Past Exam Questions

Answer: \(14-\sqrt{10}\).

Work through the given conditions step by step, keeping exact values until the final answer is required.

First combine the \(+1\) with the fraction:

\(\frac{16+11\sqrt{10}}{2+\sqrt{10}}+1=\frac{16+11\sqrt{10}+2+\sqrt{10}}{2+\sqrt{10}}=\frac{18+12\sqrt{10}}{2+\sqrt{10}}.\)

Rationalise the denominator:

\(\frac{18+12\sqrt{10}}{2+\sqrt{10}}\cdot\frac{2-\sqrt{10}}{2-\sqrt{10}}\).

The numerator is

\(36-18\sqrt{10}+24\sqrt{10}-120=-84+6\sqrt{10}\).

The denominator is \(4-10=-6\).

So the expression is

\(\frac{-84+6\sqrt{10}}{-6}=14-\sqrt{10}\).

This gives the required answer: \(14-\sqrt{10}\).

Answer: \(\displaystyle \frac38p^{-2}q^{3/2}r^{-16/5}\), so \(k=\frac38\), \(a=-2\), \(b=\frac32\), \(c=-\frac{16}{5}\).

Answer: \(\displaystyle \frac38p^{-2}q^{3/2}r^{-16/5}\), so \(k=\frac38\), \(a=-2\), \(b=\frac32\), \(c=-\frac{16}{5}\).

Rewrite the expression using index notation:

Now divide powers with the same base:

\(\frac{3p}{8p^3}=\frac38p^{-2},\)

\(\frac{q^{1/2}}{q^{-1}}=q^{1/2-(-1)}=q^{3/2},\)

and

\(\frac{r^{-3}}{r^{1/5}}=r^{-3-\frac15}=r^{-16/5}.\)

Therefore the expression is

\(\frac38p^{-2}q^{3/2}r^{-16/5}.\)

Answer: \(\frac{13+5\sqrt5}{4}\).

Multiply the numerator and denominator by the conjugate of the denominator:

The denominator is

\(7^2-(3\sqrt5)^2=49-45=4.\)

The numerator is

\((4-\sqrt5)(7+3\sqrt5) =28+12\sqrt5-7\sqrt5-15.\)

So the numerator simplifies to

\(13+5\sqrt5.\)

Therefore

Answer: (a) \(\frac43\). (b) \(\frac{3\sqrt3-5}{2}\).

(a)

\(\frac{\sqrt{128}}{\sqrt{72}}=\sqrt{\frac{128}{72}}=\sqrt{\frac{16}{9}}=\frac43.\)

(b) First rationalise each term:

\(\frac{1}{1+\sqrt3}=\frac{1-\sqrt3}{1-3}=\frac{\sqrt3-1}{2}.\)

Also

\(\frac{\sqrt3}{3+2\sqrt3} =\frac{\sqrt3(3-2\sqrt3)}{(3+2\sqrt3)(3-2\sqrt3)}.\)

The denominator is \(9-12=-3\), so

\(\frac{\sqrt3}{3+2\sqrt3}=\frac{3\sqrt3-6}{-3}=2-\sqrt3.\)

Therefore

\(\frac{1}{1+\sqrt3}-\frac{\sqrt3}{3+2\sqrt3} =\frac{\sqrt3-1}{2}-(2-\sqrt3).\)

Putting this over a common denominator gives

\(\frac{\sqrt3-1-4+2\sqrt3}{2}=\frac{3\sqrt3-5}{2}.\)

Hence

\(\boxed{\frac{3\sqrt3-5}{2}}.\)

Answer: \(5\sqrt5-11\).

