Past Exam Questions

Answer: \(m=6\), \(n=-10\), so \(x-2\) is a factor.

Since \(3x-1\) is a factor, \(x=\frac13\) is a root. Therefore

\(\mathrm p\left(\frac13\right)=0.\)

This gives

\(\frac{m}{27}-\frac{29}{9}+13+n=0. \tag{1}\)

The remainder when divided by \(x-1\) is \(6\), so

\(\mathrm p(1)=6.\)

Thus

\(m-29+39+n=6,\)

or

\(m+n=-4. \tag{2}\)

Solving (1) and (2) gives

\(m=6,\qquad n=-10.\)

Now substitute \(x=2\):

\(\mathrm p(2)=6(2)^3-29(2)^2+39(2)-10.\)

So

\(\mathrm p(2)=48-116+78-10=0.\)

Since \(\mathrm p(2)=0\), \(x-2\) is a factor of \(\mathrm p(x)\).

Answer: \(a=-3\), \(b=2\); remainder \(4\).

Since \(\mathrm p(x)\) is divisible by \(x-2\),

\(\mathrm p(2)=0.\)

So

\(8a+4b+12+4=0,\)

which simplifies to

\(2a+b=-4.\)

Now

\(\mathrm p'(x)=3ax^2+2bx+6.\)

When \(\mathrm p'(x)\) is divided by \(x+1\), the remainder is \(\mathrm p'(-1)\). Hence

\(3a-2b+6=-7,\)

so

\(3a-2b=-13.\)

Using \(b=-4-2a\) in \(3a-2b=-13\),

\(3a-2(-4-2a)=-13.\)

Therefore

\(7a=-21,\)

so \(a=-3\). Then

\(b=-4-2(-3)=2.\)

Using these values,

\(\mathrm p''(x)=6ax+2b=-18x+4.\)

The remainder when \(\mathrm p''(x)\) is divided by \(x\) is \(\mathrm p''(0)\), which is

\(4.\)

Answer: \(a=-13\), \(b=1\); remainder \(0\); \(\mathrm{p}(x)=(x-2)(2x-1)(3x+1)\).

Since \(x-2\) is a factor,

\(\mathrm{p}(2)=0.\)

Therefore

\(6(2)^3+a(2)^2+b(2)+2=0,\)

so

\(48+4a+2b+2=0.\)

This gives

\(2a+b=-25.\)

Also \(\mathrm{p}(0)=2\), so \(\mathrm{p}(1)=-2\mathrm{p}(0)=-4\). Hence

\(6+a+b+2=-4,\)

which gives

\(a+b=-12.\)

Solving

\(2a+b=-25,\qquad a+b=-12,\)

gives

\(a=-13,\qquad b=1.\)

So

\(\mathrm{p}(x)=6x^3-13x^2+x+2.\)

The remainder when dividing by \(2x-1\) is \(\mathrm{p}\left(\frac12\right)\):

\(\mathrm{p}\left(\frac12\right) =6\left(\frac18\right)-13\left(\frac14\right)+\frac12+2 =0.\)

For the factorisation, since \(x-2\) and \(2x-1\) are factors, the remaining factor is \(3x+1\). Therefore

\(\mathrm{p}(x)=(x-2)(2x-1)(3x+1).\)

First, find the angle \(\theta\) of the sector using the formula for the area of a sector:

\(\frac{1}{2} \times 8^2 \times \theta = \frac{16}{3} \pi\)

Solving for \(\theta\):

\(32 \theta = \frac{16}{3} \pi\)

\(\theta = \frac{\pi}{6}\)

Next, calculate the arc length \(BC\):

\(\text{Arc length} = 8 \times \frac{\pi}{6} = \frac{4\pi}{3} \approx 4.1887 \text{ cm}\)

Now, find the length of \(BC\) using the sine rule or cosine rule. Using the cosine rule:

\(BC^2 = 8^2 + 8^2 - 2 \times 8 \times 8 \times \cos\left(\frac{\pi}{6}\right)\)

\(BC^2 = 128 - 64 \sqrt{3}\)

\(BC \approx 4.1411 \text{ cm}\)

Finally, the perimeter of segment \(BCD\) is:

\(\text{Perimeter} = BC + \text{Arc length} = 4.1411 + 4.1887 \approx 8.33 \text{ cm}\)

(i) To find the perimeter of the shaded region, we first calculate the length of the major arc. The angle of the major arc is given by:

\(2\pi - 2.2 = 4.083\) radians.

