(a) Differentiate \(y = k \sqrt{4x + 1} - x + 5\) with respect to \(x\):

\(\frac{dy}{dx} = k \cdot \frac{1}{2}(4x + 1)^{-\frac{1}{2}} \cdot 4 - 1\)

\(\frac{dy}{dx} = \frac{2k}{\sqrt{4x+1}} - 1\)

(b) Set \(\frac{dy}{dx} = 0\) for stationary points:

\(\frac{2k}{\sqrt{4x+1}} - 1 = 0\)

\(\frac{2k}{\sqrt{4x+1}} = 1\)

\(\sqrt{4x+1} = 2k\)

\(4x + 1 = 4k^2\)

\(x = \frac{4k^2 - 1}{4}\)

(c) Given \(k = 10.5\), substitute into \(\frac{dy}{dx} = 2\):

\(\frac{2 \times 10.5}{\sqrt{4x+1}} - 1 = 2\)

\(\frac{21}{\sqrt{4x+1}} = 3\)

\(\sqrt{4x+1} = 7\)

\(4x + 1 = 49\)

\(x = 12\)

Find \(y\) at \(x = 12\):

\(y = 10.5 \sqrt{4 \times 12 + 1} - 12 + 5\)

\(y = 66.5\)

The slope of the normal is \(-\frac{1}{2}\) (perpendicular to \(2\)).

Equation of the normal: \(y - 66.5 = -\frac{1}{2}(x - 12)\)

The equation of the curve is \(y = 3x - \frac{4}{x}\).

The derivative is \(\frac{dy}{dx} = 3 + \frac{4}{x^2}\).

The gradient of \(AB\) is \(m_{AB} = 4\).

Set \(\frac{dy}{dx} = 4\) to find \(C\) and \(D\):

\(3 + \frac{4}{x^2} = 4\)

\(\frac{4}{x^2} = 1\)

\(x^2 = 4\)

\(x = \pm 2\)

For \(x = 2\), \(y = 3(2) - \frac{4}{2} = 6 - 2 = 4\), so \(C(2, 4)\).

For \(x = -2\), \(y = 3(-2) - \frac{4}{-2} = -6 + 2 = -4\), so \(D(-2, -4)\).

The midpoint \(M\) of \(CD\) is \((0, 0)\).

The gradient of \(CD\) is \(m_{CD} = 2\).

The perpendicular gradient is \(-\frac{1}{2}\).

The equation of the perpendicular bisector is \(y = -\frac{1}{2}x\).

First, find the derivative \(\frac{dy}{dx}\) of the curve \(y = \frac{4}{(3x + 1)^2}\).

Using the chain rule, let \(u = 3x + 1\), then \(y = 4u^{-2}\).

The derivative \(\frac{dy}{du} = -8u^{-3}\).

Since \(u = 3x + 1\), \(\frac{du}{dx} = 3\).

Thus, \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -8u^{-3} \cdot 3 = -24(3x + 1)^{-3}\).

When \(x = -1\), \(y = \frac{4}{(3(-1) + 1)^2} = \frac{4}{1} = 1\).

Also, \(\frac{dy}{dx} = -24(3(-1) + 1)^{-3} = 3\).

The equation of the tangent line at \(x = -1\) is given by the point-slope form: \(y - y_1 = m(x - x_1)\).

Substitute \(y_1 = 1\), \(m = 3\), and \(x_1 = -1\):

\(y - 1 = 3(x + 1)\).

Simplifying gives \(y = 3x + 4\).

(i) To find the equation of the tangent, we first differentiate the curve \(y = \sqrt{x^4 + 4x + 4}\).

The derivative is \(\frac{dy}{dx} = \left[ \frac{1}{2} (x^4 + 4x + 4)^{-\frac{1}{2}} \right] \times [4x^3 + 4]\).

At \(x = 0\), \(\frac{dy}{dx} = \frac{1}{2} \times 1 \times 4 = 2\).

The equation of the tangent is \(y - 2 = 2(x - 0)\), which simplifies to \(y - 2 = x\).

(ii) For the intersection, set \(x + 2 = \sqrt{x^4 + 4x + 4}\).

Squaring both sides gives \((x + 2)^2 = x^4 + 4x + 4\).

