(a) Calculate \(\cot \frac{1}{2}x - 3x\) at \(x = 0.5\) and \(x = 1\).

For \(x = 0.5\), \(\cot \frac{1}{2} \times 0.5 = \cot 0.25 \approx 3.92\) and \(3 \times 0.5 = 1.5\). So, \(3.92 - 1.5 > 0\).

For \(x = 1\), \(\cot \frac{1}{2} \times 1 = \cot 0.5 \approx 1.83\) and \(3 \times 1 = 3\). So, \(1.83 - 3 < 0\).

Since there is a change of sign, \(\alpha\) lies between 0.5 and 1.

(b) Rearrange \(\cot \frac{x}{2} = 3x\) to \(x = \frac{1}{3} \left( x + 4 \arctan \left( \frac{1}{3x} \right) \right)\).

The iterative formula \(x_{n+1} = \frac{1}{3} \left( x_n + 4 \arctan \left( \frac{1}{3x_n} \right) \right)\) is derived from this rearrangement.

If the sequence converges, it must converge to the root \(\alpha\).

(c) Start with an initial guess, say \(x_0 = 0.5\).

Calculate successive iterations:

\(x_1 = \frac{1}{3} \left( 0.5 + 4 \arctan \left( \frac{1}{3 \times 0.5} \right) \right) \approx 0.7623\)

\(x_2 = \frac{1}{3} \left( 0.7623 + 4 \arctan \left( \frac{1}{3 \times 0.7623} \right) \right) \approx 0.8037\)

\(x_3 = \frac{1}{3} \left( 0.8037 + 4 \arctan \left( \frac{1}{3 \times 0.8037} \right) \right) \approx 0.7921\)

\(x_4 = \frac{1}{3} \left( 0.7921 + 4 \arctan \left( \frac{1}{3 \times 0.7921} \right) \right) \approx 0.7951\)

\(x_5 = \frac{1}{3} \left( 0.7951 + 4 \arctan \left( \frac{1}{3 \times 0.7951} \right) \right) \approx 0.7943\)

\(x_6 = \frac{1}{3} \left( 0.7943 + 4 \arctan \left( \frac{1}{3 \times 0.7943} \right) \right) \approx 0.7945\)

The value converges to \(0.79\) to 2 decimal places.

(i) Consider the function \(f(x) = x - \frac{10}{e^{2x} - 1}\).

Calculate \(f(1) = 1 - \frac{10}{e^2 - 1}\) and \(f(2) = 2 - \frac{10}{e^4 - 1}\).

Since \(f(1) < 0\) and \(f(2) > 0\), by the Intermediate Value Theorem, \(\alpha\) lies between 1 and 2.

(ii) Assume the sequence \(x_n\) converges to \(L\). Then \(L = \frac{1}{2} \ln \left( 1 + \frac{10}{L} \right)\).

Rearranging gives \(L = \frac{10}{e^{2L} - 1}\), so \(L = \alpha\).

(iii) Use the iterative formula:

Start with \(x_1 = 1\).

Calculate \(x_2 = \frac{1}{2} \ln \left( 1 + \frac{10}{1} \right) = 1.1513\).

Calculate \(x_3 = \frac{1}{2} \ln \left( 1 + \frac{10}{1.1513} \right) = 1.1394\).

Calculate \(x_4 = \frac{1}{2} \ln \left( 1 + \frac{10}{1.1394} \right) = 1.1427\).

Calculate \(x_5 = \frac{1}{2} \ln \left( 1 + \frac{10}{1.1427} \right) = 1.1414\).

Calculate \(x_6 = \frac{1}{2} \ln \left( 1 + \frac{10}{1.1414} \right) = 1.1418\).

Since the values are converging, \(\alpha \approx 1.14\) to 2 decimal places.

(i) Start with \(x_1 = 3.5\).

Calculate \(x_2 = \frac{3.5(3.5^3 + 100)}{2(3.5^3 + 25)} = 3.683943\).

Calculate \(x_3 = \frac{3.683943(3.683943^3 + 100)}{2(3.683943^3 + 25)} = 3.684002\).

Calculate \(x_4 = \frac{3.684002(3.684002^3 + 100)}{2(3.684002^3 + 25)} = 3.684000\).

Calculate \(x_5 = \frac{3.684000(3.684000^3 + 100)}{2(3.684000^3 + 25)} = 3.684000\).

The sequence converges to \(\alpha = 3.6840\) correct to 4 decimal places.

(ii) The equation satisfied by \(\alpha\) is \(x = \frac{x(x^3 + 100)}{2(x^3 + 25)}\).

Solving for \(x\), we find \(\alpha = 3\sqrt{50}\).

(i) To find \(\alpha\), substitute \(x = -2\) into the equation \(y = x^4 + 2x^3 + 2x^2 - 4x - 16\):

\((-2)^4 + 2(-2)^3 + 2(-2)^2 - 4(-2) - 16 = 16 - 16 + 8 + 8 - 16 = 0\).

Thus, \(\alpha = -2\).

(ii) To show \(\beta\) satisfies \(x = \sqrt[3]{8 - 2x}\), factor the quartic equation:

\((x + 2)(x^3 + 2x - 8) = 0\).

Rearrange \(x^3 + 2x - 8 = 0\) to \(x = \sqrt[3]{8 - 2x}\).

(iii) Use the iterative formula \(x_{n+1} = \sqrt[3]{8 - 2x_n}\) starting with an initial guess, say \(x_0 = 1.5\):

\(x_1 = \sqrt[3]{8 - 2(1.5)} = 1.709975947\).

\(x_2 = \sqrt[3]{8 - 2(1.709975947)} = 1.671585\).

\(x_3 = \sqrt[3]{8 - 2(1.671585)} = 1.667680\).

\(x_4 = \sqrt[3]{8 - 2(1.667680)} = 1.667056\).

Thus, \(\beta \approx 1.67\) to 2 decimal places.

(i) Using the identity \(\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}\), substitute \(\tan x = t\) into the equation:

\(2 \left(\frac{2t}{1-t^2}\right) + 5t^2 = 0\)

Simplifying gives:

\(4t + 5t^2 - 5t^4 = 0\)

Factoring out \(t\), we get:

\(t(4 + 5t - 5t^3) = 0\)

Thus, \(t = 0\) or \(t = \sqrt[3]{(t + 0.8)}\).

(ii) Check the sign of \(t - \sqrt[3]{(t + 0.8)}\) at \(t = 1.2\) and \(t = 1.3\):

At \(t = 1.2\), \(1.2 - \sqrt[3]{(1.2 + 0.8)} = -0.06\)

At \(t = 1.3\), \(1.3 - \sqrt[3]{(1.3 + 0.8)} = 0.02\)

There is a change of sign, confirming a root between 1.2 and 1.3.

(iii) Using the iterative formula \(t_{n+1} = \sqrt[3]{(t_n + 0.8)}\):

Start with \(t_0 = 1.2\):

\(t_1 = \sqrt[3]{(1.2 + 0.8)} = 1.2765\)

\(t_2 = \sqrt[3]{(1.2765 + 0.8)} = 1.2760\)

\(t_3 = \sqrt[3]{(1.2760 + 0.8)} = 1.2760\)

The value of \(t\) correct to 3 decimal places is 1.276.

(iv) Solve \(2 \tan 2x + 5 \tan^2 x = 0\) using \(t = 1.276\):

\(\tan x = 1.276\), so \(x = \arctan(1.276)\)

\(x \approx 0.906\)

Considering the periodicity of \(\tan x\), the solutions are:

\(x = -2.24, 0.906, \pi\)

(i) Evaluate the function \(f(x) = x^3 - 8x - 13\) at \(x = 3\) and \(x = 4\):

\(f(3) = 3^3 - 8 \times 3 - 13 = 27 - 24 - 13 = -10\)

\(f(4) = 4^3 - 8 \times 4 - 13 = 64 - 32 - 13 = 19\)

Since \(f(3) < 0\) and \(f(4) > 0\), the root lies between 3 and 4.

(ii) Use the iterative formula \(x_{n+1} = (8x_n + 13)^{\frac{1}{3}}\):

Start with an initial guess, \(x_0 = 3.5\).

\(x_1 = (8 \times 3.5 + 13)^{\frac{1}{3}} = (41)^{\frac{1}{3}} \approx 3.4482\)

\(x_2 = (8 \times 3.4482 + 13)^{\frac{1}{3}} = (40.5856)^{\frac{1}{3}} \approx 3.4325\)

\(x_3 = (8 \times 3.4325 + 13)^{\frac{1}{3}} = (40.46)^{\frac{1}{3}} \approx 3.4301\)

\(x_4 = (8 \times 3.4301 + 13)^{\frac{1}{3}} = (40.441)^{\frac{1}{3}} \approx 3.4297\)

The root is approximately 3.43 to 2 decimal places.

(i) Start with \(x_1 = 3\).

Calculate \(x_2 = \frac{3 \times 3}{4} + \frac{15}{3^3} = 2.75\).

Calculate \(x_3 = \frac{3 \times 2.75}{4} + \frac{15}{2.75^3} = 2.7816\).

Calculate \(x_4 = \frac{3 \times 2.7816}{4} + \frac{15}{2.7816^3} = 2.7794\).

Calculate \(x_5 = \frac{3 \times 2.7794}{4} + \frac{15}{2.7794^3} = 2.7800\).

Calculate \(x_6 = \frac{3 \times 2.7800}{4} + \frac{15}{2.7800^3} = 2.7798\).

Since the values are converging, \(\alpha \approx 2.78\) to 2 decimal places.

(ii) The equation satisfied by \(\alpha\) is \(x = \frac{3}{4}x + \frac{15}{x^3}\).

Rearrange to find \(x^4 = 60\), so \(\alpha = \sqrt[4]{60}\).

