Past Exam Questions

Answer: (a) \(a=6,\ b=1\); (b) \(\mathrm P(x)=(2x+1)(3x^2-x+2)\), and the quadratic factor has negative discriminant.

Use the remainder theorem: when a polynomial \(P(x)\) is divided by \(x-a\), the remainder is \(P(a)\). If \(x-a\) is a factor, then \(P(a)=0\).

(a) Since \(2x+1\) is a factor, \(x=-\frac12\) is a root.

So

\(\mathrm P\left(-\frac12\right)=0.\)

This gives

\(-\frac{a}{8}+\frac{b}{4}-\frac32+2=0.\)

Hence

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

The remainder when divided by \(x+1\) is \(\mathrm P(-1)\), so

\(\mathrm P(-1)=-6.\)

Therefore

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

which gives

\(-a+b+5=0.\)

Solving

\(-a+2b+4=0,\quad -a+b+5=0\)

gives

\(b=1,\quad a=6.\)

(b) With these values,

\(\mathrm P(x)=6x^3+x^2+3x+2.\)

Since \(2x+1\) is a factor, divide by \(2x+1\):

\(\mathrm P(x)=(2x+1)(3x^2-x+2).\)

The quadratic factor has discriminant

\((-1)^2-4(3)(2)=1-24=-23.\)

Since this is negative, \(3x^2-x+2\) has no real roots.

The only real root therefore comes from \(2x+1=0\), so \(\mathrm P(x)=0\) has only one real root.

Answer: The remainder is \(-24\).

Answer: The remainder is \(-24\).

Since \(x-3\) is a factor, the factor theorem gives

\(p(3)=0\).

So

\(27m-17(9)+3n+6=0\),

which simplifies to

\(27m+3n=147\),

or

\(9m+n=49\). \(\quad (1)\)

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

\(p(-1)=-12\).

Therefore

\(-m-17-n+6=-12\),

so

\(m+n=1\). \(\quad (2)\)

Subtracting (2) from (1) gives

\(8m=48\),

so \(m=6\). Then \(n=1-6=-5\).

The remainder when \(p(x)\) is divided by \(x-2\) is \(p(2)\):

\(p(2)=6(8)-17(4)-5(2)+6\).

Thus

\(p(2)=48-68-10+6=-24\).

Answer: \(a=6,\ b=-3\), \(\mathrm p(x)=(x-2)(6x^2+3x+3)\), one real solution only.

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

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

So

\(8a-36+2b-6=0,\)

which gives

\(8a+2b=42. \tag{1}\)

The remainder when divided by \(x-3\) is \(66\), so

\(\mathrm p(3)=66.\)

Hence

\(27a-81+3b-6=66,\)

so

\(27a+3b=153. \tag{2}\)

Divide (1) by \(2\) and (2) by \(3\):

\(4a+b=21,\qquad 9a+b=51.\)

Subtracting gives

\(5a=30,\)

so

\(a=6.\)

Then

\(4(6)+b=21,\)

so

\(b=-3.\)

Thus

\(\mathrm p(x)=6x^3-9x^2-3x-6.\)

Since \(x-2\) is a factor, divide by \(x-2\):

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

For the quadratic factor, the discriminant is

\(3^2-4(6)(3)=9-72=-63.\)

This is negative, so \(6x^2+3x+3=0\) has no real solutions. Therefore the only real solution of \(\mathrm p(x)=0\) is \(x=2\).

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

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

Answer: \(p(-k)=-12\), \(k=-2\), and \(p(x)=-25\) has no real solutions.

When \(x=-k\), the factor \(x+k\) is zero. Hence

\(p(-k)=(2(-k)-1)(0)-12=-12.\)

By the remainder theorem, the remainder on division by \(x+3\) is \(p(-3)\). So

\(p(-3)=23.\)

Now

\(p(-3)=(-7)(k-3)-12.\)

Therefore

\(-7(k-3)-12=23.\)

So

\(-7k+9=23,\qquad k=-2.\)

Using \(k=-2\),

\(p(x)=(2x-1)(x-2)-12=2x^2-5x-10.\)

The equation \(p(x)=-25\) becomes

\(2x^2-5x-10=-25,\)

or

\(2x^2-5x+15=0.\)

Its discriminant is

\((-5)^2-4(2)(15)=25-120=-95.\)

Since the discriminant is negative, there are no real solutions.

Answer: \(a=10\), \(b=-12\).

