Let

\[ u=2^x. \]

Then

\[ 2^{x+2}=4u \]

and

\[ 2^{2-x}=\frac{4}{u}. \]

So the equation becomes

\[ 4(4u)-5\left(\frac{4}{u}\right)=3. \]

Hence

\[ 16u-\frac{20}{u}=3. \]

Multiply by \(u\):

\[ 16u^2-20=3u. \]

So

\[ 16u^2-3u-20=0. \]

Using the quadratic formula,

\[ u=\frac{3\pm\sqrt{(-3)^2-4(16)(-20)}}{2(16)} =\frac{3\pm\sqrt{1289}}{32}. \]

Since \(u=2^x>0\),

\[ u=\frac{3+\sqrt{1289}}{32}. \]

Therefore

\[ 2^x=\frac{3+\sqrt{1289}}{32}. \]

Taking logarithms,

\[ x=\frac{\ln\left(\frac{3+\sqrt{1289}}{32}\right)}{\ln2}. \]

\[ x=0.281796\ldots \]

Answer: \(x=0.282\).

(a) The condition

\[ |z|=9 \]

means the points lie on a circle with centre \(O\) and radius \(9\).

The condition

\[ \frac12\pi\leq \arg z<\pi \]

restricts the locus to the part of the circle in the second quadrant.

So the locus is the arc of the circle \(|z|=9\) from \(9i\) to \(-9\), with \(9i\) included and \(-9\) not included.

(b) The transformation \(z\mapsto z^*\) reflects the locus in the real axis.

The transformation \(z^*\mapsto z^*+3\) then translates the reflected locus \(3\) units in the positive real direction.

So the locus is an arc of a circle with centre \(3\) and radius \(9\), from \(3-9i\) to \(-6\), with \(3-9i\) included and \(-6\) not included.

Use

\[ \sin2x=2\sin x\cos x. \]

Then

\[ 3\sin x\sin2x=3\sin x(2\sin x\cos x)=6\sin^2x\cos x. \]

So

\[ \int 3\sin x\sin2x\,dx=\int 6\sin^2x\cos x\,dx. \]

Let \(u=\sin x\), so \(du=\cos x\,dx\).

Then

\[ \int 6\sin^2x\cos x\,dx=\int 6u^2\,du=2u^3=2\sin^3x. \]

Therefore

\[ \int_{\frac14\pi}^{\frac13\pi} 3\sin x\sin2x\,dx =\left[2\sin^3x\right]_{\frac14\pi}^{\frac13\pi}. \]

\[ =2\left(\frac{\sqrt3}{2}\right)^3-2\left(\frac{\sqrt2}{2}\right)^3. \]

\[ =\frac{3\sqrt3}{4}-\frac{\sqrt2}{2}. \]

Answer: \(\frac34\sqrt3-\frac12\sqrt2\).

(a) In the expansion of \((1-ax)^{\frac25}\), the coefficient of \(x^3\) is

\[ \frac{\frac25\left(\frac25-1\right)\left(\frac25-2\right)}{3!}(-a)^3. \]

So

\[ \frac{\frac25\left(-\frac35\right)\left(-\frac85\right)}{6}(-a)^3=1. \]

\[ \frac{8}{125}(-a)^3=1. \]

\[ -a=\frac52. \]

Therefore

\[ a=-\frac52. \]

(b) Since \(a=-\frac52\),

\[ (1-ax)^{\frac25}=\left(1+\frac52x\right)^{\frac25}. \]

The coefficient of \(x^3\) in this expansion is \(1\).

The coefficient of \(x^4\) in \(\left(1+\frac52x\right)^{\frac25}\) is

\[ \frac{\frac25\left(-\frac35\right)\left(-\frac85\right)\left(-\frac{13}{5}\right)}{4!}\left(\frac52\right)^4. \]

\[ =-\frac{13}{8}. \]

To get the coefficient of \(x^4\) in

\[ (2x+1)(1-ax)^{\frac25}, \]

we need:

\[ 1\times\text{coefficient of }x^4 \]

and

\[ 2x\times\text{coefficient of }x^3. \]

Hence the required coefficient is

\[ -\frac{13}{8}+2(1)=\frac38. \]

Answer: \(\frac38\).