The length of the major arc is:

\(s = r\theta = 6 \times 4.083 = 24.5\) cm.

The perimeter of the shaded region is the sum of the lengths of the two radii and the major arc:

\(6 + 6 + 24.5 = 36.5\) cm.

(ii) The area of the major sector is given by:

\(\frac{1}{2} r^2 \theta = \frac{1}{2} \times 6^2 \times 4.083 = 73.49\) cm².

The area of triangle AOB is given by:

\(\frac{1}{2} \times 6^2 \times \sin 2.2 = 14.55\) cm².

The ratio of the area of the shaded region to the area of triangle AOB is:

\(\frac{73.49}{14.55} = 5.05\).

Thus, the ratio is 5.05 : 1.

(i) The area of sector OAB is \(\frac{1}{2} \times 11^2 \times \alpha\).

The area of sector OCD is \(\frac{1}{2} \times 5^2 \times \alpha\).

The area of the shaded region ABDC is \(\frac{1}{2} \times 11^2 \times \alpha - \frac{1}{2} \times 5^2 \times \alpha\).

Thus, \(k = \frac{\frac{1}{2} \times 11^2 \times \alpha - \frac{1}{2} \times 5^2 \times \alpha}{\frac{1}{2} \times 5^2 \times \alpha} = \frac{11^2 - 5^2}{5^2} = \frac{96}{25}\) or 3.84.

(ii) The perimeter of the shaded region ABDC is \(11\alpha + 5\alpha + 6 + 6 = 16\alpha + 12\).

The perimeter of the unshaded region OCD is \(5\alpha + 5 + 5 = 5\alpha + 10\).

Setting the perimeter of the shaded region equal to twice the perimeter of the unshaded region gives:

\(16\alpha + 12 = 2(5\alpha + 10)\)

\(16\alpha + 12 = 10\alpha + 20\)

\(6\alpha = 8\)

\(\alpha = \frac{4}{3}\) or 1.33.

(i) Since angle AOC and angle BOC are both 2.3 radians, angle AOB is calculated as:

\(\angle AOB = 2\pi - 2 \times 2.3 = 1.683 \text{ radians (correct to 4 significant figures)}\)

(ii) The area of the shaded region ACBP can be found by calculating the area of the sector AOB and subtracting the areas of triangles AOC and BOC.

The area of triangle AOC is given by:

\(\frac{1}{2} \times 3^2 \times \sin(2.3)\)

The area of triangle BOC is the same as triangle AOC.

The area of sector AOB is:

\(\frac{1}{2} \times 3^2 \times 1.683\)

Thus, the area of the shaded region ACBP is:

\(2 \times \frac{1}{2} \times 3^2 \times \sin(2.3) + \frac{1}{2} \times 3^2 \times 1.683 = 14.3 \text{ (correct to 3 significant figures)}\)

(i) The area of the shaded region is the difference between the areas of sectors OAB and OCD.

The area of sector OAB is \(\frac{1}{2} \times 16^2 \times 0.8\).

The area of sector OCD is \(\frac{1}{2} \times 10^2 \times 0.8\).

Thus, the area of the shaded region is \(\frac{1}{2} \times 16^2 \times 0.8 - \frac{1}{2} \times 10^2 \times 0.8 = 62.4 \text{ cm}^2\).

(ii) The perimeter of the shaded region consists of the arcs AC and BD and the straight lines AB and CD.

The length of arc AC is \(10 \alpha\) and the length of arc BD is \(16 \alpha\).

The lengths of AB and CD are both 6 cm.

Thus, the perimeter is \(10 \alpha + 16 \alpha + 6 + 6 = 28.9\).

Solving \(26 \alpha + 12 = 28.9\) gives \(\alpha = 0.65\).

(a) Since the total angle around point O is 2π radians, and angle AOB is 2.8 radians, angle ACO is given by:

\(\angle ACO = \pi - 2.8 = 0.7 \text{ radians}\)

(b) Using the sine rule in triangle ACO, we have:

\(\frac{R}{\sin(2.8)} = \frac{r}{\sin(0.7)}\)

Solving for R gives:

\(R = \frac{r \cdot \sin(2.8)}{\sin(0.7)} \approx 1.53r\)

(c) The area of sector OAB is:

\(\text{Area of sector } OAB = \frac{1}{2} r^2 \times 2.8 = 1.4r^2\)

The area of sector CAB is:

\(\text{Area of sector } CAB = \frac{1}{2} R^2 \times 2 \times 0.7 = 1.638r^2\)

The area of triangle OAC is:

\(\text{Area of } \triangle OAC = \frac{1}{2} r^2 \sin(\pi - 2.8) = 0.4927r^2\)

The area of the shaded region is:

\(1.4r^2 - (1.638r^2 - 0.985r^2) = 0.747r^2 \text{ to } 0.748r^2\)

(a) To find the perimeter of the shaded segment, we first determine the radius of the circle. The perimeter of the sector is given by the sum of the arc length and twice the radius:

\(2r + 8 = 20\)

Solving for \(r\), we get \(r = 6\) cm.