First expand the numerator:

\((\sqrt5-3)^2=5-6\sqrt5+9=14-6\sqrt5.\)

So

\(\frac{(\sqrt5-3)^2}{\sqrt5+1} =\frac{14-6\sqrt5}{\sqrt5+1}.\)

Rationalise the denominator by multiplying by \(\sqrt5-1\):

\(\frac{14-6\sqrt5}{\sqrt5+1}\cdot\frac{\sqrt5-1}{\sqrt5-1}.\)

The denominator is

\((\sqrt5+1)(\sqrt5-1)=5-1=4.\)

The numerator is

\((14-6\sqrt5)(\sqrt5-1)=14\sqrt5-14-30+6\sqrt5=20\sqrt5-44.\)

Therefore

\(\frac{20\sqrt5-44}{4}=5\sqrt5-11.\)

So the expression is

\(\boxed{5\sqrt5-11}.\)

Answer: (a) \(2+2\sqrt5\); (b) \(2\sqrt{30}\).

For part (a), first simplify the numerator:

\((3+2\sqrt5)(6-2\sqrt5)=18-6\sqrt5+12\sqrt5-20=6\sqrt5-2.\)

So the expression is

\(\dfrac{6\sqrt5-2}{4-\sqrt5}.\)

Rationalise the denominator:

\(\dfrac{6\sqrt5-2}{4-\sqrt5}\cdot\dfrac{4+\sqrt5}{4+\sqrt5}=\dfrac{(6\sqrt5-2)(4+\sqrt5)}{16-5}.\)

The numerator is

\(24\sqrt5+30-8-2\sqrt5=22+22\sqrt5.\)

Therefore the simplified value is

\(\dfrac{22+22\sqrt5}{11}=2+2\sqrt5.\)

For part (b), use the cosine rule:

\(AC^2=AB^2+BC^2-2(AB)(BC)\cos ABC.\)

Since \(\cos ABC=-\dfrac12\), this becomes

\(AC^2=(6-2\sqrt3)^2+(6+2\sqrt3)^2+(6-2\sqrt3)(6+2\sqrt3).\)

Now

\((6-2\sqrt3)^2+(6+2\sqrt3)^2=96,\)

and

\((6-2\sqrt3)(6+2\sqrt3)=36-12=24.\)

So

\(AC^2=120.\)

Hence

\(\boxed{AC=\sqrt{120}=2\sqrt{30}}.\)

Answer: (a) \(\sqrt5\). (b) \(8\sqrt6\). (c) \(x=\sqrt2\) or \(x=-2\sqrt2\).

(a) Rationalise the denominator:

\(\frac{5+6\sqrt5}{6+\sqrt5} \cdot \frac{6-\sqrt5}{6-\sqrt5}.\)

The denominator is

\(36-5=31.\)

The numerator is

\((5+6\sqrt5)(6-\sqrt5) = 30-5\sqrt5+36\sqrt5-30.\)

So the numerator is

\(31\sqrt5.\)

Therefore

\(\frac{5+6\sqrt5}{6+\sqrt5} = \sqrt5.\)

(b) Since \(3^{0.5}=\sqrt3\),

\(3^{0.5}\times(\sqrt2)^7 = \sqrt3\times(\sqrt2)^7.\)

Now

\((\sqrt2)^7=(\sqrt2)^6\sqrt2=2^3\sqrt2=8\sqrt2.\)

Hence

\(\sqrt3\times8\sqrt2=8\sqrt6.\)

(c) Multiply both sides of

\(x+\sqrt2=\frac4x\)

by \(x\):

\(x^2+\sqrt2x=4.\)

So

\(x^2+\sqrt2x-4=0.\)

Use the quadratic formula:

\(x=\frac{-\sqrt2\pm\sqrt{(\sqrt2)^2-4(1)(-4)}}{2}.\)

This gives

\(x=\frac{-\sqrt2\pm\sqrt{18}}{2}.\)

Since \(\sqrt{18}=3\sqrt2\),

\(x=\frac{-\sqrt2\pm3\sqrt2}{2}.\)

Therefore

\(\boxed{x=\sqrt2\quad\text{or}\quad x=-2\sqrt2}.\)

Answer: (a) \(\sqrt5\). (b) \(8\sqrt6\). (c) \(x=\sqrt2\) or \(x=-2\sqrt2\).