This simplifies to \(x^2 - x^4 = 0\), or \(x^2(1 - x^2) = 0\).

Thus, \(x = 0, \pm 1\).

(i) Differentiate the curve \(y = (6x + 2)^{\frac{1}{3}}\) to find the gradient at \(A\):

\(\frac{dy}{dx} = 6 \times \frac{1}{3}(6x + 2)^{-\frac{2}{3}} = 2(6x + 2)^{-\frac{2}{3}}\)

At \(x = 1\), \(\frac{dy}{dx} = 2(8)^{-\frac{2}{3}} = \frac{1}{2}\).

The equation of the tangent is \(y - 2 = \frac{1}{2}(x - 1)\).

The gradient of the normal is \(-2\) (negative reciprocal of \(\frac{1}{2}\)).

The equation of the normal is \(y - 2 = -2(x - 1)\).

(ii) The coordinates of \(B\) are \((0, \frac{1}{2})\) and \(C\) are \((2, 0)\).

Distance \(BC = \sqrt{(2 - 0)^2 + (0 - \frac{1}{2})^2} = \frac{5}{2}\).

(iii) The equation of \(BC\) is \(y - \frac{1}{2} = -\frac{3}{4}(x - 0)\) or \(y = -\frac{3}{4}x + \frac{1}{2}\).

Intersection \(E\) is found by solving \(-\frac{3}{4}x + \frac{1}{2} = 2x\).

\(x = \frac{6}{11}, \; y = \frac{12}{11}\).

The mid-point of \(OA\) is \(\left( \frac{1}{2}, 1 \right)\), so \(E\) is not the mid-point.

(i) To find the equation of the tangent at \(A\), we first find the derivative of the curve:

\(\frac{dy}{dx} = \frac{-10}{(2x+1)^2} \times 2\).

At \(A\), \(y = 0\), so \(x = 2\).

The slope \(m\) at \(x = 2\) is \(\frac{-4}{5}\).

The equation of the tangent is \(y = \frac{-4}{5}(x - 2)\).

Rearranging gives \(5y + 4x = 8\).

(ii) The point \(C\) is \((0, 1.6)\).

The distance \(AC\) is \(\sqrt{(1.6)^2 + 2^2} = 2.56\).

(i) Differentiate the curve: \(\frac{dy}{dx} = 4 - 2x\).

At \(x = 3\), the gradient \(m = -2\).

The gradient of the normal is \(\frac{1}{2}\) (since \(m_1 m_2 = -1\)).

Equation of the normal: \(y - 6 = \frac{1}{2}(x - 3)\).

Simplify to get \(2y = x + 9\).

(ii) The normal meets the axes at \((0, \frac{9}{2})\) and \((-9, 0)\).

Mid-point formula: \(\left(\frac{-9 + 0}{2}, \frac{0 + \frac{9}{2}}{2}\right) = \left(-\frac{9}{2}, \frac{9}{4}\right)\).

(iii) Set \(2y = x + 9\) equal to \(y = 4x - x^2 + 3\).

Substitute to get \(2(4x - x^2 + 3) = x + 9\).

Simplify to \(2x^2 - 7x + 3 = 0\).

Solving gives \(x = \frac{1}{2}\).

Substitute back to find \(y = \frac{43}{4}\).

Coordinates are \(\left(\frac{1}{2}, \frac{43}{4}\right)\).

(i) To find the equation of the normal at \(A\), first find the derivative of the curve \(y = 2 - \frac{18}{2x+3}\).

\(\frac{dy}{dx} = 18(2x+3)^{-2} \times 2\).

At \(x = 3\), the slope \(m\) of the tangent is \(\frac{4}{9}\).

The slope of the normal is \(-\frac{9}{4}\) (since \(m_1 m_2 = -1\)).

The equation of the normal is \(y = -\frac{9}{4}(x-3)\).

Rearranging gives \(4y + 9x = 27\).

(ii) The normal meets the y-axis at \((0, 6\frac{3}{4})\).

The curve meets the y-axis at \((0, -4)\).

Thus, the length of \(BC\) is \(10\frac{3}{4}\).