(i) Calculate \(f(x) = x^3 - 2x - 2\) at \(x = 1\) and \(x = 2\):

\(f(1) = 1^3 - 2 \times 1 - 2 = -3\)

\(f(2) = 2^3 - 2 \times 2 - 2 = 2\)

Since \(f(1) < 0\) and \(f(2) > 0\), the root lies between \(x = 1\) and \(x = 2\).

(ii) The iterative formula is \(x_{n+1} = \frac{2x_n^3 + 2}{3x_n^2 - 2}\). Rearrange to show it satisfies \(x^3 - 2x - 2 = 0\):

Multiply both sides by \(3x_n^2 - 2\):

\(x_{n+1}(3x_n^2 - 2) = 2x_n^3 + 2\)

Rearrange to \(x_{n+1} = x_n\) to show convergence to the root.

(iii) Use the iterative formula:

Start with \(x_1 = 1.5\):

\(x_2 = \frac{2(1.5)^3 + 2}{3(1.5)^2 - 2} = 1.8421\)

\(x_3 = \frac{2(1.8421)^3 + 2}{3(1.8421)^2 - 2} = 1.7549\)

\(x_4 = \frac{2(1.7549)^3 + 2}{3(1.7549)^2 - 2} = 1.7693\)

\(x_5 = \frac{2(1.7693)^3 + 2}{3(1.7693)^2 - 2} = 1.7669\)

The root correct to 2 decimal places is 1.77.

(i) To show that \(\alpha\) lies between 1 and 2, evaluate the function \(f(x) = x^3 - x - 3\) at \(x = 1\) and \(x = 2\):

\(f(1) = 1^3 - 1 - 3 = -3\)

\(f(2) = 2^3 - 2 - 3 = 3\)

Since \(f(1) < 0\) and \(f(2) > 0\), by the Intermediate Value Theorem, there is a root \(\alpha\) between 1 and 2.

(ii) Apply the iterative formulae:

Using formula (A):

\(x_1 = 1.5\)

\(x_2 = (1.5)^3 - 3 = 0.375\)

\(x_3 = (0.375)^3 - 3 = -2.9473\)

The sequence diverges.

Using formula (B):

\(x_1 = 1.5\)

\(x_2 = (1.5 + 3)^{\frac{1}{3}} = 1.6509\)

\(x_3 = (1.6509 + 3)^{\frac{1}{3}} = 1.6684\)

\(x_4 = (1.6684 + 3)^{\frac{1}{3}} = 1.6715\)

\(x_5 = (1.6715 + 3)^{\frac{1}{3}} = 1.6719\)

The sequence converges to \(\alpha = 1.67\) correct to 2 decimal places.

(i) Evaluate the cubic at \(x = -1\) and \(x = 0\):

\(f(-1) = (-1)^3 + (-1) + 1 = -1\)

\(f(0) = 0^3 + 0 + 1 = 1\)

Since \(f(-1) < 0\) and \(f(0) > 0\), there is a root between \(-1\) and \(0\).

(ii) Given \(x_{n+1} = \frac{2x_n^3 - 1}{3x_n^2 + 1}\), rearrange to:

\(x(3x^2 + 1) = 2x^3 - 1\)

\(3x^3 + x = 2x^3 - 1\)

\(x^3 + x + 1 = 0\)

Thus, if the sequence converges, it converges to the root of \(x^3 + x + 1 = 0\).

(iii) Using the iterative formula with \(x_1 = -0.5\):

\(x_2 = \frac{2(-0.5)^3 - 1}{3(-0.5)^2 + 1} = -0.6667\)

\(x_3 = \frac{2(-0.6667)^3 - 1}{3(-0.6667)^2 + 1} = -0.6825\)

\(x_4 = \frac{2(-0.6825)^3 - 1}{3(-0.6825)^2 + 1} = -0.6849\)

\(x_5 = \frac{2(-0.6849)^3 - 1}{3(-0.6849)^2 + 1} = -0.6855\)

The root correct to 2 decimal places is \(-0.68\).

(i) Start with \(x_1 = 1\).

Calculate \(x_2 = \frac{2}{3} \left( 1 + \frac{1}{1^2} \right) = \frac{2}{3} \times 2 = \frac{4}{3} \approx 1.3333\).

Calculate \(x_3 = \frac{2}{3} \left( 1.3333 + \frac{1}{(1.3333)^2} \right) \approx 1.2639\).

Calculate \(x_4 = \frac{2}{3} \left( 1.2639 + \frac{1}{(1.2639)^2} \right) \approx 1.2600\).

Calculate \(x_5 = \frac{2}{3} \left( 1.2600 + \frac{1}{(1.2600)^2} \right) \approx 1.2599\).

Since the values are converging, \(\alpha \approx 1.26\) to 2 decimal places.

(ii) The equation satisfied by \(\alpha\) is \(\alpha = \frac{2}{3} \left( \alpha + \frac{1}{\alpha^2} \right)\).

Solving \(\alpha = \frac{2}{3} \left( \alpha + \frac{1}{\alpha^2} \right)\) gives \(\alpha^3 = 2\).

Thus, \(\alpha = \sqrt[3]{2}\) or \(2^{1/3}\).

(a) To show that \(p\) satisfies \(\tan p = \frac{1}{1+p}\), start by equating the functions at \(x = p\):

\(\cos p = \frac{k}{1+p}\).

Differentiate both functions:

\(\frac{d}{dx}(\cos x) = -\sin x\)

\(\frac{d}{dx}\left(\frac{k}{1+x}\right) = -\frac{k}{(1+x)^2}\).

Equate the derivatives at \(x = p\):

\(-\sin p = -\frac{k}{(1+p)^2}\).

Substitute \(k = (1+p)\cos p\) into the derivative equation:

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

Simplify to obtain:

\(\tan p = \frac{1}{1+p}\).

(b) Use the iterative formula:

\(p_{n+1} = \arctan\left(\frac{1}{1+p_n}\right)\).

Start with an initial guess, say \(p_0 = 0.5\), and iterate:

\(p_1 = \arctan\left(\frac{1}{1+0.5}\right) = 0.54042\)

\(p_2 = \arctan\left(\frac{1}{1+0.54042}\right) = 0.56609\)

\(p_3 = \arctan\left(\frac{1}{1+0.56609}\right) = 0.56777\)

\(p_4 = \arctan\left(\frac{1}{1+0.56777}\right) = 0.56799\)

\(p_5 = \arctan\left(\frac{1}{1+0.56799}\right) = 0.56800\)

The value of \(p\) correct to 3 decimal places is \(0.568\).

(c) Substitute \(p = 0.568\) back into \(\cos p = \frac{k}{1+p}\):

\(\cos 0.568 = \frac{k}{1+0.568}\).

Calculate \(\cos 0.568 \approx 0.842\).

\(0.842 = \frac{k}{1.568}\).

\(k = 0.842 \times 1.568 \approx 1.32\).

(i) To find the value of \(b\), we need to find the integer root of the polynomial \(x^4 - 2x^3 - 7x - 6 = 0\). By inspection or using the Rational Root Theorem, we find that \(b = 3\) is a root.

(ii) To show that \(a\) satisfies the equation \(a = -\frac{1}{3}(2 + a^2 + a^3)\), perform polynomial division of \(x^4 - 2x^3 - 7x - 6\) by \(x - 3\). The quotient is \(x^3 + x^2 + 3x + 2\). Setting this equal to zero gives \(x^3 + x^2 + 3x + 2 = 0\). Rearranging gives \(a = -\frac{1}{3}(2 + a^2 + a^3)\).

(iii) Using the iterative formula \(a_{n+1} = -\frac{1}{3}(2 + a_n^2 + a_n^3)\), start with an initial guess, say \(a_0 = -1\). Calculate successive iterations:

\(a_1 = -\frac{1}{3}(2 + (-1)^2 + (-1)^3) = -0.66667\)

\(a_2 = -\frac{1}{3}(2 + (-0.66667)^2 + (-0.66667)^3) = -0.71605\)

\(a_3 = -\frac{1}{3}(2 + (-0.71605)^2 + (-0.71605)^3) = -0.71484\)

\(a_4 = -\frac{1}{3}(2 + (-0.71484)^2 + (-0.71484)^3) = -0.71506\)

\(a_5 = -\frac{1}{3}(2 + (-0.71506)^2 + (-0.71506)^3) = -0.71498\)

Continue until the value stabilizes to \(a \approx -0.715\) to 3 decimal places.

(i) Start with \(x_1 = 2\).

Calculate \(x_2 = \frac{2(2)^6 + 12(2)}{3(2)^5 + 8} = \frac{128 + 24}{96 + 8} = \frac{152}{104} = 1.461538\).

Calculate \(x_3 = \frac{2(1.461538)^6 + 12(1.461538)}{3(1.461538)^5 + 8} \approx 1.336405\).

Calculate \(x_4 = \frac{2(1.336405)^6 + 12(1.336405)}{3(1.336405)^5 + 8} \approx 1.320756\).

Calculate \(x_5 = \frac{2(1.320756)^6 + 12(1.320756)}{3(1.320756)^5 + 8} \approx 1.319545\).

Calculate \(x_6 = \frac{2(1.319545)^6 + 12(1.319545)}{3(1.319545)^5 + 8} \approx 1.319505\).

Calculate \(x_7 = \frac{2(1.319505)^6 + 12(1.319505)}{3(1.319505)^5 + 8} \approx 1.319505\).

Thus, \(\alpha \approx 1.3195\) to 4 decimal places.

(ii) The equation satisfied by \(\alpha\) is \(\alpha = \frac{2\alpha^6 + 12\alpha}{3\alpha^5 + 8}\).

Solving \(\alpha = \frac{2\alpha^6 + 12\alpha}{3\alpha^5 + 8}\) gives \(\alpha^5 = 4\).

Thus, \(\alpha = \sqrt[5]{4}\).

