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