(i) Differentiate \(y = 5 - \frac{8}{x}\) to find the gradient of the tangent: \(\frac{dy}{dx} = \frac{8}{x^2}\).

At \(x = 2\), \(\frac{dy}{dx} = \frac{8}{4} = 2\). The gradient of the normal is \(-\frac{1}{2}\) (since \(m_1 m_2 = -1\)).

The equation of the normal is \(y - 1 = -\frac{1}{2}(x - 2)\), which simplifies to \(2y + x = 4\).

(ii) Solve the simultaneous equations \(2y + x = 4\) and \(y = 5 - \frac{8}{x}\).

Substitute \(y = \frac{4 - x}{2}\) into \(y = 5 - \frac{8}{x}\) to get \(\frac{4 - x}{2} = 5 - \frac{8}{x}\).

Multiply through by \(2x\) to clear fractions: \((4 - x)x = 10x - 16\).

Simplify to \(x^2 + 6x - 16 = 0\). Solving gives \(x = -8\) or \(x = 2\). Since \(x = 2\) is point \(P\), \(Q\) is \((-8, 6)\).

(iii) Use the distance formula to find \(PQ\): \(\sqrt{(2 - (-8))^2 + (1 - 6)^2} = \sqrt{10^2 + 5^2} = \sqrt{125}\).

The length of \(PQ\) is approximately \(11.2\).

(i) Differentiate \(y = 2x + \frac{8}{x^2}\) with respect to \(x\):

\(\frac{dy}{dx} = 2 - \frac{16}{x^3}\).

Differentiate again to find \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = \frac{48}{x^4}\).

(ii) Set \(\frac{dy}{dx} = 0\) to find the stationary point:

\(2 - \frac{16}{x^3} = 0 \Rightarrow x = 2\).

Substitute \(x = 2\) into the original equation:

\(y = 2(2) + \frac{8}{2^2} = 4 + 2 = 6\).

Stationary point is \((2, 6)\).

Since \(\frac{d^2y}{dx^2}\) is positive, the stationary point is a minimum.

(iii) At \(x = -2\), \(\frac{dy}{dx} = 4\), so the gradient of the tangent is 4.

The gradient of the normal is \(-\frac{1}{4}\).

Equation of the normal: \(y + 2 = -\frac{1}{4}(x + 2)\).

Set \(y = 0\) to find the x-intercept:

\(0 + 2 = -\frac{1}{4}(x + 2) \Rightarrow x = -10\).

The normal intersects the x-axis at \((-10, 0)\).

First, find the derivative of the curve \(y = \frac{4}{\sqrt{x}}\). Rewrite it as \(y = 4x^{-0.5}\).

The derivative is \(\frac{dy}{dx} = -2x^{-1.5}\).

At the point \((4, 2)\), the gradient of the tangent is \(\frac{dy}{dx} = -\frac{1}{4}\).

The gradient of the normal is the negative reciprocal, so \(m = 4\).

The equation of the normal is \(y - 2 = 4(x - 4)\).

Simplify to get \(y = 4x - 14\).

Find where the normal meets the \(x\)-axis by setting \(y = 0\):

\(0 = 4x - 14\)

\(x = 3.5\)

So, \(P(3.5, 0)\).

Find where the normal meets the \(y\)-axis by setting \(x = 0\):

\(y = -14\)

So, \(Q(0, -14)\).

The length of \(PQ\) is given by:

\(\sqrt{(3.5 - 0)^2 + (0 + 14)^2} = \sqrt{3.5^2 + 14^2}\)

\(= \sqrt{12.25 + 196} = \sqrt{208.25}\)

\(\approx 14.4\)

(a) To find the coordinates of \(P\), equate the line and curve equations: \(mx + c = -\frac{m}{x}\). Rearrange to get \(mx^2 + cx + m = 0\). For the line to be tangent, the discriminant must be zero: \(b^2 - 4ac = 0\), giving \(c^2 - 4m^2 = 0\). Solving gives \(c = \pm 2m\). Substitute \(c = 2m\) into the quadratic: \(mx^2 + 2mx + m = 0\), which simplifies to \((x+1)^2 = 0\), so \(x = -1\). Substituting back, \(y = m\), so \(P = (-1, m)\).