(i) Calculate \(f(x) = x^3 - 3x - 7\) at \(x = 2\) and \(x = 3\):

\(f(2) = 2^3 - 3 \times 2 - 7 = 8 - 6 - 7 = -5\)

\(f(3) = 3^3 - 3 \times 3 - 7 = 27 - 9 - 7 = 11\)

Since \(f(2) < 0\) and \(f(3) > 0\), by the Intermediate Value Theorem, \(\alpha\) lies between 2 and 3.

(ii) Using formula \(A\):

\(x_1 = 2.5\)

\(x_2 = (3 \times 2.5 + 7)^{\frac{1}{3}} = 2.4190\)

\(x_3 = (3 \times 2.4190 + 7)^{\frac{1}{3}} = 2.4320\)

\(x_4 = (3 \times 2.4320 + 7)^{\frac{1}{3}} = 2.4268\)

\(x_5 = (3 \times 2.4268 + 7)^{\frac{1}{3}} = 2.4295\)

\(x_6 = (3 \times 2.4295 + 7)^{\frac{1}{3}} = 2.4282\)

\(x_7 = (3 \times 2.4282 + 7)^{\frac{1}{3}} = 2.4288\)

Using formula \(B\):

\(x_1 = 2.5\)

\(x_2 = \frac{2.5^3 - 7}{3} = 2.9583\)

\(x_3 = \frac{2.9583^3 - 7}{3} = 7.3878\)

\(x_4 = \frac{7.3878^3 - 7}{3} = 131.1805\)

Formula \(B\) fails to converge. Formula \(A\) converges to \(\alpha = 2.43\) correct to 2 decimal places.

(i) Calculate \(\cot(2.5)\) and \(1 - 2.5 = -1.5\). Since \(\cot(2.5) > -1.5\), \(\alpha > 2.5\).

(ii) Assume \(x = \pi + \arctan \left( \frac{1}{1-x} \right)\). Rearrange to \(\cot x = 1 - x\), showing convergence to \(\alpha\).

(iii) Using the iterative formula:

1. \(x_1 = 3.00000\)

2. \(x_2 = 2.57732\)

3. \(x_3 = 2.57605\)

4. \(x_4 = 2.57600\)

5. \(x_5 = 2.57600\)

Thus, \(\alpha = 2.576\) correct to 3 decimal places.

(i) Differentiate both equations: \(y = x \cos x\) gives \(\frac{dy}{dx} = \cos x - x \sin x\) and \(y = \frac{k}{x}\) gives \(\frac{dy}{dx} = -\frac{k}{x^2}\). Equate the derivatives at \(x = a\):

\(\cos a - a \sin a = -\frac{k}{a^2}\).

Since the curves touch, \(a \cos a = \frac{k}{a}\). Substitute \(k = a^2 \cos a\) into the derivative equation:

\(\cos a - a \sin a = -\frac{a^2 \cos a}{a^2} = -\cos a\).

Rearrange to get \(\tan a = \frac{2}{a}\).

(ii) Use the iterative formula \(a_{n+1} = \arctan \left( \frac{2}{a_n} \right)\) starting with an initial guess, say \(a_0 = 1\):

\(a_1 = \arctan \left( \frac{2}{1} \right) = 1.10715\)

\(a_2 = \arctan \left( \frac{2}{1.10715} \right) = 1.07989\)

\(a_3 = \arctan \left( \frac{2}{1.07989} \right) = 1.07735\)

\(a_4 = \arctan \left( \frac{2}{1.07735} \right) = 1.07708\)

\(a_5 = \arctan \left( \frac{2}{1.07708} \right) = 1.07703\)

Thus, \(a \approx 1.077\) to 3 decimal places.

(iii) Use \(a \cos a = \frac{k}{a}\) with \(a = 1.077\):

\(k = a^2 \cos a = (1.077)^2 \cos(1.077) \approx 0.55\).

(i) Evaluate the function at \(x = 1\) and \(x = 2\):

\(f(1) = 1^5 - 3(1)^3 + 1^2 - 4 = 1 - 3 + 1 - 4 = -5\)

\(f(2) = 2^5 - 3(2)^3 + 2^2 - 4 = 32 - 24 + 4 - 4 = 8\)

Since \(f(1) < 0\) and \(f(2) > 0\), there is a root between 1 and 2.

(ii) Rearrange the equation:

Start with \(x^5 - 3x^3 + x^2 - 4 = 0\).

Add 4 to both sides: \(x^5 - 3x^3 + x^2 = 4\).

Divide by \(x^2\): \(x^3 - 3x + \frac{1}{x} = \frac{4}{x^2}\).

Rearrange to: \(x^3 = 3x + \frac{4}{x^2} - 1\).

Take the cube root: \(x = \sqrt[3]{3x + \frac{4}{x^2} - 1}\).

(iii) Use the iterative formula \(x_{n+1} = \sqrt[3]{3x_n + \frac{4}{x_n^2} - 1}\).

Start with \(x_0 = 1.5\).

\(x_1 = \sqrt[3]{3(1.5) + \frac{4}{(1.5)^2} - 1} = 1.8041\)

\(x_2 = \sqrt[3]{3(1.8041) + \frac{4}{(1.8041)^2} - 1} = 1.7817\)

\(x_3 = \sqrt[3]{3(1.7817) + \frac{4}{(1.7817)^2} - 1} = 1.7800\)

\(x_4 = \sqrt[3]{3(1.7800) + \frac{4}{(1.7800)^2} - 1} = 1.7800\)

The root correct to 2 decimal places is 1.78.

(i) Evaluate \(x^3 - x^2 - 6\) for \(x = 2\) and \(x = 3\):

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

For \(x = 3\), \(3^3 - 3^2 - 6 = 27 - 9 - 6 = 12\).

Since the sign changes between \(x = 2\) and \(x = 3\), \(\alpha\) lies between 2 and 3.

(ii) The iterative formula is \(x_{n+1} = \sqrt{x_n + \frac{6}{x_n}}\).

If the sequence converges, then \(x_{n+1} = x_n = \alpha\).

Substitute into the formula: \(\alpha = \sqrt{\alpha + \frac{6}{\alpha}}\).

Square both sides: \(\alpha^2 = \alpha + \frac{6}{\alpha}\).

Multiply through by \(\alpha\): \(\alpha^3 = \alpha^2 + 6\).

Rearrange to \(\alpha^3 - \alpha^2 - 6 = 0\), which is the original equation, confirming convergence to \(\alpha\).

(iii) Use the iterative formula starting with an initial guess, say \(x_0 = 2.5\):

\(x_1 = \sqrt{2.5 + \frac{6}{2.5}} = 2.31662\)

\(x_2 = \sqrt{2.31662 + \frac{6}{2.31662}} = 2.22192\)

\(x_3 = \sqrt{2.22192 + \frac{6}{2.22192}} = 2.21936\)

\(x_4 = \sqrt{2.21936 + \frac{6}{2.21936}} = 2.21900\)

\(x_5 = \sqrt{2.21900 + \frac{6}{2.21900}} = 2.21900\)

The value converges to \(\alpha = 2.219\) correct to 3 decimal places.

(a) To find the maximum point, we need to differentiate \(y = x^2 \cos 3x\) and set the derivative to zero. Using the product rule:

\(\frac{d}{dx}(x^2 \cos 3x) = \frac{d}{dx}(x^2) \cdot \cos 3x + x^2 \cdot \frac{d}{dx}(\cos 3x)\).

\(= 2x \cos 3x - 3x^2 \sin 3x\).

Setting the derivative to zero for maximum point:

\(2x \cos 3x - 3x^2 \sin 3x = 0\).

\(2x \cos 3x = 3x^2 \sin 3x\).

\(\frac{2}{3x} = \tan 3x\).

\(x = \frac{1}{3} \arctan \left( \frac{2}{3x} \right)\).

At maximum point \(x = a\), so \(a = \frac{1}{3} \arctan \left( \frac{2}{3a} \right)\).

(b) Using the iterative formula \(a_{n+1} = \frac{1}{3} \arctan \left( \frac{2}{3a_n} \right)\), start with an initial guess and iterate:

1. \(a_1 = 0.5\)

2. \(a_2 = \frac{1}{3} \arctan \left( \frac{2}{3 \times 0.5} \right) \approx 0.3435\)

3. \(a_3 = \frac{1}{3} \arctan \left( \frac{2}{3 \times 0.3435} \right) \approx 0.3826\)

4. \(a_4 = \frac{1}{3} \arctan \left( \frac{2}{3 \times 0.3826} \right) \approx 0.4264\)

5. \(a_5 = \frac{1}{3} \arctan \left( \frac{2}{3 \times 0.4264} \right) \approx 0.4740\)

6. \(a_6 = \frac{1}{3} \arctan \left( \frac{2}{3 \times 0.4740} \right) \approx 0.3017\)

7. \(a_7 = \frac{1}{3} \arctan \left( \frac{2}{3 \times 0.3017} \right) \approx 0.3989\)

Continue iterating until the value stabilizes to 2 decimal places: \(a \approx 0.36\).

(i) Differentiate \(y = x \ln(8 - x)\) using the product rule: \(\frac{dy}{dx} = \ln(8 - x) - \frac{x}{8 - x}\).

Set \(\frac{dy}{dx} = 1\) and solve: \(\ln(8 - x) - \frac{x}{8 - x} = 1\).

Rearrange to find \(x = 8 - \frac{8}{\ln(8 - x)}\).

(ii) Calculate \(8 - \frac{8}{\ln(8 - 2.9)} = 3.09\) and \(8 - \frac{8}{\ln(8 - 3.1)} = 2.97\).

Since \(3.09 > 2.9\) and \(2.97 < 3.1\), \(a\) lies between 2.9 and 3.1.

(iii) Use the iterative formula \(x_{n+1} = 8 - \frac{8}{\ln(8 - x_n)}\).

