(a) The gradient of \(PQ\) is

\( \frac{6-p}{8-0}=\frac{6-p}{8} \).

The gradient of \(QR\) is

\( \frac{10-6}{r-8}=\frac4{r-8} \).

Since \(PQ\) is perpendicular to \(QR\), the product of the gradients is \(-1\):

\( \frac{6-p}{8}\cdot\frac4{r-8}=-1 \).

\( 6-p=-2(r-8) \).

\( 6-p=-2r+16 \).

Therefore \( p=2r-10 \).

(b) Since \(PR=\sqrt{85}\),

\( r^2+(10-p)^2=85 \).

Use \(p=2r-10\):

\( r^2+(10-(2r-10))^2=85 \).

\( r^2+(20-2r)^2=85 \).

\( r^2+4(r-10)^2=85 \).

\( 5r^2-80r+315=0 \).

Divide by \(5\):

\( r^2-16r+63=0 \).

\( (r-7)(r-9)=0 \).

So \( r=7 \) or \( r=9 \).

If \( r=7 \), then \( p=2(7)-10=4 \).

If \( r=9 \), then \( p=2(9)-10=8 \).

Answer: \( (p,r)=(4,7) \) or \( (p,r)=(8,9) \).

(a) Differentiate:

\( \frac{dy}{dx}=-\frac32x^{-\frac32}+3x^{-\frac52} \).

At a stationary point, \( \frac{dy}{dx}=0 \):

\( -\frac32x^{-\frac32}+3x^{-\frac52}=0 \).

Multiply by \(2x^{\frac52}\):

\( -3x+6=0 \).

So \( x=2 \). Therefore \( k=2 \).

(b)

\( \frac{d^2y}{dx^2}=\frac94x^{-\frac52}-\frac{15}{2}x^{-\frac72} \).

At \(x=2\),

\( \frac{d^2y}{dx^2}=2^{-\frac72}\left(\frac94(2)-\frac{15}{2}\right)<0 \).

Therefore the stationary point is a maximum.

(c) Since \(k=2\), the required area is

\( \int_2^4 \left(3x^{-\frac12}-2x^{-\frac32}\right)\,dx \).

\( \int \left(3x^{-\frac12}-2x^{-\frac32}\right)\,dx=6x^{\frac12}+4x^{-\frac12} \).

So the area is

\( \left[6\sqrt{x}+\frac4{\sqrt{x}}\right]_2^4 \).

\( =\left(12+2\right)-\left(6\sqrt2+\frac4{\sqrt2}\right) \).

\( =14-8\sqrt2 \).

Answer: \( 14-8\sqrt2 \).

(a) The third term of the arithmetic progression is \(20+2d\), and the third term of the geometric progression is \(20r^2\).

So

\( 20r^2=20+2d+5 \).

\( 20r^2=25+2d \). (1)

The fifth term of the arithmetic progression is \(20+4d\), and the fifth term of the geometric progression is \(20r^4\).

So

\( 20r^4=20+4d+30 \).

\( 20r^4=50+4d \). (2)

From (1), \( d=10r^2-\frac{25}{2} \).

From (2), \( d=5r^4-\frac{25}{2} \).

Equating these:

\( 10r^2-\frac{25}{2}=5r^4-\frac{25}{2} \).

\( 10r^2=5r^4 \).

Since \(r>0\), \( r^2=2 \), so \( r=\sqrt2 \).

Then

\( d=10(2)-\frac{25}{2}=\frac{15}{2} \).

Answer: \( r=\sqrt2,\ d=\frac{15}{2} \).

(b) The ninth term of the geometric progression is

\( 20r^8=20(\sqrt2)^8=20(16)=320 \).

The ninth term of the arithmetic progression is

\( 20+8d=20+8\left(\frac{15}{2}\right)=80 \).

Since \(320=4(80)\), the ninth term of the geometric progression is \(4\) times the ninth term of the arithmetic progression.

(a) Since \(AB=5\) and \(BC=p\),

\( AC=5+p \).

When \(p=4\), \( AC=9 \).

In right-angled triangle \(ACD\), \( \angle A=\frac{\pi}{6} \), so

\( AD=AC\cos\frac{\pi}{6}=9\cdot\frac{\sqrt3}{2}=\frac{9\sqrt3}{2} \),

and

\( CD=AC\sin\frac{\pi}{6}=9\cdot\frac12=\frac92 \).

Also \( AE=5 \), so

\( ED=AD-AE=\frac{9\sqrt3}{2}-5 \).

The arc length \(BE\) is

\( 5\cdot\frac{\pi}{6}=\frac{5\pi}{6} \).

The perimeter of the shaded region is

\( ED+DC+CB+\text{arc }BE \).

\( =\left(\frac{9\sqrt3}{2}-5\right)+\frac92+4+\frac{5\pi}{6} \).

Answer: \( \frac{9\sqrt3}{2}+\frac72+\frac{5\pi}{6} \).

(b) The shaded area is the area of triangle \(ACD\) minus the area of sector \(ABE\).

\( AC=5+p \).

The area of triangle \(ACD\) is

\( \frac12(AC\cos\frac{\pi}{6})(AC\sin\frac{\pi}{6}) \).

\( =\frac12(5+p)^2\cdot\frac{\sqrt3}{2}\cdot\frac12 \).

\( =\frac{\sqrt3}{8}(5+p)^2 \).

The area of sector \(ABE\) is

\( \frac12(5^2)\left(\frac{\pi}{6}\right)=\frac{25\pi}{12} \).

So

\( \frac{\sqrt3}{8}(5+p)^2-\frac{25\pi}{12}=8\sqrt3-\frac{25\pi}{12} \).

\( \frac{\sqrt3}{8}(5+p)^2=8\sqrt3 \).

\( (5+p)^2=64 \).

Since \(p>0\), \(5+p=8\).

Answer: \( p=3 \).

(a) Since \(x>0\),

\( f(x)=3x-6>-6 \).

Answer: the range is \( f(x)>-6 \).

(b) Let \(y=g(x)\).

\( y=\frac4{(ax-3)^2} \).

Since \(x>\frac3a\), we have \( ax-3>0 \).

Therefore

\( ax-3=\frac2{\sqrt y} \).

\( x=\frac{3+\frac2{\sqrt y}}{a} \).

Hence

Answer: \( g^{-1}(x)=\frac{3+\frac2{\sqrt x}}{a} \), for \(x>0\).

(c) If \(a=2\), then

\( g^{-1}(x)=\frac{3+\frac2{\sqrt x}}{2} \).

Solve \(g^{-1}(x)=4\):

\( \frac{3+\frac2{\sqrt x}}{2}=4 \).

\( 3+\frac2{\sqrt x}=8 \).

\( \frac2{\sqrt x}=5 \).

\( \sqrt x=\frac25 \).

Answer: \( x=\frac4{25} \).

(d) \(fg(8)=6\) means \( f(g(8))=6 \).

Since \(f(x)=3x-6\),

\( 3g(8)-6=6 \).

\( g(8)=4 \).

Now

\( g(8)=\frac4{(8a-3)^2} \).

So

\( \frac4{(8a-3)^2}=4 \).

\( (8a-3)^2=1 \).

\( 8a-3=1 \) or \( 8a-3=-1 \).

So \( a=\frac12 \) or \( a=\frac14 \).

But \(g(8)\) is defined only if \(8>\frac3a\), which means \(a>\frac38\).

Therefore \( a=\frac14 \) is rejected.

Answer: \( a=\frac12 \).