Integrate \( \frac{dy}{dx} \):

\( y=\int \left(2x-6x^{\frac12}\right)\,dx \).

\( y=x^2-4x^{\frac32}+C \).

Use the point \( (4,-9) \):

\( -9=4^2-4(4^{\frac32})+C \).

\( -9=16-4(8)+C=-16+C \).

So \( C=7 \).

Answer: \( y=x^2-4x^{\frac32}+7 \).

(a) Since \( f(4-x)=f(-(x-4)) \), the graph may be transformed by:

reflection in the \(y\)-axis, followed by a translation \(4\) units in the positive \(x\)-direction.

(b) A stretch with scale factor \( \frac12 \) in the \(x\)-direction changes \(x\) to \(2x\):

\( y=(2x)^3-3(2x)-4=8x^3-6x-4 \).

Then translate down by \(3\):

\( y=8x^3-6x-4-3 \).

Answer: \( y=8x^3-6x-7 \).

At points of intersection,

\( kx^2-5x-6=3x-7k \).

So

\( kx^2-8x+7k-6=0 \).

For the line to intersect the curve, this equation must have real roots, so the discriminant is non-negative:

\( (-8)^2-4(k)(7k-6)\geq 0 \).

\( 64-28k^2+24k\geq 0 \).

\( 7k^2-6k-16\leq 0 \).

Factorise:

\( (7k+8)(k-2)\leq 0 \).

Answer: \( -\frac87\leq k\leq 2 \).

The coefficient of \(x^2\) in \( (2-qx)^4 \) is

\( \binom42(2)^2(-q)^2=24q^2 \).

The coefficient of \(x^2\) in \( \left(1+\frac8q x\right)^6 \) is

\( \binom62\left(\frac8q\right)^2=\frac{960}{q^2} \).

Since the second expansion is subtracted,

\( 24q^2-\frac{960}{q^2}=324 \).

Multiply by \(q^2\):

\( 24q^4-324q^2-960=0 \).

Divide by \(12\):

\( 2q^4-27q^2-80=0 \).

Let \(u=q^2\). Then

\( 2u^2-27u-80=0 \).

\( (2u+5)(u-16)=0 \).

Since \(u=q^2\), \(u=16\).

Answer: \( q=4 \) or \( q=-4 \).

(a) Start with the left-hand side:

\( \frac{1-\sin\theta}{\cos\theta}+\frac{\cos\theta}{1-\sin\theta} \).

Use a common denominator:

\( \frac{(1-\sin\theta)^2+\cos^2\theta}{\cos\theta(1-\sin\theta)} \).

Expand the numerator:

\( (1-2\sin\theta+\sin^2\theta)+\cos^2\theta \).

Since \( \sin^2\theta+\cos^2\theta=1 \), the numerator becomes

\( 2-2\sin\theta=2(1-\sin\theta) \).

Therefore

\( \frac{2(1-\sin\theta)}{\cos\theta(1-\sin\theta)}=\frac2{\cos\theta} \).

Hence the identity is proved.

(b) Using part (a),

\( \frac2{\cos\theta}=\frac{\tan^3\theta}{\sin\theta} \).

Now

\( \frac{\tan^3\theta}{\sin\theta}=\frac{\sin^2\theta}{\cos^3\theta} \).

So

\( \frac2{\cos\theta}=\frac{\sin^2\theta}{\cos^3\theta} \).

Multiply by \( \cos^3\theta \):

\( 2\cos^2\theta=\sin^2\theta \).

Using \( \sin^2\theta=1-\cos^2\theta \):

\( 2\cos^2\theta=1-\cos^2\theta \).

\( 3\cos^2\theta=1 \), so \( \cos\theta=\pm\frac1{\sqrt3} \).

Answer: \( \theta=54.7^\circ,\ 125.3^\circ,\ 234.7^\circ,\ 305.3^\circ \).

(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: \( a=\frac12 \).

(a) The gradient of the tangent through \( (1,5) \) and \( (4,4) \) is

\( \frac{4-5}{4-1}=-\frac13 \).

Using \( (1,5) \),

\( y-5=-\frac13(x-1) \).

So

\( x+3y=16 \).

The other tangent is

\( x-3y=16 \).

Solve simultaneously:

\( x+3y=16 \),

\( x-3y=16 \).

Adding gives \(2x=32\), so \(x=16\).

Then \(y=0\).

Answer: \( (16,0) \).

(b) The tangent through \(A\) has gradient \(-\frac13\), so the normal through \(A\) has gradient \(3\).

The normal also passes through the centre \( (a,0) \).

Therefore

Answer: \( y=3(x-a) \).

(c) Consider the right-angled triangle formed by the centre of the circle, the point of contact, and the point of intersection of the two tangents.

The tangent has gradient \( -\frac13 \), so the radius to the point of contact is perpendicular to the tangent and has gradient \(3\).

Therefore, for the angle between the radius and the horizontal axis,

\[ \tan\theta=3. \]

The radius of the circle is \( \sqrt{40} \). Hence the other perpendicular side of the right-angled triangle is

\[ 3\sqrt{40}. \]

By Pythagoras, the hypotenuse is

\[ \sqrt{(\sqrt{40})^2+(3\sqrt{40})^2} =\sqrt{40+360} =\sqrt{400} =20. \]

This hypotenuse is the distance from the centre \( (a,0) \) to the point of intersection of the tangents \( (16,0) \). Therefore

\[ 16-a=20. \]

So

\[ a=-4. \]

The centre is \( (-4,0) \), and \(r^2=40\). Therefore the equation of the circle is

\[ (x+4)^2+y^2=40. \]

(c) Method(Using FM formula) The centre is \( (a,0) \) and the radius is \( \sqrt{40} \).

The distance from the centre \( (a,0) \) to the tangent \(x+3y=16\) is equal to the radius:

\( \frac{|a-16|}{\sqrt{1^2+3^2}}=\sqrt{40} \).

Since \(a<0\), \( |a-16|=16-a \).

So

\( \frac{16-a}{\sqrt{10}}=\sqrt{40} \).

\( 16-a=\sqrt{400}=20 \).

Thus \(a=-4\).

The centre is \( (-4,0) \) and the radius squared is \(40\).

Answer: \( (x+4)^2+y^2=40 \).