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