Starting with \(x_0 = 3\), iterate: \(x_1 = 3.0293\), \(x_2 = 3.0111\), \(x_3 = 3.0223\), \(x_4 = 3.0154\), \(x_5 = 3.0198\).

Consecutive values \(3.0223\) and \(3.0154\) round to 3.02, confirming \(a = 3.02\).

(i) Differentiate \(y = x^2 \cos \frac{1}{2}x\) using the product rule:

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

Set \(\frac{dy}{dx} = 0\) for a stationary point:

\(2x \cos \frac{1}{2}x = \frac{1}{2}x^2 \sin \frac{1}{2}x\).

Rearrange to get \(\tan \frac{1}{2}x = \frac{4}{x}\).

Thus, \(\tan \frac{1}{2}p = \frac{4}{p}\).

(ii) Calculate \(\tan \frac{1}{2}x\) at \(x = 2\) and \(x = 2.5\):

For \(x = 2\), \(\tan 1 = 1.5574\) and \(\frac{4}{2} = 2\).

For \(x = 2.5\), \(\tan 1.25 = 1.9207\) and \(\frac{4}{2.5} = 1.6\).

Since \(1.5574 < 2\) and \(1.9207 > 1.6\), \(p\) is between 2 and 2.5.

(iii) Use the iterative formula:

Start with \(p_0 = 2\).

\(p_1 = 2 \arctan \left( \frac{4}{2} \right) = 2 \arctan(2) = 2.0344\).

\(p_2 = 2 \arctan \left( \frac{4}{2.0344} \right) = 2.1489\).

\(p_3 = 2 \arctan \left( \frac{4}{2.1489} \right) = 2.1455\).

\(p_4 = 2 \arctan \left( \frac{4}{2.1455} \right) = 2.1456\).

\(p_5 = 2 \arctan \left( \frac{4}{2.1456} \right) = 2.1456\).

Thus, \(p \approx 2.15\) to 2 decimal places.

(i) To differentiate \(y = \csc x = \frac{1}{\sin x}\), use the quotient rule or chain rule. The derivative of \(\sin x\) is \(\cos x\), so:

\(\frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{\sin x} \right) = -\frac{\cos x}{\sin^2 x} = -\csc x \cot x.\)

(ii) The gradients of the curves are equal when \(x = a\). For \(y = \csc x\), the gradient is \(-\csc a \cot a\). For \(y = e^{-x}\), the gradient is \(-e^{-a}\). Equating these gives:

\(-e^{-a} = -\csc a \cot a.\)

Rearranging gives:

\(a = \arctan \left( \frac{e^a}{\sin a} \right).\)

(iii) To verify \(a\) lies between 1 and 1.5, calculate the expression for \(a = 1\) and \(a = 1.5\). Check that the values satisfy the inequality.

(iv) Use the iterative formula \(a_{n+1} = \arctan \left( \frac{e^{a_n}}{\sin a_n} \right)\). Start with an initial guess, such as \(a_0 = 1\), and iterate:

\(a_1 = \arctan \left( \frac{e^{1}}{\sin 1} \right) \approx 1.317\)

Continue iterating until the value stabilizes to 3 decimal places:

\(a_2 \approx 1.317\)

\(a_3 \approx 1.317\)

The final answer is \(a \approx 1.317\).

(i) Differentiate \(x = t^2 + 3t + 1\) to get \(\frac{dx}{dt} = 2t + 3\).

Differentiate \(y = t^4 + 1\) to get \(\frac{dy}{dt} = 4t^3\).

The gradient \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4t^3}{2t + 3}\).

Set \(\frac{dy}{dx} = 4\) and solve: \(\frac{4t^3}{2t + 3} = 4\).

Simplify to get \(4t^3 = 4(2t + 3)\).

\(t^3 = 2t + 3\).

Thus, \(p = \sqrt[3]{2p + 3}\).

(ii) Evaluate \(p - \sqrt[3]{2p + 3}\) at \(p = 1.8\) and \(p = 2.0\).

At \(p = 1.8\), \(1.8 - \sqrt[3]{2(1.8) + 3} \approx -0.076\).

At \(p = 2.0\), \(2.0 - \sqrt[3]{2(2.0) + 3} \approx 0.087\).

Since the sign changes, \(p\) is between 1.8 and 2.0.

(iii) Use the iterative formula \(p_{n+1} = \sqrt[3]{2p_n + 3}\).

Start with \(p_0 = 1.9\).

\(p_1 = \sqrt[3]{2(1.9) + 3} \approx 1.8854\).

\(p_2 = \sqrt[3]{2(1.8854) + 3} \approx 1.8895\).

\(p_3 = \sqrt[3]{2(1.8895) + 3} \approx 1.8899\).

\(p_4 = \sqrt[3]{2(1.8899) + 3} \approx 1.8899\).

The value of \(p\) correct to 2 decimal places is 1.89.

(i) To find the gradient \(\frac{dy}{dx}\), we first find \(\frac{dx}{dt} = \frac{2}{t+2}\) and \(\frac{dy}{dt} = 3t^2 + 2\).

Then, \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^2 + 2}{\frac{2}{t+2}} = \frac{1}{2}(3t^2 + 2)(t+2)\).

At the origin, \(x = 0\) and \(y = 0\), which gives \(t = -1\).

Substitute \(t = -1\) into \(\frac{dy}{dx}\) to get \(\frac{5}{2}\) as the gradient at the origin.

(ii)(a) Given \(\frac{dy}{dx} = \frac{1}{2}\), equate to \(\frac{1}{2}(3p^2 + 2)(p+2)\) and solve to confirm \(p = \frac{1}{3p^2 + 2} - 2\).

(ii)(b) Use the iterative formula based on \(p = \frac{1}{3p^2 + 2} - 2\) to find \(p\).

Start with an initial guess, iterate to find \(p = -1.924\) or better \(-1.92367\).

Calculate coordinates using \(x = 2 \ln(p + 2)\) and \(y = p^3 + 2p + 3\).

Coordinates of \(P\) are \((-5.15, -7.97)\).

(i) The derivatives of the curves are \(2e^{2x-3}\) and \(\frac{2}{x}\). Setting these equal for parallel tangents gives:

\(2e^{2a-3} = \frac{2}{a}\)

\(e^{2a-3} = \frac{1}{a}\)

Taking natural logarithms on both sides:

\(2a - 3 = -\ln a\)

\(2a = 3 - \ln a\)

\(a = \frac{1}{2}(3 - \ln a)\)

(ii) Consider \(f(a) = a - \frac{1}{2}(3 - \ln a)\). Calculate \(f(1)\) and \(f(2)\):

\(f(1) = 1 - \frac{1}{2}(3 - \ln 1) = 1 - \frac{3}{2} = -0.5\)

\(f(2) = 2 - \frac{1}{2}(3 - \ln 2) = 2 - \frac{1}{2}(3 - 0.693) = 0.1535\)

Since \(f(1) < 0\) and \(f(2) > 0\), there is a root between 1 and 2.

(iii) Using the iterative formula \(a_{n+1} = \frac{1}{2}(3 - \ln a_n)\):

Start with \(a_1 = 1.5\).

\(a_2 = \frac{1}{2}(3 - \ln 1.5) = 1.3466\)

\(a_3 = \frac{1}{2}(3 - \ln 1.3466) = 1.3539\)

\(a_4 = \frac{1}{2}(3 - \ln 1.3539) = 1.3519\)

\(a_5 = \frac{1}{2}(3 - \ln 1.3519) = 1.3525\)

\(a_6 = \frac{1}{2}(3 - \ln 1.3525) = 1.3522\)

\(a_7 = \frac{1}{2}(3 - \ln 1.3522) = 1.3523\)

The value of \(a\) correct to 2 decimal places is 1.35.

(i) To find the \(x\)-coordinate of \(M\), we need to find the maximum point of the function \(y = e^{-\frac{1}{2}x^2} \sqrt{(1 + 2x^2)}\). Differentiate \(y\) with respect to \(x\) using the product and chain rules. Set the derivative to zero and solve for \(x\). The solution is \(x = \frac{1}{\sqrt{2}}\).

(ii) The iterative formula \(x_{n+1} = \sqrt{(\ln(4 + 8x_n^2))}\) converges to \(\alpha\). Therefore, \(\alpha = \sqrt{(\ln(4 + 8\alpha^2))}\). Rearrange to find \(e^{\alpha^2} = 4 + 8\alpha^2\). Substitute \(y = 0.5\) into the original curve equation and show that \(\alpha\) satisfies this equation.

(iii) Use the iterative formula starting with \(x_1 = 2\) and calculate successive values until the result stabilizes to 2 decimal places. The iterations yield \(\alpha = 1.86\).

(i) To find the stationary point, we first find the derivative of \(y = \frac{\ln x}{x + 1}\) using the quotient rule:

\(\frac{d}{dx} \left( \frac{\ln x}{x + 1} \right) = \frac{(x + 1) \cdot \frac{1}{x} - \ln x \cdot 1}{(x + 1)^2} = \frac{1 - \ln x}{x(x + 1)^2}\).

Setting the derivative to zero gives:

\(1 - \ln x = 0\) or \(\ln x = 1\).

Solving \(\ln x = 1\) gives \(x = e\), but we need to verify the given equation:

Rearrange \(x = \frac{x + 1}{\ln x}\) to \(x \ln x = x + 1\), which simplifies to \(\ln x = 1 + \frac{1}{x}\).

Check the interval \(x = 3\) and \(x = 4\):

For \(x = 3\), \(\ln 3 \approx 1.0986\) and \(\frac{3 + 1}{3} \approx 1.3333\).

For \(x = 4\), \(\ln 4 \approx 1.3863\) and \(\frac{4 + 1}{4} = 1.25\).

Since \(\ln x\) changes from greater than \(\frac{x + 1}{x}\) to less than \(\frac{x + 1}{x}\) between 3 and 4, there is a root in this interval.