The angle \(\angle AOB\) in radians is given by:

\(\frac{8}{6} = \frac{4}{3} \text{ radians}\)

Using the cosine rule in triangle \(OAB\), the length of \(AB\) is:

\(AB = \sqrt{6^2 + 6^2 - 2 \times 6 \times 6 \times \cos \left( \frac{4}{3} \right)}\)

Calculating this gives \(AB \approx 7.42\) cm.

Thus, the perimeter of the shaded segment is:

\(7.42 + 8 = 15.4 \text{ cm}\)

(b) To find the area of the shaded segment, we calculate the area of the sector and subtract the area of triangle \(OAB\).

The area of the sector is:

\(\frac{1}{2} \times 6^2 \times \frac{4}{3}\)

The area of triangle \(OAB\) is:

\(\frac{1}{2} \times 6 \times 6 \times \sin \left( \frac{4}{3} \right)\)

Calculating these gives the sector area as \(24\) cm² and the triangle area as \(17.49\) cm².

Thus, the area of the shaded segment is:

\(24 - 17.49 = 6.51 \text{ cm}^2\)

First, find the length of BC using the tangent of angle 60 degrees (\(\frac{\pi}{3}\) radians):

\(\tan 60 = \frac{BC}{6}\)

\(BC = 6\sqrt{3}\)

Next, calculate the area of triangle OBC:

\(\text{Area of } \triangle OBC = \frac{1}{2} \times 6 \times 6\sqrt{3} = 18\sqrt{3}\)

Calculate the area of sector AOC:

\(\text{Area of sector } AOC = \frac{1}{2} \times 6^2 \times \frac{\pi}{3} = 6\pi\)

Finally, find the area of the shaded region by subtracting the area of the sector from the area of the triangle:

\(\text{Area of shaded region} = 18\sqrt{3} - 6\pi\)

(i) To find the area of the shaded region, we calculate the area of the sector and subtract the area of triangle OCD.

The area of the sector is given by \(\frac{1}{2} r^2 \theta\), where \(r = 6 \text{ cm}\) and \(\theta = 0.8 \text{ radians}\). Thus, the area of the sector is \(\frac{1}{2} \times 6^2 \times 0.8 = 14.4 \text{ cm}^2\).

The area of triangle OCD is given by \(\frac{1}{2} ab \sin C\), where \(a = b = 10 \text{ cm}\) and \(C = 0.8 \text{ radians}\). Thus, the area of the triangle is \(\frac{1}{2} \times 10^2 \times \sin 0.8 = 35.9 \text{ cm}^2\).

The shaded area is \(14.4 - 35.9 = 21.5 \text{ cm}^2\).

(ii) To find the perimeter of the shaded region, we calculate the arc length and the length of CD.

The arc length is given by \(s = r \theta\), where \(r = 6 \text{ cm}\) and \(\theta = 0.8 \text{ radians}\). Thus, the arc length is \(6 \times 0.8 = 4.8 \text{ cm}\).

The length of CD can be found using the cosine rule or \(2 \times 10 \times \sin 0.4\). Thus, \(CD = 7.8 \text{ cm}\).

The perimeter of the shaded region is \(8 + 4.8 + 7.8 = 20.6 \text{ cm}\).

(i) The perimeter of the sector is given by the sum of the two radii and the arc length: \(2r + r\theta = 20\). Solving for \(\theta\), we have:

\(20 = 2r + r\theta\)

\(r\theta = 20 - 2r\)

\(\theta = \frac{20}{r} - 2\)

(ii) The area of the sector is given by \(\frac{1}{2}r^2\theta\). Substituting \(\theta = \frac{20}{r} - 2\), we get:

\(A = \frac{1}{2}r^2\left(\frac{20}{r} - 2\right)\)

\(A = \frac{1}{2}(20r - 2r^2)\)

\(A = 10r - r^2\)