(a) Rationalise the denominator:

\(\frac{5+6\sqrt5}{6+\sqrt5} \cdot \frac{6-\sqrt5}{6-\sqrt5}.\)

The denominator is

\(36-5=31.\)

The numerator is

\((5+6\sqrt5)(6-\sqrt5) = 30-5\sqrt5+36\sqrt5-30.\)

So the numerator is

\(31\sqrt5.\)

Therefore

\(\frac{5+6\sqrt5}{6+\sqrt5} = \sqrt5.\)

(b) Since \(3^{0.5}=\sqrt3\),

\(3^{0.5}\times(\sqrt2)^7 = \sqrt3\times(\sqrt2)^7.\)

Now

\((\sqrt2)^7=(\sqrt2)^6\sqrt2=2^3\sqrt2=8\sqrt2.\)

Hence

\(\sqrt3\times8\sqrt2=8\sqrt6.\)

(c) Multiply both sides of

\(x+\sqrt2=\frac4x\)

by \(x\):

\(x^2+\sqrt2x=4.\)

So

\(x^2+\sqrt2x-4=0.\)

Use the quadratic formula:

\(x=\frac{-\sqrt2\pm\sqrt{(\sqrt2)^2-4(1)(-4)}}{2}.\)

This gives

\(x=\frac{-\sqrt2\pm\sqrt{18}}{2}.\)

Since \(\sqrt{18}=3\sqrt2\),

\(x=\frac{-\sqrt2\pm3\sqrt2}{2}.\)

Therefore

\(\boxed{x=\sqrt2\quad\text{or}\quad x=-2\sqrt2}.\)

Answer: (a) \(-22+5\sqrt{10}\). (b) \(10\sqrt3-13\sqrt2\).

(a) Expand each term:

\((\sqrt2+2\sqrt5)(4\sqrt2-3\sqrt5) =8-3\sqrt{10}+8\sqrt{10}-30.\)

Collect like terms:

\(8-30+5\sqrt{10}=-22+5\sqrt{10}.\)

(b) Rationalise the denominator by multiplying by \(\sqrt3-\sqrt2\):

\(\frac{4-3\sqrt6}{\sqrt3+\sqrt2} \cdot \frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}.\)

The denominator is

\(3-2=1.\)

The numerator is

\(4\sqrt3-4\sqrt2-3\sqrt{18}+3\sqrt{12}.\)

Since \(\sqrt{18}=3\sqrt2\) and \(\sqrt{12}=2\sqrt3\), this becomes

\(4\sqrt3-4\sqrt2-9\sqrt2+6\sqrt3 =10\sqrt3-13\sqrt2.\)

Answer: (a) \(27\sqrt2\). (b) \(x=15+8\sqrt3\).

(a) First simplify each term:

\(\sqrt{24}\times\sqrt{27}=(2\sqrt6)(3\sqrt3)=6\sqrt{18}=18\sqrt2,\)

and

\(\frac{9\sqrt{30}}{\sqrt{15}}=9\sqrt{\frac{30}{15}}=9\sqrt2.\)

So the whole expression is

\(18\sqrt2+9\sqrt2=27\sqrt2.\)

Hence it can be written in the form \(a\sqrt2\) with

\(a=27.\)

(b) Expand and collect the \(x\)-terms:

\(\sqrt3+\sqrt3\,x=2x-6\)

so

\(x(\sqrt3-2)=-(6+\sqrt3).\)

Therefore

\(x=\frac{6+\sqrt3}{2-\sqrt3}.\)

Rationalise the denominator:

\(x=\frac{6+\sqrt3}{2-\sqrt3}\cdot\frac{2+\sqrt3}{2+\sqrt3} =\frac{(6+\sqrt3)(2+\sqrt3)}{4-3}.\)

Expand the numerator:

\((6+\sqrt3)(2+\sqrt3)=12+6\sqrt3+2\sqrt3+3=15+8\sqrt3.\)

Hence

\(\boxed{x=15+8\sqrt3}.\)

Answer: \(9-4\sqrt5\).