(ii) Using the iterative formula \(x_{n+1} = \frac{x_n + 1}{\ln x_n}\):

Start with \(x_1 = 3.5\):

\(x_2 = \frac{3.5 + 1}{\ln 3.5} \approx 3.5718\)

\(x_3 = \frac{3.5718 + 1}{\ln 3.5718} \approx 3.5911\)

\(x_4 = \frac{3.5911 + 1}{\ln 3.5911} \approx 3.5900\)

\(x_5 = \frac{3.5900 + 1}{\ln 3.5900} \approx 3.5900\)

The \(x\)-coordinate correct to 2 decimal places is 3.59.

(i) To find the \(x\)-coordinate of the minimum point \(M\), we need to find the derivative of \(y = \frac{\sin x}{x}\) and set it to zero.

Using the quotient rule, the derivative is:

\(y' = \frac{x \cos x - \sin x}{x^2}\).

Setting \(y' = 0\), we have:

\(x \cos x - \sin x = 0\).

Rearranging gives:

\(x \cos x = \sin x\).

Dividing both sides by \(\cos x\), we get:

\(x = \tan x\).

(ii) Using the iterative formula \(x_{n+1} = \arctan(x_n) + \pi\), we start with an initial guess and iterate:

1. \(x_1 = \arctan(x_0) + \pi\)

2. \(x_2 = \arctan(x_1) + \pi\)

Continue iterating until the value stabilizes to 4 decimal places.

The final \(x\)-coordinate of \(M\) is 4.49, correct to 2 decimal places.

(i) To find the stationary point, we first find the derivative of \(y = \ln x + \frac{2}{x}\). The derivative is \(\frac{dy}{dx} = \frac{1}{x} - \frac{2}{x^2}\).

Setting \(\frac{dy}{dx} = 0\), we have \(\frac{1}{x} - \frac{2}{x^2} = 0\).

Multiplying through by \(x^2\), we get \(x - 2 = 0\), so \(x = 2\).

Substituting \(x = 2\) back into the original equation, \(y = \ln 2 + \frac{2}{2} = \ln 2 + 1\).

To determine if it is a minimum or maximum, consider the second derivative: \(\frac{d^2y}{dx^2} = -\frac{1}{x^2} + \frac{4}{x^3}\).

At \(x = 2\), \(\frac{d^2y}{dx^2} = -\frac{1}{4} + \frac{4}{8} = \frac{1}{4} > 0\), indicating a minimum point.

(ii) The iterative formula is \(x_{n+1} = \frac{2}{3 - \ln x_n}\). As \(n \to \infty\), \(x_n \to \alpha\), so \(\alpha = \frac{2}{3 - \ln \alpha}\).

Rearranging gives \(3 = \ln \alpha + \frac{2}{\alpha}\), which matches the equation \(y = 3\).

(iii) Using the iterative formula:

\(x_1 = 1\)

\(x_2 = \frac{2}{3 - \ln 1} = \frac{2}{3} \approx 0.67\)

\(x_3 = \frac{2}{3 - \ln 0.67} \approx 0.58\)

\(x_4 = \frac{2}{3 - \ln 0.58} \approx 0.56\)

\(x_5 = \frac{2}{3 - \ln 0.56} \approx 0.56\)

Thus, \(\alpha \approx 0.56\) to 2 decimal places.

(a) To find the stationary point, we need to differentiate \(y = \frac{x^3}{e^x - 1}\) and set the derivative to zero.

Using the quotient rule, \(\frac{d}{dx} \left( \frac{x^3}{e^x - 1} \right) = \frac{(e^x - 1)(3x^2) - x^3(e^x)}{(e^x - 1)^2}\).

Setting the numerator to zero for a stationary point: \((e^x - 1)(3x^2) - x^3(e^x) = 0\).

Simplifying gives \(3x^2 e^x - 3x^2 - x^3 e^x = 0\).

Rearranging, \(x^2 e^x (3 - x) = 3x^2\).

Dividing by \(x^2\) (assuming \(x \neq 0\)), \(e^x (3 - x) = 3\).

Thus, \(3 - x = 3e^{-x}\) or \(x = 3(1 - e^{-x})\).

Therefore, \(p = 3(1 - e^{-p})\).

(b) To verify \(p\) lies between 2.5 and 3, calculate \(f(p) = 3(1 - e^{-p}) - p\).

For \(p = 2.5\), \(f(2.5) = 3(1 - e^{-2.5}) - 2.5 \approx 0.351\).

For \(p = 3\), \(f(3) = 3(1 - e^{-3}) - 3 \approx -0.950\).

Since \(f(2.5) > 0\) and \(f(3) < 0\), by the Intermediate Value Theorem, \(p\) lies between 2.5 and 3.

(c) Using the iterative formula \(p_{n+1} = 3(1 - e^{-p_n})\), start with an initial guess, say \(p_0 = 2.5\).

Calculate successive iterations:

\(p_1 = 3(1 - e^{-2.5}) \approx 2.7639\)

\(p_2 = 3(1 - e^{-2.7639}) \approx 2.8113\)

\(p_3 = 3(1 - e^{-2.8113}) \approx 2.8212\)

\(p_4 = 3(1 - e^{-2.8212}) \approx 2.8231\)

\(p_5 = 3(1 - e^{-2.8231}) \approx 2.8235\)

The iterations converge to \(p = 2.82\) to 2 decimal places.

(a) Differentiate \(y = \frac{x}{\cos^2 x}\) using the quotient rule:

\(\frac{dy}{dx} = \frac{\cos^2 x + 2x \sin x \cos x}{\cos^4 x}\).

At \(x = a\), set \(\frac{dy}{dx} = 12\):

\(12 = \frac{\cos^2 a + 2a \sin a \cos a}{\cos^4 a}\).

Simplify to obtain:

\(12 \cos^3 a = \cos a + 2a \sin a\).

\(\cos^3 a = \frac{\cos a + 2a \sin a}{12}\).

\(a = \cos^{-1} \left( \sqrt[3]{\frac{\cos a + 2a \sin a}{12}} \right)\).

(b) Calculate \(\cos 0.9 \approx 0.622\) and \(\cos 1 \approx 0.540\).

\(\cos 0.9 > 0.553\) and \(\cos 1 < 0.570\), so \(0.9 < a < 1\).

(c) Use the iterative formula:

\(a_{n+1} = \cos^{-1} \left( \sqrt[3]{\frac{\cos a_n + 2a_n \sin a_n}{12}} \right)\).

Starting with \(a_0 = 0.9\), iterate to find:

\(a_1 \approx 0.95\), \(a_2 \approx 0.9743\), \(a_3 \approx 0.9694\), \(a_4 \approx 0.9704\).

Final answer: \(a \approx 0.97\).

(a) To find the stationary point, we need to differentiate \(y = x \sqrt{\sin x}\) and set the derivative to zero. Using the product rule:

\(\frac{dy}{dx} = \sqrt{\sin x} + x \cdot \frac{1}{2\sqrt{\sin x}} \cdot \cos x\)

\(= \sqrt{\sin x} + \frac{x \cos x}{2\sqrt{\sin x}}\)

Setting \(\frac{dy}{dx} = 0\):

\(\sqrt{\sin x} + \frac{x \cos x}{2\sqrt{\sin x}} = 0\)

\(2\sin x + x \cos x = 0\)

\(\tan x = -\frac{1}{2}x\)

(b) Calculate \(\tan 2\) and \(\tan 2.5\):

\(\tan 2 \approx -2.18\) and \(\tan 2.5 \approx -0.747\)

Since \(-2.18 < -1 < -0.747\), \(a\) lies between 2 and 2.5.

(c) The iterative formula is \(x_{n+1} = \pi - \arctan\left(\frac{1}{2}x_n\right)\). If it converges, it converges to \(a\) where \(\tan a = -\frac{1}{2}a\).

(d) Using the iterative formula:

Start with \(x_0 = 2.25\):

\(x_1 = \pi - \arctan\left(\frac{1}{2} \times 2.25\right) \approx 2.2974\)

\(x_2 = \pi - \arctan\left(\frac{1}{2} \times 2.2974\right) \approx 2.2871\)

\(x_3 = \pi - \arctan\left(\frac{1}{2} \times 2.2871\right) \approx 2.2893\)

\(x_4 = \pi - \arctan\left(\frac{1}{2} \times 2.2893\right) \approx 2.2888\)

Thus, \(a \approx 2.29\) to 2 decimal places.

(a) To find \(\frac{dy}{dx}\), use the chain rule. Let \(u = \tan x\), then \(y = \sqrt{u}\). The derivative \(\frac{dy}{du} = \frac{1}{2\sqrt{u}}\) and \(\frac{du}{dx} = \sec^2 x\). Therefore, \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{\tan x}} \cdot \sec^2 x = \frac{1 + \tan^2 x}{2\sqrt{\tan x}}\).

Verify at \(x = \frac{1}{4}\pi\): \(\tan \frac{1}{4}\pi = 1\), so \(\frac{dy}{dx} = \frac{1 + 1^2}{2\sqrt{1}} = 1\).

(b) Substitute \(t = \tan a\) into the equation from part (a): \(\frac{1 + t^2}{2\sqrt{t}} = 1\). Rearrange to get \((1 + t^2)^2 = 4t\), leading to \(t^4 + 2t^2 - 4t + 1 = 0\). Divide by \(t - 1\) to obtain \(t^3 + t^2 + 3t - 1 = 0\).

(c) Use the iterative formula \(a_{n+1} = \arctan \left( \frac{1}{3} (1 - \tan^2 a_n - \tan^3 a_n) \right)\). Start with an initial guess, e.g., \(a_0 = 0.3\), and iterate: \(a_1 = 0.2854\), \(a_2 = 0.2894\), \(a_3 = 0.2883\), etc. Continue until \(a\) stabilizes to 2 decimal places: \(a = 0.29\).