(iii) Using the cosine rule in triangle \(OPQ\), where \(r = 8\) and \(\theta = 0.5\) radians:

\(PQ^2 = 8^2 + 8^2 - 2 \times 8 \times 8 \times \cos(0.5)\)

\(PQ^2 = 64 + 64 - 128 \times \cos(0.5)\)

\(PQ = \sqrt{128 - 128 \times \cos(0.5)}\)

Alternatively, using the sine rule:

\(PQ = 2 \times 8 \times \sin(0.25)\)

\(PQ \approx 3.96\) (allow 3.95)

(i) For \(\theta = 1\), angle BOC = \(\pi - \theta\). The area of sector BOC is given by \(\frac{1}{2} r^2 (\pi - \theta)\). Substituting \(r = 8\) and \(\theta = 1\), the area is \(\frac{1}{2} \times 8^2 \times (\pi - 1) = 32(\pi - 1)\) or approximately 68.5.

(ii) The perimeter of sector AOB is \(8 + 8 + 8\theta\) and the perimeter of sector BOC is \(8 + 8 + 8(\pi - \theta)\). Setting \(8 + 8 + 8\theta = \frac{1}{2}(8 + 8 + 8(\pi - \theta))\), solve for \(\theta\):

\(16 + 8\theta = 8 + 8 + 4\pi - 4\theta\)

\(12\theta = 4\pi - 16\)

\(\theta = \frac{1}{3}(\pi - 2)\) or approximately 0.381.

(iii) For \(\theta = \frac{1}{3}\pi\), AB = 8 cm. Using the cosine rule or trigonometry, BC = \(2 \times 8 \sin \frac{\pi}{3} = 8\sqrt{3}\). The perimeter of triangle ABC is \(8 + 8 + 8\sqrt{3} = 24 + 8\sqrt{3}\) cm.

(i) To find the area of the shaded region, we need to calculate the area of triangle OQR and subtract the area of sector OPQ.

The length QR is given by QR = r \tan \theta.

The area of triangle OQR is \(\frac{1}{2} \times r \times r \tan \theta = \frac{1}{2}r^2 \tan \theta\).

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

Thus, the area of the shaded region is \(\frac{1}{2}r^2 \tan \theta - \frac{1}{2}r^2 \theta = \frac{1}{2}r^2(\tan \theta - \theta)\).

(ii) Given θ = 0.8 and r = 15, we calculate the perimeter of the shaded region.

The arc length PQ is \(r \theta = 15 \times 0.8 = 12 \text{ cm}\).

The length OR is \(\frac{r}{\cos \theta} = \frac{15}{\cos 0.8} \approx 21.53 \text{ cm}\).

The perimeter of the shaded region is \(r \tan \theta + \text{arc } PQ + (r - r \cos \theta)\).

Substituting the values, we get \(15 \tan 0.8 + 12 + (15 - 15 \cos 0.8) \approx 34.0 \text{ cm}\).

(i) To find angle AOB, use the sine rule in the right triangle formed by the radius and half of AB. The sine of half the angle is given by:

\(\sin(\frac{1}{2} \text{angle}) = \frac{16}{20}\)

Solving for the angle, we find:

\(\text{Required angle} = 1.855 \text{ radians}\)

(ii) The area of sector AXBO is calculated using the formula:

\(\frac{1}{2} r^2 \theta\)

Substituting the values, we get:

\(\frac{1}{2} \times 20^2 \times 1.855 = 371 \text{ cm}^2\)

(iii) The area of the new cross-section is found by subtracting the area of the rectangle and the sector from the area of the circle, and then adding the area of the triangle:

\(\pi r^2 - l \times b - \frac{1}{2} r^2 \theta + \frac{1}{2} bh\)

Substituting the values, we get:

\(\pi \times 20^2 - 32 \times 18 - \frac{1}{2} \times 20^2 \times 1.855 + \frac{1}{2} \times 32 \times 18 = 502 \text{ cm}^2\) (accept 501)

(a) To find the perimeter of the shaded region, we first need to find the length of AC. Using the cosine rule:

\(AC^2 = 6^2 + 6^2 - 2 \times 6 \times 6 \times \cos(1.8)\)

\(AC = 9.40 \text{ cm}\)

Next, calculate the angle CAB:

\(\text{Angle } CAB = \frac{1}{2}(\pi - 1.8)\)

\(\text{Angle } CAB = 0.6708 \text{ radians}\)

Now, find the length of arc CD:

\(\text{Arc } CD = 9.40 \times 0.6708 = 6.306 \text{ cm}\)