First deal with the negative index:

\(\left(\frac{1+\sqrt5}{3-\sqrt5}\right)^{-2} = \left(\frac{3-\sqrt5}{1+\sqrt5}\right)^2.\)

Now rationalise the fraction inside the bracket:

\(\frac{3-\sqrt5}{1+\sqrt5}\cdot\frac{1-\sqrt5}{1-\sqrt5} = \frac{(3-\sqrt5)(1-\sqrt5)}{1-5}.\)

Expand the numerator:

\((3-\sqrt5)(1-\sqrt5)=3-3\sqrt5-\sqrt5+5=8-4\sqrt5.\)

So

\(\frac{3-\sqrt5}{1+\sqrt5} = \frac{8-4\sqrt5}{-4} = -2+\sqrt5.\)

Hence the required expression is

\((-2+\sqrt5)^2 =4-4\sqrt5+5 =9-4\sqrt5.\)

Therefore

\(\boxed{9-4\sqrt5}.\)

Answer: \(a=2\), \(b=20\).

Square the expression for \(z\):

\(z^2=\left(a+(a+3)\sqrt3\right)^2.\)

Expand:

\(z^2=a^2+2a(a+3)\sqrt3+3(a+3)^2.\)

Since \(z^2=79+b\sqrt3\), compare rational parts and surd parts.

The rational part gives

\(a^2+3(a+3)^2=79.\)

Expand:

\(a^2+3(a^2+6a+9)=79.\)

So

\(4a^2+18a+27=79,\)

which gives

\(2a^2+9a-26=0.\)

Factor:

\((a-2)(2a+13)=0.\)

Since \(a\) is an integer,

\(a=2.\)

The coefficient of \(\sqrt3\) is

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

So

\(b=2(2)(5)=20.\)

Therefore

\(\boxed{a=2,\quad b=20}.\)

Answer: \(a=1\), \(b=-3\).

First calculate \(z^2\):

\(z^2=(2+\sqrt3)^2=4+4\sqrt3+3=7+4\sqrt3.\)

Now substitute into \(az^2+bz=1+\sqrt3\):

\(a(7+4\sqrt3)+b(2+\sqrt3)=1+\sqrt3.\)

Collect the rational and \(\sqrt3\) parts:

\((7a+2b)+(4a+b)\sqrt3=1+\sqrt3.\)

Equating corresponding parts gives

\(7a+2b=1,\)

and

\(4a+b=1.\)

From \(4a+b=1\),

\(b=1-4a.\)

Substitute into \(7a+2b=1\):

\(7a+2(1-4a)=1.\)

So

\(-a+2=1,\)

giving

\(a=1.\)

Then

\(b=1-4(1)=-3.\)

Therefore

\(\boxed{a=1,\quad b=-3}.\)

Answer: \(p=5\), \(q=2\).

Put the left hand side over a common denominator:

\(\frac{p}{\sqrt3-1}+\frac{1}{\sqrt3+1} = \frac{p(\sqrt3+1)+(\sqrt3-1)}{(\sqrt3-1)(\sqrt3+1)}.\)

The denominator is

\((\sqrt3-1)(\sqrt3+1)=3-1=2.\)

So

\(\frac{p(\sqrt3+1)+(\sqrt3-1)}{2}=q+3\sqrt3.\)

Multiply by \(2\):

\(p(\sqrt3+1)+(\sqrt3-1)=2q+6\sqrt3.\)

Collect rational and surd parts on the left:

\((p-1)+(p+1)\sqrt3=2q+6\sqrt3.\)

Equating the coefficients of \(\sqrt3\),

\(p+1=6\), so \(p=5\).

Equating the rational parts,

\(p-1=2q\). Since \(p=5\),

\(4=2q\), so \(q=2\).

Therefore

\(\boxed{p=5,\quad q=2}.\)

Answer: (a) \(\mathrm q(x)=3x^2-10x-8,\ r=5\); (b) \((3x+2)(x-4)(2x-5)\); (c) \(x=-\frac23,\ 4,\ \frac52\).