(a) To find the maximum point, we differentiate \(y = \frac{\arctan x}{\sqrt{x}}\) using the quotient rule:

\(\frac{dy}{dx} = \frac{\sqrt{x} \cdot \frac{1}{1+x^2} - \arctan x \cdot \frac{1}{2\sqrt{x}}}{x}\).

Setting \(\frac{dy}{dx} = 0\) gives:

\(\sqrt{x} \cdot \frac{1}{1+x^2} = \arctan x \cdot \frac{1}{2\sqrt{x}}\).

Rearranging and simplifying leads to:

\(a = \tan \left( \frac{2a}{1 + a^2} \right)\).

(b) Calculate \(a = \tan \left( \frac{2a}{1 + a^2} \right)\) for \(a = 1.3\) and \(a = 1.5\):

For \(a = 1.3\), \(\tan \left( \frac{2 \times 1.3}{1 + 1.3^2} \right) \approx 1.448\).

For \(a = 1.5\), \(\tan \left( \frac{2 \times 1.5}{1 + 1.5^2} \right) \approx 1.322\).

Since \(1.3 < 1.448\) and \(1.5 > 1.322\), \(a\) lies between 1.3 and 1.5.

(c) Using the iterative formula \(a_{n+1} = \tan \left( \frac{2a_n}{1 + a_n^2} \right)\):

Start with \(a_0 = 1.4\).

\(a_1 = \tan \left( \frac{2 \times 1.4}{1 + 1.4^2} \right) \approx 1.3914\).

\(a_2 = \tan \left( \frac{2 \times 1.3914}{1 + 1.3914^2} \right) \approx 1.3900\).

\(a_3 = \tan \left( \frac{2 \times 1.3900}{1 + 1.3900^2} \right) \approx 1.3900\).

Thus, \(a \approx 1.39\) correct to 2 decimal places.

(a) Calculate \(f(1) = \frac{e^{2} + 1}{e^{2} - 1}\) and \(f(1.5) = \frac{e^{3} + 1}{e^{3} - 1}\). Verify that \(f(1) < 1\) and \(f(1.5) > 1.5\), confirming \(a\) is between 1 and 1.5.

(b) Use the iterative formula \(x_{n+1} = \frac{e^{2x_n} + 1}{e^{2x_n} - 1}\). Start with \(x_1 = 1\):

\(x_2 = 1.2311\)

\(x_3 = 1.2032\)

\(x_4 = 1.2002\)

\(x_5 = 1.2000\)

Thus, \(a = 1.20\) to 2 decimal places.

(c) Use the quotient rule to find \(f'(x) = \frac{(e^{2x} - 1)(2e^{2x}) - (e^{2x} + 1)(2e^{2x})}{(e^{2x} - 1)^2}\).

Simplify to \(f'(x) = \frac{-4e^{2x}}{(e^{2x} - 1)^2}\).

Set \(f'(x) = -8\):

\(\frac{-4e^{2x}}{(e^{2x} - 1)^2} = -8\)

\(2(e^{2x})^2 - 5e^{2x} + 2 = 0\)

Solve the quadratic equation for \(e^{2x}\):

\(e^{2x} = 2\) or \(e^{2x} = \frac{1}{2}\)

Thus, \(x = \frac{1}{2} \ln 2\).

(i) Differentiate \(y = e^{-2x} \ln(x-1)\) using the product rule:

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

Set \(\frac{dy}{dx} = 0\):

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

Simplify to find \(x = 1 + e^{\frac{1}{2(x-1)}}\).

(ii) Calculate \(f(x) = \ln(x-1) - \frac{1}{2(x-1)}\):

\(f(2.2) = -0.234\), \(f(2.6) = 0.317\).

Calculate \(f(x) = 2e^{-2x} \ln(x-1) + \frac{e^{-2x}}{x-1}\):

\(f(2.2) = 0.005\), \(f(2.6) = -0.0017\).

Since \(f(x)\) changes sign, \(p\) is between 2.2 and 2.6.

(iii) Use the iterative formula \(p_{n+1} = 1 + \exp\left( \frac{1}{2(p_n-1)} \right)\):

Start with \(p_0 = 2.4\).

\(p_1 = 2.4152\)

\(p_2 = 2.4213\)

\(p_3 = 2.4198\)

\(p_4 = 2.4204\)

\(p_5 = 2.4201\)

\(p_6 = 2.4203\)

\(p_7 = 2.4202\)

Thus, \(p \approx 2.42\) to 2 decimal places.

(i) Differentiate the curves: \(y = 4 \cos \frac{1}{2} x\) gives \(\frac{dy}{dx} = -2 \sin \frac{1}{2} x\) and \(y = \frac{1}{4-x}\) gives \(\frac{dy}{dx} = \frac{1}{(4-x)^2}\).

For perpendicular tangents, the product of the derivatives is \(-1\):

\(-2 \sin \frac{1}{2} a \cdot \frac{1}{(4-a)^2} = -1\)

Rearrange to find \(a\):

\(2 \sin \frac{1}{2} a = (4-a)^2\)

\(a = 4 - \sqrt{2 \sin \frac{1}{2} a}\)

(ii) Calculate values at \(a = 2\) and \(a = 3\):

For \(a = 2\), \(2 < 2.7027 \ldots\)

For \(a = 3\), \(3 > 2.587 \ldots\)

Thus, \(a\) lies between 2 and 3.

(iii) Use the iterative formula \(a_{n+1} = 4 - \sqrt{2 \sin \frac{1}{2} a_n}\):

Start with \(a_1 = 2\):

\(a_2 = 2.70272\)

\(a_3 = 2.60285\)

\(a_4 = 2.61152\)

\(a_5 = 2.61070\)

\(a_6 = 2.61077\)

Final answer: \(a = 2.611\)

(a) To solve \(\int_0^a xe^{-2x} \, dx = \frac{1}{8}\), we start by integrating by parts. Let \(u = x\) and \(dv = e^{-2x} \, dx\). Then \(du = dx\) and \(v = -\frac{1}{2}e^{-2x}\).

Using integration by parts, \(\int xe^{-2x} \, dx = -\frac{1}{2}xe^{-2x} + \frac{1}{2} \int e^{-2x} \, dx\).

Integrating \(\int e^{-2x} \, dx\) gives \(-\frac{1}{2}e^{-2x}\), so \(\int xe^{-2x} \, dx = -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x}\).

Evaluating from 0 to a, we have:

\(\left[ -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} \right]_0^a = -\frac{1}{2}ae^{-2a} - \frac{1}{4}e^{-2a} + \frac{1}{4}\).

Setting this equal to \(\frac{1}{8}\), we solve:

\(-\frac{1}{2}ae^{-2a} - \frac{1}{4}e^{-2a} + \frac{1}{4} = \frac{1}{8}\).

Rearranging gives \(-\frac{1}{2}ae^{-2a} - \frac{1}{4}e^{-2a} = -\frac{1}{8}\).

Multiplying through by -1 and simplifying, we find \(a = \frac{1}{2} \ln(4a + 2)\).

(b) To verify a lies between 0.5 and 1, calculate:

For \(a = 0.5\), \(\frac{1}{2} \ln(4 \times 0.5 + 2) = \frac{1}{2} \ln(4) = 0.693...\).

For \(a = 1\), \(\frac{1}{2} \ln(4 \times 1 + 2) = \frac{1}{2} \ln(6) = 0.895...\).

Since \(0.5 < 0.693 < 1\), a is between 0.5 and 1.

(c) Using the iterative formula \(a_{n+1} = \frac{1}{2} \ln(4a_n + 2)\), start with \(a_0 = 0.75\):

\(a_1 = \frac{1}{2} \ln(4 \times 0.75 + 2) = 0.8047\)

\(a_2 = \frac{1}{2} \ln(4 \times 0.8047 + 2) = 0.8261\)

\(a_3 = \frac{1}{2} \ln(4 \times 0.8261 + 2) = 0.8343\)

\(a_4 = \frac{1}{2} \ln(4 \times 0.8343 + 2) = 0.8373\)

\(a_5 = \frac{1}{2} \ln(4 \times 0.8373 + 2) = 0.8385\)

Continuing this process, we find \(a \approx 0.84\) to 2 decimal places.

(i) To solve \(\int_{1}^{a} \ln(2x) \, dx = 1\), we first integrate:

\(\int \ln(2x) \, dx = x \ln(2x) - \int x \cdot \frac{1}{x} \, dx = x \ln(2x) - x + C\).

Evaluating from 1 to \(a\):

\([a \ln(2a) - a] - [1 \ln(2) - 1] = 1\).

Simplifying gives:

\(a \ln(2a) - a - \ln(2) + 1 = 1\).

\(a \ln(2a) - a = \ln(2)\).

\(a(\ln 2 + \ln a) - a = \ln(2)\).

\(a \ln a + a \ln 2 - a = \ln(2)\).

\(a \ln a = a - a \ln 2 + \ln(2)\).

\(a \ln a = a(1 - \ln 2) + \ln(2)\).

Rearranging gives:

\(a = \frac{1}{2} \exp \left( 1 + \frac{\ln 2}{a} \right)\).

(ii) Using the iterative formula:

Start with an initial guess, say \(a_1 = 2\).

\(a_2 = \frac{1}{2} \exp \left( 1 + \frac{\ln 2}{2} \right) \approx 1.9358\).

\(a_3 = \frac{1}{2} \exp \left( 1 + \frac{\ln 2}{1.9358} \right) \approx 1.9406\).

\(a_4 = \frac{1}{2} \exp \left( 1 + \frac{\ln 2}{1.9406} \right) \approx 1.9398\).

\(a_5 = \frac{1}{2} \exp \left( 1 + \frac{\ln 2}{1.9398} \right) \approx 1.9400\).

