frac{dy}{dx} = 0\).

Rearrange to solve for \(\frac{dy}{dx}\):

\(2y \frac{dy}{dx} + 3 \frac{dy}{dx} = -3x^2 - 6x\).

\(\frac{dy}{dx} (2y + 3) = -3x^2 - 6x\).

\(\frac{dy}{dx} = -\frac{3x^2 + 6x}{2y + 3}\).

(b) For the tangent to be parallel to the x-axis, \(\frac{dy}{dx} = 0\).

Set the numerator of \(\frac{dy}{dx} = -\frac{3x^2 + 6x}{2y + 3}\) to zero:

\(3x^2 + 6x = 0\).

Factorize:

\(3x(x + 2) = 0\).

\(x = 0\) or \(x = -2\).

Substitute \(x = 0\) into the original equation:

\(y^2 + 3y - 4 = 0\).

Factorize:

\((y - 1)(y + 4) = 0\).

\(y = 1\) or \(y = -4\).

Coordinates: \((0, 1)\) and \((0, -4)\).

Substitute \(x = -2\) into the original equation:

\(y^2 + 3y = 0\).

Factorize:

\(y(y + 3) = 0\).

\(y = 0\) or \(y = -3\).

Coordinates: \((-2, 0)\) and \((-2, -3)\).

To find the gradient of the curve \(x^3 + 3xy^2 - y^3 = 1\), we need to differentiate implicitly with respect to \(x\).

The derivative of \(x^3\) is \(3x^2\).

For \(3xy^2\), use the product rule: \(3(y^2 + 2xy\frac{dy}{dx}) = 3y^2 + 6xy\frac{dy}{dx}\).

The derivative of \(y^3\) is \(3y^2\frac{dy}{dx}\).

Substituting these into the equation, we get:

\(3x^2 + 3y^2 + 6xy\frac{dy}{dx} - 3y^2\frac{dy}{dx} = 0\).

Simplify to find \(\frac{dy}{dx}\):

\(6xy\frac{dy}{dx} - 3y^2\frac{dy}{dx} = -3x^2 - 3y^2\).

\((6xy - 3y^2)\frac{dy}{dx} = -3x^2 - 3y^2\).

\(\frac{dy}{dx} = \frac{-3x^2 - 3y^2}{6xy - 3y^2}\).

Substitute \(x = 1\) and \(y = 3\):

\(\frac{dy}{dx} = \frac{-3(1)^2 - 3(3)^2}{6(1)(3) - 3(3)^2}\).

\(\frac{dy}{dx} = \frac{-3 - 27}{18 - 27}\).

\(\frac{dy}{dx} = \frac{-30}{-9} = \frac{10}{3}\).