By the remainder theorem, the remainder on division by \(x+3\) is \(p(-3)\), and the remainder on division by \(x-2\) is \(p(2)\).

Now

\(p(-3)=-27+9a+9+b=9a+b-18\),

and

\(p(2)=8+4a-6+b=4a+b+2\).

Since \(p(-3)=2p(2)\),

\(9a+b-18=2(4a+b+2)\).

So

\(9a+b-18=8a+2b+4\),

and hence

\(a-b=22.\)

Also, \(p(x)\) is exactly divisible by \(x+1\), so \(p(-1)=0\). Therefore

\(-1+a+3+b=0\),

which gives

\(a+b=-2.\)

Solving

\(a-b=22,\qquad a+b=-2,\)

gives

\(\boxed{a=10,\quad b=-12}.\)

Answer: The remainder is \(60\).

Since \(x+3\) is a factor, \(p(-3)=0\).

\(2(-3)^3+a(-3)^2-24(-3)+b=0.\)

So

\(-54+9a+72+b=0, \qquad 9a+b=-18.\)

The remainder when \(p(x)\) is divided by \(x-2\) is \(p(2)\), so \(p(2)=-15\):

\(2(2)^3+a(2)^2-24(2)+b=-15.\)

Hence

\(16+4a-48+b=-15, \qquad 4a+b=17.\)

Subtracting the two equations gives

\(5a=-35, \qquad a=-7.\)

Then

\(4(-7)+b=17, \qquad b=45.\)

The remainder when \(p(x)\) is divided by \(x+1\) is \(p(-1)\):

\(p(-1)=2(-1)^3-7(-1)^2-24(-1)+45.\)

Therefore

\(p(-1)=-2-7+24+45=60.\)

Answer: \(a=3\), \(b=2\); remainder \(=-\dfrac{35}{27}\).

Since \(p(x)\) has remainder 2 when divided by \(2x+1\), use \(x=-\frac12\):

\(p\left(-\frac12\right)=2.\)

Now

\(2\left(-\frac12\right)^3+5\left(-\frac12\right)^2+4\left(-\frac12\right)+a=a-1.\)

So \(a-1=2\), giving

\(a=3.\)

Since \(q(x)\) is divisible by \(x+2\),

\(q(-2)=0.\)

Thus

\(4(-2)^2+3(3)(-2)+b=0,\)

so

\(16-18+b=0,\qquad b=2.\)

Then

\(r(x)=p(x)-q(x)=2x^3+x^2-5x+1.\)

The remainder when \(r(x)\) is divided by \(3x-2\) is \(r\left(\frac23\right)\):

\(r\left(\frac23\right) =2\left(\frac23\right)^3+\left(\frac23\right)^2-5\left(\frac23\right)+1 =-\frac{35}{27}.\)

Answer: \(a=10\), \(b=43\), \(c=36\); remainder \(21\).

Since \(p(x)\) is divisible by \(x+3\),

\(p(-3)=0.\)

This gives

\(-27a+9b-3c-9=0.\)

Differentiate:

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

Since \(p'(0)=36\),

\(c=36.\)

Differentiate again:

\(p''(x)=6ax+2b.\)

Since \(p''(0)=86\),

\(2b=86,\qquad b=43.\)

Substitute \(b=43\) and \(c=36\) into the equation from \(p(-3)=0\):

\(-27a+9(43)-3(36)-9=0.\)

So

\(-27a+270=0,\qquad a=10.\)

The remainder when dividing by \(2x-1\) is \(p\left(\frac12\right)\). Hence

\(p\left(\frac12\right) =10\left(\frac12\right)^3+43\left(\frac12\right)^2+36\left(\frac12\right)-9 =21.\)

Answer: \(a=12\), \(b=-27\).

By the remainder theorem, the remainder when \(p(x)\) is divided by \(x-3\) is \(p(3)\).

So

\(p(3)=6.\)

Substitute \(x=3\):

\(27+9a+3b-48=6.\)

Hence

\(9a+3b=27,\)

or

\(3a+b=9. \tag{1}\)

Now differentiate:

\(p'(x)=3x^2+2ax+b.\)

Given \(p'(1)=0\),

\(3+2a+b=0.\)

So

\(2a+b=-3. \tag{2}\)

Subtract equation (2) from equation (1):

\(a=12.\)

Then from \(2a+b=-3\),

\(24+b=-3,\)

so

\(b=-27.\)

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:

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