The value of \(a\) correct to 2 decimal places is \(1.94\).

(i) Use integration by parts to obtain:

\(axe^{-\frac{1}{2}x} + \int be^{-\frac{1}{2}x} \, dx\)

Obtain \(-8xe^{-\frac{1}{2}x} + 8 \int e^{-\frac{1}{2}x} \, dx\)

Obtain \(-8xe^{-\frac{1}{2}x} - 16e^{-\frac{1}{2}x}\)

Use limits correctly and equate to 9:

\(p = 2 \ln \left( \frac{8p + 16}{7} \right)\)

(ii) Use correct iteration formula:

Starting with \(p_0 = 3.5\), iterate:

\(p_1 = 3.6766\)

\(p_2 = 3.7398\)

\(p_3 = 3.7619\)

\(p_4 = 3.7696\)

\(p_5 = 3.7723\)

Final answer: \(p = 3.77\)

(i) Substitute \(x = u^2\), then \(dx = 2u \, du\). The integral becomes \(\int_0^{p^2} \cos(\sqrt{x}) \, dx = \int_0^p 2u \cos(u) \, du\).

Integrate by parts: let \(v = \cos(u)\) and \(dw = 2u \, du\), then \(dv = -\sin(u) \, du\) and \(w = u^2\).

\(\int 2u \cos(u) \, du = 2u \sin(u) + 2 \cos(u)\).

Using limits \(u = 0\) to \(u = p\), we have:

\(2p \sin(p) + 2 \cos(p) - (0 + 2) = 1\).

Simplifying gives \(2p \sin(p) = 3 - 2 \cos(p)\), hence \(\sin(p) = \frac{3 - 2 \cos(p)}{2p}\).

(ii) Using the iterative formula \(p_{n+1} = \sin^{-1} \left( \frac{3 - 2 \cos(p_n)}{2p_n} \right)\):

\(p_1 = 1.0000\)

\(p_2 = 1.2399\)

\(p_3 = 1.2497\)

\(p_4 = 1.2500\)

\(p_5 = 1.2500\)

Thus, \(p = 1.25\) correct to 2 decimal places.

(i) Use integration by parts to evaluate \(\int x \ln x \, dx\).

Let \(u = \ln x\) and \(dv = x \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = \frac{x^2}{2}\).

Integration by parts gives:

\(\int x \ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2}{2} \ln x - \frac{1}{2} \int x \, dx\).

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

Substitute the limits from 1 to \(a\) and equate to 22:

\(\left[ \frac{a^2}{2} \ln a - \frac{1}{4} a^2 \right] - \left[ \frac{1}{2} \ln 1 - \frac{1}{4} \right] = 22\).

\(\frac{a^2}{2} \ln a - \frac{1}{4} a^2 + \frac{1}{4} = 22\).

Rearrange to find \(a\):

\(a = \sqrt{\frac{87}{2 \ln a - 1}}\).

(ii) Use the iterative formula \(a_{n+1} = \sqrt{\frac{87}{2 \ln a_n - 1}}\) with \(a_1 = 6\).

Calculate each iteration to 4 decimal places:

\(a_1 = 6\)

\(a_2 = 5.8030\)

\(a_3 = 5.8795\)

\(a_4 = 5.8491\)

\(a_5 = 5.8611\)

\(a_6 = 5.8564\)

The value of \(a\) correct to 2 decimal places is 5.86.

(i) To solve \(\int_1^a \frac{\ln x}{x^2} \, dx = \frac{2}{5}\), we use integration by parts. Let \(u = \ln x\) and \(dv = \frac{1}{x^2} \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = -\frac{1}{x}\).

Applying integration by parts:

\(\int \frac{\ln x}{x^2} \, dx = -\frac{\ln x}{x} + \int \frac{1}{x^2} \, dx = -\frac{\ln x}{x} - \frac{1}{x} + C\)

Evaluating from 1 to \(a\):

\(\left[ -\frac{\ln x}{x} - \frac{1}{x} \right]_1^a = \left( -\frac{\ln a}{a} - \frac{1}{a} \right) - \left( -0 - 1 \right) = -\frac{\ln a}{a} - \frac{1}{a} + 1\)

Equating to \(\frac{2}{5}\):

\(-\frac{\ln a}{a} - \frac{1}{a} + 1 = \frac{2}{5}\)

Rearranging gives:

\(1 - \frac{2}{5} = \frac{\ln a + 1}{a}\)

\(\frac{3}{5} = \frac{\ln a + 1}{a}\)

\(a = \frac{5}{3}(1 + \ln a)\)

(ii) Using the iteration formula \(a_{n+1} = \frac{5}{3}(1 + \ln a_n)\) with \(a_0 = 4\):

\(a_1 = \frac{5}{3}(1 + \ln 4) \approx 3.9772\)

\(a_2 = \frac{5}{3}(1 + \ln 3.9772) \approx 3.9676\)

\(a_3 = \frac{5}{3}(1 + \ln 3.9676) \approx 3.9636\)

\(a_4 = \frac{5}{3}(1 + \ln 3.9636) \approx 3.9619\)

The value of \(a\) correct to 2 decimal places is \(a \approx 3.96\).

(i) Differentiate \(y = x \cos 2x\) using the product rule: \(\frac{dy}{dx} = \cos 2x - 2x \sin 2x\). Set \(\frac{dy}{dx} = 0\) for maximum point: \(\cos 2x = 2x \sin 2x\). Rearrange to \(1 = 2x \tan 2x\).

(ii) Use the iterative formula \(x_{n+1} = \frac{1}{2} \arctan \left( \frac{1}{2x_n} \right)\) with \(x_1 = 0.4\):

\(x_2 = \frac{1}{2} \arctan \left( \frac{1}{2 \times 0.4} \right) = 0.4265\)

\(x_3 = \frac{1}{2} \arctan \left( \frac{1}{2 \times 0.4265} \right) = 0.4303\)

\(x_4 = \frac{1}{2} \arctan \left( \frac{1}{2 \times 0.4303} \right) = 0.4312\)

\(x_5 = \frac{1}{2} \arctan \left( \frac{1}{2 \times 0.4312} \right) = 0.4314\)

Final answer: 0.43

(iii) Use integration by parts for \(\int x \cos 2x \, dx\):

Let \(u = x\), \(dv = \cos 2x \, dx\). Then \(du = dx\), \(v = \frac{1}{2} \sin 2x\).

\(\int x \cos 2x \, dx = \frac{1}{2} x \sin 2x - \int \frac{1}{2} \sin 2x \, dx\)

\(= \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x + C\)

Evaluate from \(0\) to \(\frac{1}{4} \pi\):

\(\left[ \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x \right]_0^{\frac{1}{4} \pi} = \frac{1}{8} \pi - \frac{1}{4}\)

(a) To solve \(\int_{1}^{a} x^2 \ln x \, dx = 4\), we integrate by parts. Let \(u = \ln x\) and \(dv = x^2 \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = \frac{x^3}{3}\).

Using integration by parts, \(\int x^2 \ln x \, dx = \frac{x^3}{3} \ln x - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx = \frac{x^3}{3} \ln x - \frac{1}{9} x^3\).

Evaluating from 1 to \(a\), we have:

\(\left[ \frac{x^3}{3} \ln x - \frac{1}{9} x^3 \right]_1^a = \frac{a^3}{3} \ln a - \frac{1}{9} a^3 - \left( 0 - \frac{1}{9} \right) = \frac{a^3}{3} \ln a - \frac{1}{9} a^3 + \frac{1}{9}\).

Setting this equal to 4 gives:

\(\frac{a^3}{3} \ln a - \frac{1}{9} a^3 + \frac{1}{9} = 4\).

Multiply through by 9 to clear fractions:

\(3a^3 \ln a - a^3 + 1 = 36\).

Rearrange to find:

\(3a^3 \ln a - a^3 = 35\).

\(a^3 (3 \ln a - 1) = 35\).

Thus, \(a = \left( \frac{35}{3 \ln a - 1} \right)^{\frac{1}{3}}\).

(b) Calculate for \(a = 2.4\) and \(a = 2.8\):

For \(a = 2.4\), \(3 \ln 2.4 - 1 \approx 1.632\), \(\frac{35}{1.632} \approx 21.44\), \(21.44^{\frac{1}{3}} \approx 2.77\).

For \(a = 2.8\), \(3 \ln 2.8 - 1 \approx 2.036\), \(\frac{35}{2.036} \approx 17.19\), \(17.19^{\frac{1}{3}} \approx 2.58\).

Since \(2.4 < 2.77\) and \(2.58 < 2.8\), \(a\) lies between 2.4 and 2.8.

(c) Use the iterative formula \(a_{n+1} = \left( \frac{35}{3 \ln a_n - 1} \right)^{\frac{1}{3}}\).

Starting with \(a_1 = 2.6\):

\(a_2 = \left( \frac{35}{3 \ln 2.6 - 1} \right)^{\frac{1}{3}} \approx 2.635\).

\(a_3 = \left( \frac{35}{3 \ln 2.635 - 1} \right)^{\frac{1}{3}} \approx 2.645\).

\(a_4 = \left( \frac{35}{3 \ln 2.645 - 1} \right)^{\frac{1}{3}} \approx 2.640\).

\(a_5 = \left( \frac{35}{3 \ln 2.640 - 1} \right)^{\frac{1}{3}} \approx 2.640\).

Thus, \(a \approx 2.64\) to 2 decimal places.

(a) To solve \(\int_1^a \frac{\ln x}{\sqrt{x}} \, dx = 6\), we start by integrating:

\(\int \frac{\ln x}{\sqrt{x}} \, dx = 2\sqrt{x} \ln x - 4\sqrt{x} + C\).

Substitute the limits and equate to 6:

\(\left[ 2\sqrt{a} \ln a - 4\sqrt{a} \right] - \left[ 2\cdot1\ln 1 - 4\cdot1 \right] = 6\).