(c) The binomial expansion is valid when

\[ \left|\frac52x\right|<1. \]

Therefore

\[ |x|<\frac25. \]

Answer: \(-\frac25<x<\frac25\).

(a) Rationalise the denominator:

\[ z=\frac{3+\lambda i}{\lambda+2i}\times\frac{\lambda-2i}{\lambda-2i}. \]

\[ z=\frac{(3+\lambda i)(\lambda-2i)}{\lambda^2+4}. \]

Expanding the numerator,

\[ (3+\lambda i)(\lambda-2i)=3\lambda-6i+\lambda^2 i+2\lambda. \]

\[ =5\lambda+(\lambda^2-6)i. \]

So

\[ z=\frac{5\lambda}{\lambda^2+4}+\frac{\lambda^2-6}{\lambda^2+4}i. \]

For \(\arg z=\frac14\pi\), the real and imaginary parts must be equal and positive.

Therefore

\[ 5\lambda=\lambda^2-6. \]

\[ \lambda^2-5\lambda-6=0. \]

\[ (\lambda-6)(\lambda+1)=0. \]

So

\[ \lambda=6\quad\text{or}\quad \lambda=-1. \]

Since the real part must be positive, \(\lambda=6\).

Answer: \(\lambda=6\).

(b) When \(\lambda=6\),

\[ z=\frac{3+6i}{6+2i}. \]

Use

\[ \left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}. \]

So

\[ |z|=\frac{|3+6i|}{|6+2i|}. \]

\[ |z|=\frac{\sqrt{3^2+6^2}}{\sqrt{6^2+2^2}} =\frac{\sqrt{45}}{\sqrt{40}}. \]

\[ |z|=\sqrt{\frac{9}{8}}=\frac{3\sqrt2}{4}. \]

Answer: \(\frac{3\sqrt2}{4}\).

(a) Since \((x^2+x+1)\) is a factor of \(p(x)\), divide \(p(x)\) by \(x^2+x+1\).

\[ p(x)=2x^4+ax^3+4x^2+bx-3. \]

Using polynomial division, the remainder is

\[ (b-2)x+(a-7). \]

For \(x^2+x+1\) to be a factor, the remainder must be zero.

Therefore

\[ b-2=0 \]

and

\[ a-7=0. \]

So

\[ a=7,\quad b=2. \]

Answer: \(a=7\), \(b=2\).

(b) With \(a=7\) and \(b=2\),

\[ p(x)=2x^4+7x^3+4x^2+2x-3. \]

Substitute \(x=-3\):

\[ p(-3)=2(-3)^4+7(-3)^3+4(-3)^2+2(-3)-3. \]

\[ =162-189+36-6-3=0. \]

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

(a) Sketch the graphs

\[ y=\ln x \]

and

\[ y=\operatorname{cosec}\frac12x \]

for \(0<x<\pi\).

On this interval, \(y=\ln x\) is increasing.

Also, \(\sin\frac12x\) is increasing on \(0<x<\pi\), so \(y=\operatorname{cosec}\frac12x\) is decreasing.

Therefore the two graphs can meet at most once.

Since the sketch shows one intersection, the equation has exactly one root in \(0<x<\pi\).

(b) Let

\[ f(x)=\ln x-\operatorname{cosec}\frac12x. \]

At \(x=2.6\),

\[ f(2.6)=\ln(2.6)-\operatorname{cosec}(1.3)=-0.0823\ldots \]

At \(x=2.9\),

\[ f(2.9)=\ln(2.9)-\operatorname{cosec}(1.45)=0.0574\ldots \]

Since there is a change of sign, the root lies between \(2.6\) and \(2.9\).

(c) Using

\[ x_{n+1}=\exp\left(\operatorname{cosec}\frac12x_n\right), \]

and starting with \(x_0=2.6\):

\[ x_1=2.82306 \]

\[ x_2=2.75335 \]

\[ x_3=2.77082 \]

\[ x_4=2.76609 \]

\[ x_5=2.76734 \]

\[ x_6=2.76701 \]

\[ x_7=2.76710 \]

Therefore the root correct to \(3\) decimal places is

\[ x=2.767. \]

Answer: \(2.767\).