(b) The equation of the normal at \(P\) is \(y - m = -\frac{1}{m}(x + 1)\). Simplifying gives \(y = -\frac{1}{m}x + \frac{1}{m} + m\). Equate this to the curve equation \(y = -\frac{m}{x}\) to find \(x\): \(-\frac{1}{m}x + \frac{1}{m} + m = -\frac{m}{x}\). Rearrange to get \(x^2(1 - m^2) - m^2 = 0\). Solving gives \(x = m^2\). Substitute back to find \(y\): \(y = -\frac{1}{m}\), so \(Q = (m^2, -\frac{1}{m})\).

(ii) To find the stationary point, set \(\frac{dy}{dx} = 0\):

\(\frac{2}{\sqrt{x}} - 1 = 0\)

\(\frac{2}{\sqrt{x}} = 1\)

\(\sqrt{x} = 2\)

\(x = 4\)

Substitute \(x = 4\) into the curve equation to find \(y\):

\(y = 6\)

Thus, the coordinates of the stationary point are \((4, 6)\).

(iii) Differentiate \(\frac{dy}{dx} = \frac{2}{\sqrt{x}} - 1\) to find \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = -\frac{3}{x^2}\)

Since \(\frac{d^2y}{dx^2} < 0\), the stationary point is a maximum.

(iv) The gradient of the tangent at \(P\) is \(\frac{dy}{dx} = -\frac{1}{3}\).

The gradient of the normal is \(m = 3\) (since \(m_1 m_2 = -1\)).

The angle \(\theta\) is given by \(\tan \theta = 3\), so \(k = 3\).

(i) To find the equation of the tangent, we first need the derivative \(\frac{dy}{dx}\) of the curve \(y = \frac{4}{3x-4}\).

The derivative is \(\frac{dy}{dx} = -4(3x-4)^{-2} \times 3\).

Substituting \(x = 2\) into the derivative gives the gradient \(m = -3\).

The equation of the tangent at \(P(2, 2)\) is \(y - 2 = -3(x - 2)\).

(ii) The tangent makes an angle \(\theta\) with the \(x\)-axis, where \(\tan \theta = \pm (-3)\).

This gives \(\theta = \pm 108.4^\circ\) (or \(\pm 71.6^\circ\)).

(a) To find the \(x\)-coordinates of \(A\) and \(B\), solve the simultaneous equations:

\(2y = x + 5\)

\(y = x^2 - 4x + 7\)

Substitute \(y = \frac{x + 5}{2}\) into the quadratic equation:

\(\frac{x + 5}{2} = x^2 - 4x + 7\)

Multiply through by 2 to eliminate the fraction:

\(x + 5 = 2x^2 - 8x + 14\)

Rearrange to form a quadratic equation:

\(2x^2 - 9x + 9 = 0\)

Factorize or use the quadratic formula to solve:

\(x = 3 \text{ or } x = \frac{1}{2}\)

(b) To find the equation of the tangent to the curve at \(B\), first find the derivative:

\(\frac{dy}{dx} = 2x - 4\)

At \(x = 3\), the gradient \(m\) is:

\(m = 2(3) - 4 = 2\)

The equation of the tangent at \(x = 3\) is:

\(y - 4 = 2(x - 3)\)

(c) To find the acute angle between the tangent and the line \(2y = x + 5\), first find the gradients:

The gradient of the line \(2y = x + 5\) is \(\frac{1}{2}\).

The angle \(\theta_1\) with the \(x\)-axis for the tangent is:

\(\arctan(2) \approx 63.4^\circ\)

The angle \(\theta_2\) with the \(x\)-axis for the line is:

\(\arctan\left(\frac{1}{2}\right) \approx 26.6^\circ\)

The acute angle between them is:

\(63.4^\circ - 26.6^\circ = 36.8^\circ\)

Rounded to 1 decimal place, the angle is \(37^\circ\).

First, find the derivative \(\frac{dy}{dx}\) of the curve \(y = 2\sqrt{3x+4} - x\).

Using the chain rule, \(\frac{dy}{dx} = 3(3x+4)^{-0.5} - 1\).

Substitute \(x = 4\) into \(\frac{dy}{dx}\) to find the gradient of the tangent at the point (4, 4):

\(\frac{dy}{dx} = \frac{1}{4}\).