We start with the main method. Use polynomial division or comparison of coefficients, then apply the factor theorem to finish the factorisation.

Divide \(6x^3-35x^2+34x+45\) by \(2x-5\).

We look for

\(\mathrm p(x)=(2x-5)\mathrm q(x)+r.\)

The quotient is

\(\mathrm q(x)=3x^2-10x-8,\)

and the remainder is

\(r=5.\)

Therefore

\(\mathrm p(x)-5=(2x-5)(3x^2-10x-8).\)

Now factorise the quadratic:

\(3x^2-10x-8=(3x+2)(x-4).\)

Hence

\(\mathrm p(x)-5=(3x+2)(x-4)(2x-5).\)

For \(\mathrm p(x)=5\), set \(\mathrm p(x)-5=0\):

\((3x+2)(x-4)(2x-5)=0.\)

Thus

\(x=-\frac23,\quad x=4,\quad x=\frac52.\)

This completes the solution and gives the required result.

Answer: \(a=-2\), \(b=8\), \(\mathrm{p}(x)=(x-2)(x+1)(3x-4)\), and \(y=\frac12\ln2\) or \(y=\frac12\ln\frac43\).

Differentiate:

\(\mathrm{p}'(x)=9x^2-14x+a.\)

Using \(\mathrm{p}'(-1)=21\),

\(9+14+a=21,\)

so

\(a=-2.\)

Since \(x-2\) is a factor, \(\mathrm{p}(2)=0\):

\(24-28+2a+b=0.\)

Substitute \(a=-2\):

\(24-28-4+b=0,\)

so

\(b=8.\)

Thus

\(\mathrm{p}(x)=3x^3-7x^2-2x+8.\)

Dividing by \(x-2\) gives

\(\mathrm{p}(x)=(x-2)(3x^2-x-4).\)

Then

\(3x^2-x-4=(x+1)(3x-4),\)

so

\(\mathrm{p}(x)=(x-2)(x+1)(3x-4).\)

For the exponential equation, let \(u=\mathrm{e}^{2y}\). Then \(u\gt 0\) and

\(3u^3-7u^2-2u+8=0.\)

Using the factorisation above,

\((u-2)(u+1)(3u-4)=0.\)

Since \(u\gt 0\), we take \(u=2\) or \(u=\frac43\). Hence

\(\mathrm{e}^{2y}=2\quad\text{or}\quad \mathrm{e}^{2y}=\frac43.\)

Therefore

\(y=\frac12\ln2\quad\text{or}\quad y=\frac12\ln\frac43.\)

Answer: \(\mathrm{p}(x)=(x-1)(x-5)(x+6)\).

By the remainder theorem, the remainder when \(\mathrm{p}(x)\) is divided by \(x+2\) is \(\mathrm{p}(-2)\).

The remainder is given as \(84\), so

\(\mathrm{p}(-2)=84.\)

Now

\(\mathrm{p}(-2)=(-2)^3+A(-2)+30.\)

Thus

\(-8-2A+30=84.\)

So

\(22-2A=84,\)

which gives

\(A=-31.\)

Therefore

\(\mathrm{p}(x)=x^3-31x+30.\)

Use the factor theorem to find a linear factor. Testing \(x=1\),

\(1^3-31(1)+30=0,\)

so \((x-1)\) is a factor. Dividing gives

\(x^3-31x+30=(x-1)(x^2+x-30).\)

Finally,

\(x^2+x-30=(x-5)(x+6).\)

Hence

\(\mathrm{p}(x)=(x-1)(x-5)(x+6).\)

Answer: \(a=4\), \(\mathrm{p}(x)=2(2x+1)(4x^2+1)\), and the only real root is \(x=-\frac12\).

Use the factor theorem and the remainder theorem to turn the given information into equations for the unknown constants. After finding the constants, factorise the polynomial if required.

Since \(2x+1\) is a factor, \(x=-\frac12\) is a root. Therefore \(\mathrm{p}(-\frac12)=0\).