Start with

\[ ye^{3x}\frac{dy}{dx}=x(y+5). \]

Separate the variables:

\[ \frac{y}{y+5}\frac{dy}{dx}=xe^{-3x}. \]

So

\[ \int \frac{y}{y+5}\,dy=\int xe^{-3x}\,dx. \]

For the left-hand side, write

\[ \frac{y}{y+5}=1-\frac{5}{y+5}. \]

Hence

\[ \int \frac{y}{y+5}\,dy=y-5\ln(y+5). \]

For the right-hand side, integrate by parts:

\[ \int xe^{-3x}\,dx=-\frac13xe^{-3x}-\frac19e^{-3x}. \]

Therefore

\[ y-5\ln(y+5)=-\frac13xe^{-3x}-\frac19e^{-3x}+C. \]

Use \(y=0\) when \(x=0\):

\[ 0-5\ln5=-\frac19+C. \]

So

\[ C=\frac19-5\ln5. \]

Therefore

\[ y-5\ln(y+5)=-\frac13xe^{-3x}-\frac19e^{-3x}+\frac19-5\ln5. \]

Equivalently,

\[ y-5\ln\left(\frac{y+5}{5}\right)=\frac19-\left(\frac{x}{3}+\frac19\right)e^{-3x}. \]

(a) Use

\[ x=\sqrt3\tan u. \]

Then

\[ dx=\sqrt3\sec^2u\,du. \]

Also,

\[ 3+x^2=3+3\tan^2u=3\sec^2u. \]

and

\[ x^3=(\sqrt3\tan u)^3=3\sqrt3\tan^3u. \]

Therefore

\[ \frac{x^3}{3+x^2}\,dx = \frac{3\sqrt3\tan^3u}{3\sec^2u}\cdot\sqrt3\sec^2u\,du. \]

\[ =3\tan^3u\,du. \]

When \(x=1\),

\[ 1=\sqrt3\tan u \]

so

\[ \tan u=\frac1{\sqrt3} \]

and

\[ u=\frac{\pi}{6}. \]

When \(x=3\),

\[ 3=\sqrt3\tan u \]

so

\[ \tan u=\sqrt3 \]

and

\[ u=\frac{\pi}{3}. \]

Hence

\[ I=\int_{\frac16\pi}^{\frac13\pi}3\tan^3u\,du. \]

(b) It is easier to divide first:

\[ \frac{x^3}{3+x^2}=x-\frac{3x}{x^2+3}. \]

So

\[ I=\int_1^3\left(x-\frac{3x}{x^2+3}\right)\,dx. \]

\[ I=\left[\frac{x^2}{2}-\frac32\ln(x^2+3)\right]_1^3. \]

\[ I=\left(\frac92-\frac32\ln12\right)-\left(\frac12-\frac32\ln4\right). \]

\[ I=4-\frac32\ln3. \]

Answer: \(4-\frac32\ln3\).

(a) Start with

\[ y^2=k\frac{x-2}{x+2}. \]

Differentiate implicitly:

\[ 2y\frac{dy}{dx} = k\frac{(x+2)-(x-2)}{(x+2)^2}. \]

So

\[ 2y\frac{dy}{dx}=\frac{4k}{(x+2)^2}. \]

From the original equation,

\[ k=\frac{y^2(x+2)}{x-2}. \]

Substitute this into the derivative equation:

\[ 2y\frac{dy}{dx} = \frac{4}{(x+2)^2}\cdot\frac{y^2(x+2)}{x-2}. \]

\[ 2y\frac{dy}{dx}=\frac{4y^2}{(x+2)(x-2)}. \]

\[ 2y\frac{dy}{dx}=\frac{4y^2}{x^2-4}. \]

Therefore

\[ \frac{dy}{dx}=\frac{2y}{x^2-4}. \]

(b) When \(k=5\) and \(x=3\),

\[ y^2=5\frac{3-2}{3+2}=1. \]

So

\[ y=1\quad\text{or}\quad y=-1. \]

Using

\[ \frac{dy}{dx}=\frac{2y}{x^2-4}, \]

at \(x=3\):

\[ \frac{dy}{dx}=\frac{2y}{9-4}=\frac{2y}{5}. \]

So the two gradients are

\[ m_1=\frac25,\quad m_2=-\frac25. \]

The tangents make angles

\[ \tan^{-1}\left(\frac25\right) \]

and

\[ -\tan^{-1}\left(\frac25\right) \]

with the positive \(x\)-axis.