The gradient of the normal is the negative reciprocal of the gradient of the tangent:

\(m = -\frac{1}{\frac{1}{4}} = 4\).

Using the point-slope form of a line, \(y - y_1 = m(x - x_1)\), with \(m = 4\) and point (4, 4):

\(y - 4 = 4(x - 4)\).

Simplify to find the equation of the normal:

\(y = 4x - 16 + 4\).

Thus, \(y = 4x - 12\).

(i) To find the points of intersection, set \(4x^{\frac{1}{2}} = x + 3\). Rearrange to get \((x^{\frac{1}{2}})^2 - 4x^{\frac{1}{2}} + 3 = 0\) or \(x^2 - 10x + 9 = 0\). Solving gives \(x = 1\) or \(x = 9\). For \(x = 1\), \(y = 4\); for \(x = 9\), \(y = 12\). The points are \(A(1, 4)\) and \(B(9, 12)\). The length \(AB = \sqrt{(9-1)^2 + (12-4)^2} = \sqrt{128} = 8\sqrt{2} \approx 11.3\).

(ii) The gradient of \(AB\) is \(1\). The derivative of the curve is \(\frac{dy}{dx} = 2x^{-\frac{1}{2}}\). Set \(2x^{-\frac{1}{2}} = 1\) to find \(x = 4\). Then \(y = 4 \times 4^{\frac{1}{2}} = 8\). So, \(T(4, 8)\).

(iii) The equation of the normal at \(T\) is \(y - 8 = -1(x - 4)\) or \(y = -x + 12\). Solve \(-x + 12 = x + 3\) to find \(x = \frac{9}{2}\). Substitute into \(y = x + 3\) to get \(y = \frac{15}{2}\). The intersection point is \(\left( \frac{9}{2}, \frac{15}{2} \right)\).

(i) To find the equation of line \(AB\), calculate the gradient: \(\text{Gradient of } AB = \frac{2 - 0}{5 - 1} = \frac{1}{2}\).

The equation of the line is \(y = \frac{1}{2}x - \frac{1}{2}\).

(ii) The derivative of the curve \(y = (x - 1)^{\frac{1}{2}}\) is \(\frac{dy}{dx} = \frac{1}{2}(x - 1)^{-\frac{1}{2}}\).

Set \(\frac{dy}{dx} = \frac{1}{2}\) to find the tangent parallel to \(AB\):

\(\frac{1}{2}(x - 1)^{-\frac{1}{2}} = \frac{1}{2}\) gives \(x = 2\), \(y = 1\).

The equation of the tangent is \(y - 1 = \frac{1}{2}(x - 2)\), simplifying to \(y = \frac{1}{2}x\).

(iii) The perpendicular distance \(d\) between the line \(AB\) and the tangent is given by:

\(d = \sin(26.5^\circ) = 0.45\).

To find the equation of the tangent, we first need to find the derivative \(\frac{dy}{dx}\) of the curve.

The given equation is \(y = 2x^{\frac{3}{2}} - 3x - 4x^{\frac{1}{2}} + 4\).

Differentiate term by term:

\(\frac{dy}{dx} = \frac{d}{dx}(2x^{\frac{3}{2}}) - \frac{d}{dx}(3x) - \frac{d}{dx}(4x^{\frac{1}{2}}) + \frac{d}{dx}(4)\).

\(\frac{dy}{dx} = 3x^{\frac{1}{2}} - 3 - 2x^{-\frac{1}{2}}\).

Now, evaluate \(\frac{dy}{dx}\) at \(x = 4\):

\(\frac{dy}{dx} \bigg|_{x=4} = 3(4)^{\frac{1}{2}} - 3 - 2(4)^{-\frac{1}{2}}\).

\(\frac{dy}{dx} \bigg|_{x=4} = 6 - 3 - 1 = 2\).

The gradient of the tangent at \(x = 4\) is 2.

The point of tangency is (4, 0). Using the point-slope form of the equation of a line:

\(y - y_1 = m(x - x_1)\), where \(m = 2\), \(x_1 = 4\), and \(y_1 = 0\).

\(y - 0 = 2(x - 4)\).