The perimeter of the shaded region is:

\(6 + 3.40 + 6.306 = 15.7 \text{ cm}\)

(b) To find the area of the shaded region, calculate the area of sector ADC and subtract the area of triangle ABC:

\(\text{Area of sector } ADC = \frac{1}{2} \times 9.40^2 \times 0.6708\)

\(\text{Area of triangle } ABC = \frac{1}{2} \times 6^2 \times \sin(1.8)\)

\(\text{Area of shaded region} = 29.64 - 17.53 = 12.1 \text{ cm}^2\)

(a) To find angle BOC, we use the formula for arc length: \(s = r\theta\). Given the arc length AB is 2 cm and radius is 10 cm, we have:

\(\theta = \frac{s}{r} = \frac{2}{10} = 0.2 \text{ radians}\)

Angle BOC is the sum of angle AOC and angle AOB:

\(\angle BOC = \angle AOC + \angle AOB = \frac{1}{6} \pi + 0.2 \approx 0.724 \text{ radians}\)

(b) To find the area of the shaded region BPC, we first find the lengths of BP and OP using trigonometry in triangle OBP:

\(BP = 10 \sin \left( \frac{5\pi + 6}{30} \right) \approx 6.6208\)

\(OP = 10 \cos \left( \frac{5\pi + 6}{30} \right) \approx 7.494\)

The area of triangle OBP is:

\(\text{Area of } \triangle OBP = \frac{1}{2} \times 10 \times 10 \times \sin \left( \frac{5\pi + 6}{30} \right) \approx 24.809 \text{ cm}^2\)

The area of sector BOC is:

\(\text{Area of sector BOC} = \frac{1}{2} \times 10^2 \times \left( \frac{5\pi + 6}{30} \right) \approx 36.1799 \text{ cm}^2\)

The area of the shaded region BPC is:

\(\text{Area of region BPC} = \text{Area of sector BOC} - \text{Area of } \triangle OBP \approx 11.4 \text{ cm}^2\)

(a) The area of sector ABC is given by \(\frac{1}{2} r^2 \theta\). Substituting \(\theta = \frac{1}{6} \pi\), the sector area is \(\frac{1}{2} r^2 \times \frac{1}{6} \pi = \frac{1}{12} \pi r^2\).

The triangle ABD has base AD and height BD. Using trigonometry, \(BD = r \sin \frac{\pi}{6} = \frac{1}{2} r\) and \(AD = r \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} r\).

The area of triangle ABD is \(\frac{1}{2} \times \frac{1}{2} r \times \frac{\sqrt{3}}{2} r = \frac{\sqrt{3}}{8} r^2\).

Therefore, the area of BCD is \(\frac{1}{12} \pi r^2 - \frac{\sqrt{3}}{8} r^2\).

(b) Given BD =\( \frac{\sqrt{3}}{2} r\), use trigonometry to find angle BAC: \(\angle BAC = \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{3}\).

Then, \(AD = r \cos \frac{\pi}{3} = \frac{1}{2} r\) and \(CD = \frac{1}{2} r\).

The length of arc BC is \(r \times \frac{\pi}{3}\).

Thus, the perimeter of BCD is \(\frac{\sqrt{3}}{2} r + \frac{1}{2} r + \frac{\pi}{3} r\).

(a) To find the perimeter of the shaded region, we first calculate the angles at A and B using trigonometry. We have:

\(\tan A = \frac{12}{5}\) or \(\cos A = \frac{5}{13}\) or \(\sin A = \frac{12}{13}\).

This gives \(A = 1.176\) radians and \(B = 0.3948\) radians.

The length of DE is given as 4 cm.

The arc lengths are calculated as:

Arc on circle A: \(5 \times 1.176 = 5.880\) cm

Arc on circle B: \(8 \times 0.3948 = 3.158\) cm

Thus, the perimeter of the shaded region is \(5.880 + 3.158 + 4 = 13.0\) cm.

(b) To find the area of the shaded region, we calculate the area of triangle ADE and subtract the areas of the sectors:

Area of triangle ADE: \(\frac{1}{2} \times 5 \times 12 = 30\) cm²

Area of sector on circle A: \(\frac{1}{2} \times 5^2 \times 1.176 = 14.70\) cm²

Area of sector on circle B: \(\frac{1}{2} \times 8^2 \times 0.3948 = 12.63\) cm²

Thus, the area of the shaded region is \(30 - 14.70 - 12.63 = 2.67\) cm².