Therefore the angle between the tangents is

\[ 2\tan^{-1}\left(\frac25\right). \]

Answer: \(2\tan^{-1}\left(\frac25\right)\).

(a) Write

\[ \overrightarrow{OA}=\begin{pmatrix}2\\2\\-1\end{pmatrix}, \quad \overrightarrow{OB}=\begin{pmatrix}4\\2\\4\end{pmatrix}. \]

The perpendicular distance from \(A\) to the line through \(O\) and \(B\) is

\[ \frac{|\overrightarrow{OA}\times\overrightarrow{OB}|}{|\overrightarrow{OB}|}. \]

Now

\[ \overrightarrow{OA}\times\overrightarrow{OB} = \begin{pmatrix}2\\2\\-1\end{pmatrix} \times \begin{pmatrix}4\\2\\4\end{pmatrix}. \]

\[ = \begin{pmatrix} 10\\ -12\\ -4 \end{pmatrix}. \]

So

\[ |\overrightarrow{OA}\times\overrightarrow{OB}| = \sqrt{10^2+(-12)^2+(-4)^2} =\sqrt{260}=2\sqrt{65}. \]

Also,

\[ |\overrightarrow{OB}|=\sqrt{4^2+2^2+4^2}=\sqrt{36}=6. \]

Hence the perpendicular distance is

\[ \frac{2\sqrt{65}}{6}=\frac13\sqrt{65}. \]

(b) We have

\[ \overrightarrow{OC}=\begin{pmatrix}3\\p\\q\end{pmatrix}. \]

Also

\[ \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA} = \begin{pmatrix}4\\2\\4\end{pmatrix} - \begin{pmatrix}2\\2\\-1\end{pmatrix} = \begin{pmatrix}2\\0\\5\end{pmatrix}. \]

Since \(\overrightarrow{OC}\) is perpendicular to \(\overrightarrow{AB}\),

\[ \overrightarrow{OC}\cdot\overrightarrow{AB}=0. \]

So

\[ \begin{pmatrix}3\\p\\q\end{pmatrix} \cdot \begin{pmatrix}2\\0\\5\end{pmatrix} =0. \]

\[ 6+5q=0. \]

Therefore

\[ q=-\frac65. \]

Since angle \(AOC=\) angle \(COB\),

\[ \frac{\overrightarrow{OA}\cdot\overrightarrow{OC}}{|\overrightarrow{OA}|\,|\overrightarrow{OC}|} = \frac{\overrightarrow{OC}\cdot\overrightarrow{OB}}{|\overrightarrow{OC}|\,|\overrightarrow{OB}|}. \]

Cancel \(|\overrightarrow{OC}|\):

\[ \frac{\overrightarrow{OA}\cdot\overrightarrow{OC}}{|\overrightarrow{OA}|} = \frac{\overrightarrow{OC}\cdot\overrightarrow{OB}}{|\overrightarrow{OB}|}. \]

Now

\[ |\overrightarrow{OA}|=\sqrt{2^2+2^2+(-1)^2}=3 \]

and

\[ |\overrightarrow{OB}|=6. \]

So

\[ 2(\overrightarrow{OA}\cdot\overrightarrow{OC}) = \overrightarrow{OC}\cdot\overrightarrow{OB}. \]

Now

\[ \overrightarrow{OA}\cdot\overrightarrow{OC} = 2(3)+2p+(-1)q=6+2p-q. \]

Using \(q=-\frac65\),

\[ \overrightarrow{OA}\cdot\overrightarrow{OC}=6+2p+\frac65=\frac{36}{5}+2p. \]

Also

\[ \overrightarrow{OC}\cdot\overrightarrow{OB} = 3(4)+2p+4q=12+2p+4q. \]

Using \(q=-\frac65\),

\[ \overrightarrow{OC}\cdot\overrightarrow{OB}=12+2p-\frac{24}{5}=\frac{36}{5}+2p. \]

Therefore

\[ 2\left(\frac{36}{5}+2p\right)=\frac{36}{5}+2p. \]

\[ \frac{72}{5}+4p=\frac{36}{5}+2p. \]

\[ 2p=-\frac{36}{5}. \]

So

\[ p=-\frac{18}{5}. \]

Answer: \(p=-\frac{18}{5}\), \(q=-\frac65\).