Thus, the equation of the tangent is \(y = 2(x - 4)\).

The gradient of the line 3y + x = 25 is \\(-\frac{1}{3}\\). Since this line is normal to the curve, the gradient of the tangent to the curve at the point of intersection is 3.

Differentiate the curve equation y = x^2 - 5x + k to find the gradient of the tangent: \\(\frac{dy}{dx} = 2x - 5\\).

Set the gradient equal to 3: \\(2x - 5 = 3\\).

Solve for \\(x\\):

\\(2x - 5 = 3\\)

\\(2x = 8\\)

\\(x = 4\\)

Substitute \\(x = 4\\) into the line equation to find \\(y\\):

\\(3y + 4 = 25\\)

\\(3y = 21\\)

\\(y = 7\\)

Substitute \\(x = 4\\) and \\(y = 7\\) into the curve equation to find \\(k\\):

\\(7 = 4^2 - 5(4) + k\\)

\\(7 = 16 - 20 + k\\)

\\(7 = -4 + k\\)

\\(k = 11\\)

(i) To find the equation of the tangent at A (2, 2), first find the derivative of the curve \(y = x^2 - 2x + 2\), which is \(\frac{dy}{dx} = 2x - 2\).

At \(x = 2\), the gradient \(m\) is \(2(2) - 2 = 2\).

The equation of the tangent is \(y - 2 = 2(x - 2)\), which simplifies to \(y = 2x - 2\).

(ii) The equation of the normal at A is \(y - 2 = -\frac{1}{2}(x - 2)\), which simplifies to \(y = -\frac{1}{2}x + 3\).

Substitute \(y = -\frac{1}{2}x + 3\) into the curve equation \(x^2 - 2x + 2 = y\) to find \(B\):

\(x^2 - 2x + 2 = -\frac{1}{2}x + 3\)

\(x^2 - 2x + \frac{1}{2}x - 1 = 0\)

\(2x^2 - 3x - 2 = 0\)

Solving this quadratic gives \(x = -\frac{1}{2}\), and substituting back gives \(y = \frac{3}{4}\).

Thus, \(B = \left(-\frac{1}{2}, \frac{3}{4}\right)\).

(iii) At \(x = -\frac{1}{2}\), the gradient is \(2(-\frac{1}{2}) - 2 = -3\).

The equation of the tangent at B is \(y - \frac{3}{4} = -3(x + \frac{1}{2})\), which simplifies to \(y = -3x + \frac{7}{4}\).

Equate the tangents \(y = 2x - 2\) and \(y = -3x + \frac{7}{4}\):

\(2x - 2 = -3x + \frac{7}{4}\)

\(2x + 3x = \frac{7}{4} + 2\)

\(5x = \frac{15}{4}\)

\(x = \frac{3}{4}\)

Substitute \(x = \frac{3}{4}\) into \(y = 2x - 2\) to find \(y = -\frac{1}{2}\).

Thus, \(C = \left(\frac{3}{4}, -\frac{1}{2}\right)\).

(i) Differentiate the curve: \(\frac{dy}{dx} = -(x-1)^{-2} + 9(x-5)^{-2}\).

At \(x = 3\), \(\frac{dy}{dx} = \frac{1}{4} + \frac{9}{4} = 2\).

The slope of the normal is \(m = -\frac{1}{2}\).

Equation of the normal: \(y - 5 = -\frac{1}{2}(x - 3)\).

Setting \(y = 0\) to find \(x\)-intercept: \(0 - 5 = -\frac{1}{2}(x - 3)\).

Solving gives \(x = 13\).

(ii) Set \(\frac{dy}{dx} = 0\): \((x-5)^2 = 9(x-1)^2\).

Solving gives \(x - 5 = \pm 3(x-1)\) or \((x^2 - x - 2) = 0\).

Solutions are \(x = -1\) and \(x = 2\).

Second derivative: \(\frac{d^2y}{dx^2} = 2(x-1)^{-3} - 18(x-5)^{-3}\).

At \(x = -1\), \(\frac{d^2y}{dx^2} = -\frac{1}{6} < 0\), so maximum.

At \(x = 2\), \(\frac{d^2y}{dx^2} = \frac{8}{3} > 0\), so minimum.