Substitute \(x=-\frac12\):

\(a^2\left(-\frac12\right)^3+2a\left(-\frac12\right)^2+a\left(-\frac12\right)+2=0.\)

This gives

\(-\frac{a^2}{8}+\frac{a}{2}-\frac{a}{2}+2=0,\)

so

\(-\frac{a^2}{8}+2=0.\)

Hence \(a^2=16\). Since \(a\) is positive, \(a=4\).

Now

\(\mathrm{p}(x)=16x^3+8x^2+4x+2.\)

Factorising gives

\(\mathrm{p}(x)=2(2x+1)(4x^2+1).\)

The factor \(4x^2+1\) has no real roots, since \(4x^2+1=0\) would require \(x^2=-\frac14\).

Therefore the only real root comes from \(2x+1=0\), namely \(x=-\frac12\).

The result has been obtained using the exact condition from the question, so no additional solutions are introduced.

Answer: \(a=29,\ b=-9\); \(x=-3,\frac15\); \(\theta=99.7^\circ,170.3^\circ,279.7^\circ,350.3^\circ\).

We start with the main method. Apply the factor theorem: if \((x-a)\) is a factor, then \(p(a)=0\).

(a) Since \(x+3\) is a factor of both \(\mathrm p(x)\) and \(\mathrm p'(x)\),

\(\mathrm p(-3)=0\quad\text{and}\quad \mathrm p'(-3)=0.\)

Now

\(\mathrm p'(x)=15x^2+2ax+39.\)

Using \(\mathrm p'(-3)=0\):

\(135-6a+39=0.\)

So

\(174-6a=0,\)

hence

\(a=29.\)

Using \(\mathrm p(-3)=0\):

\(5(-3)^3+29(-3)^2+39(-3)+b=0.\)

This gives

\(-135+261-117+b=0,\)

so

\(9+b=0.\)

Hence

\(b=-9.\)

(b) With these values,

\(\mathrm p(x)=5x^3+29x^2+39x-9.\)

Since \(x+3\) is a factor, divide by \(x+3\):

\(\mathrm p(x)=(x+3)(5x^2+14x-3).\)

Then

\(5x^2+14x-3=(5x-1)(x+3).\)

Thus

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

So the solutions of \(\mathrm p(x)=0\) are

\(x=-3\quad\text{or}\quad x=\frac15.\)

(c) Let

\(u=\operatorname{cosec}2\theta.\)

The equation becomes

\(\mathrm p(u)=0,\)

so

\(u=-3\quad\text{or}\quad u=\frac15.\)

But \(\left|\operatorname{cosec}2\theta\right|\geq1\), so \(u=\frac15\) is impossible.

Therefore

\(\operatorname{cosec}2\theta=-3,\)

which gives

\(\sin2\theta=-\frac13.\)

For \(0^\circ\leq\theta\leq360^\circ\), we have \(0^\circ\leq2\theta\leq720^\circ\).

The solutions for \(2\theta\) are approximately

\(199.47^\circ,\quad340.53^\circ,\quad559.47^\circ,\quad700.53^\circ.\)

Hence

\(\theta=99.7^\circ,\quad170.3^\circ,\quad279.7^\circ,\quad350.3^\circ.\)

This completes the solution and gives the required result.

Answer: \(p(x)=(x+4)(2x^2+3x+10)\), so the only real root is \(x=-4\).

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) Substitute \(x=-4\):

\(p(-4)=2(-4)^3+11(-4)^2+22(-4)+40.\)

So

\(p(-4)=-128+176-88+40=0.\)

Therefore \(x=-4\) is a root.

(b) Since \(x=-4\) is a root, \(x+4\) is a factor of \(p(x)\). Dividing \(2x^3+11x^2+22x+40\) by \(x+4\) gives

\(p(x)=(x+4)(2x^2+3x+10).\)

For the quadratic factor, the discriminant is

\(b^2-4ac=3^2-4(2)(10)=9-80=-71.\)

Since the discriminant is negative, \(2x^2+3x+10=0\) has no real roots. Hence \(p(x)=0\) has no other real roots.