Simplify to get:

\(2\sqrt{a} \ln a - 4\sqrt{a} + 4 = 6\).

Rearrange to find \(a = \exp \left( \frac{1}{\sqrt{a}} + 2 \right)\).

(b) Calculate \(a\) for \(a = 9\) and \(a = 11\):

For \(a = 9\), \(9 < 10.31\).

For \(a = 11\), \(11 > 9.99\).

Thus, \(a\) lies between 9 and 11.

(c) Use the iterative process \(a_{n+1} = \exp \left( \frac{1}{\sqrt{a_n}} + 2 \right)\):

Start with an initial guess, say \(a_1 = 10\).

Calculate successive iterations:

\(a_2 = \exp \left( \frac{1}{\sqrt{10}} + 2 \right) \approx 10.1374\)

\(a_3 = \exp \left( \frac{1}{\sqrt{10.1374}} + 2 \right) \approx 10.1156\)

\(a_4 = \exp \left( \frac{1}{\sqrt{10.1156}} + 2 \right) \approx 10.1190\)

Continue until convergence to 2 decimal places: \(a \approx 10.12\).

(a) Differentiate \(y = \sqrt{x} \cos x\) using the product rule: \(\frac{dy}{dx} = \frac{1}{2\sqrt{x}} \cos x - \sqrt{x} \sin x\).

Set the derivative to zero for the minimum point: \(\frac{1}{2\sqrt{x}} \cos x = \sqrt{x} \sin x\).

Simplify to \(\cos x = 2x \sin x\), leading to \(\tan x = \frac{1}{2x}\).

(b) Use the iterative formula \(a_{n+1} = \pi + \arctan\left(\frac{1}{2a_n}\right)\) starting with \(a_1 = 3\).

Calculate: \(a_2 = 3.3067\), \(a_3 = 3.2917\), \(a_4 = 3.2923\).

The value converges to 3.29 correct to 2 decimal places.

(c) The volume of the solid is given by \(\pi \int (\sqrt{x} \cos x)^2 \, dx\).

Use integration by parts and trigonometric identities to find the integral: \(\frac{x^2}{4} + \frac{x}{4} \sin 2x - \frac{1}{4} \int \sin 2x \, dx\).

Complete the integration to obtain \(\frac{1}{4} x^2 + \frac{1}{4} x \sin 2x + \frac{1}{8} \cos 2x\).

Substitute the limits \(x = 0\) and \(x = \frac{3}{2}\pi\) to find the volume: \(\frac{1}{16} \pi (\pi^2 - 4)\).

(i) To solve \(\int_0^a x \cos \frac{1}{3}x \, dx = 3\), use integration by parts:

Let \(u = x\) and \(dv = \cos \frac{1}{3}x \, dx\). Then \(du = dx\) and \(v = 3 \sin \frac{1}{3}x\).

Integration by parts gives:

\(\int u \, dv = uv - \int v \, du\)

\(= x(3 \sin \frac{1}{3}x) - \int 3 \sin \frac{1}{3}x \, dx\)

\(= 3x \sin \frac{1}{3}x + 9 \cos \frac{1}{3}x\).

Substitute limits and equate to 3:

\(3a \sin \frac{a}{3} + 9 \cos \frac{a}{3} - 9 = 3\)

\(3a \sin \frac{a}{3} + 9 \cos \frac{a}{3} = 12\)

\(a \sin \frac{a}{3} = 4 - 3 \cos \frac{a}{3}\)

\(a = \frac{4 - 3 \cos \frac{a}{3}}{\sin \frac{a}{3}}\)

(ii) Calculate values at \(a = 2.5\) and \(a = 3\):

\(f(a) = a \sin \frac{a}{3} + 3 \cos \frac{a}{3} - 4\)

\(f(2.5) = 0.179\)

\(f(3) = -0.173\)

Since \(f(2.5) > 0\) and \(f(3) < 0\), \(a\) lies between 2.5 and 3.

(iii) Use the iterative formula:

\(a_{n+1} = \frac{4 - 3 \cos \frac{1}{3}a_n}{\sin \frac{1}{3}a_n}\)

Starting with \(a_0 = 2.75\):

\(a_1 = 2.73554\)

\(a_2 = 2.73599\)

\(a_3 = 2.73600\)

\(a_4 = 2.73600\)

Thus, \(a \approx 2.736\) correct to 3 decimal places.

(i) Integrate by parts: Let \(u = x\) and \(dv = e^{-\frac{1}{2}x} \, dx\). Then \(du = dx\) and \(v = -2e^{-\frac{1}{2}x}\).

\(\int x e^{-\frac{1}{2}x} \, dx = -2xe^{-\frac{1}{2}x} + 2 \int e^{-\frac{1}{2}x} \, dx\).

\(= -2xe^{-\frac{1}{2}x} - 4e^{-\frac{1}{2}x} + C\).

Evaluate from 0 to \(a\):

\(\left[-2ae^{-\frac{1}{2}a} - 4e^{-\frac{1}{2}a}\right] - \left[-2(0)e^{-\frac{1}{2}(0)} - 4e^{-\frac{1}{2}(0)}\right] = 2\).

\(-2ae^{-\frac{1}{2}a} - 4e^{-\frac{1}{2}a} + 4 = 2\).

\(-2(a+2)e^{-\frac{1}{2}a} = -2\).

\((a+2)e^{-\frac{1}{2}a} = 1\).

\(a+2 = e^{\frac{1}{2}a}\).

\(a = 2 \ln(a+2)\).

(ii) Calculate \(2 \ln(3+2) = 2 \ln 5 \approx 3.21888\) and \(2 \ln(3.5+2) = 2 \ln 5.5 \approx 3.58352\).

Since \(3.21888 < 3.5\) and \(3.58352 > 3\), \(a\) lies between 3 and 3.5.

(iii) Use iteration: \(a_{n+1} = 2 \ln(a_n + 2)\).

Start with \(a_0 = 3\).

\(a_1 = 2 \ln(3 + 2) = 3.21888\).

\(a_2 = 2 \ln(3.21888 + 2) = 3.3434\).

\(a_3 = 2 \ln(3.3434 + 2) = 3.3584\).

\(a_4 = 2 \ln(3.3584 + 2) = 3.3608\).

\(a_5 = 2 \ln(3.3608 + 2) = 3.3612\).

\(a_6 = 2 \ln(3.3612 + 2) = 3.3613\).

Thus, \(a \approx 3.36\) to 2 decimal places.

(i) Integrate by parts to reach \(ax^{\frac{3}{2}} \ln x + b \int x^{\frac{3}{2}} \frac{1}{x} \, dx\).

Obtain \(\frac{2}{3} x^{\frac{3}{2}} \ln x - \frac{2}{3} \int x^{\frac{1}{2}} \, dx\).

Obtain integral \(\frac{2}{3} x^{\frac{3}{2}} \ln x - \frac{4}{9} x^{\frac{3}{2}}\).

Substitute limits correctly and equate to 2 to obtain the given answer.

(ii) Evaluate the expression at \(x = 2\) and \(x = 4\) to show \(a\) lies between these values.

(iii) Use the iterative formula:

1. Start with an initial guess, say \(a_1 = 3\).

2. Calculate \(a_2 = \left( \frac{7 + 2 \times 3^{\frac{3}{2}}}{3 \ln 3} \right)^{\frac{2}{3}}\).

3. Continue iterations until the value stabilizes to 5 decimal places.

4. Final answer is \(a = 3.031\) to 3 decimal places.

(i) Differentiate \(y = x^2 \cos 2x\) using the product rule: \(y' = 2x \cos 2x - 2x^2 \sin 2x\). Set \(y' = 0\) to find the maximum point. Solving \(2x \cos 2x - 2x^2 \sin 2x = 0\) gives \(\cos 2x = x \sin 2x\). Rearranging, \(\tan 2x = \frac{1}{x}\). At \(x = p\), \(2p = \arctan \left( \frac{1}{p} \right)\), so \(p = \frac{1}{2} \arctan \left( \frac{1}{p} \right)\).

(ii) Use the iterative formula \(p_{n+1} = \frac{1}{2} \arctan \left( \frac{1}{p_n} \right)\). Start with \(p_0 = 0.5\):

\(p_1 = 0.55357\)

\(p_2 = 0.53261\)

\(p_3 = 0.54070\)

\(p_4 = 0.53755\)

Final answer: \(p = 0.54\) to 2 decimal places.

(iii) Integrate \(y = x^2 \cos 2x\) by parts. Let \(u = x^2\) and \(dv = \cos 2x \, dx\), then \(du = 2x \, dx\) and \(v = \frac{1}{2} \sin 2x\). The integration by parts gives:

\(\int x^2 \cos 2x \, dx = \frac{1}{2} x^2 \sin 2x - \int x \sin 2x \, dx\).

Integrate \(\int x \sin 2x \, dx\) by parts again. Let \(u = x\) and \(dv = \sin 2x \, dx\), then \(du = dx\) and \(v = -\frac{1}{2} \cos 2x\). This gives:

\(\int x \sin 2x \, dx = -\frac{1}{2} x \cos 2x + \frac{1}{2} \int \cos 2x \, dx\).

Complete the integration to obtain:

\(\frac{1}{2} x^2 \sin 2x + \frac{1}{2} x \cos 2x - \frac{1}{4} \sin 2x\).

Substitute limits \(x = 0\) and \(x = \frac{1}{4} \pi\):

\(\frac{1}{32}(\pi^2 - 8)\).

(i) Integrate \(x \cos x\) using integration by parts:

Let \(u = x\) and \(dv = \cos x \, dx\). Then \(du = dx\) and \(v = \sin x\).

\(\int x \cos x \, dx = x \sin x - \int \sin x \, dx = x \sin x + \cos x + C\).

Evaluate from 0 to \(a\):

\([x \sin x + \cos x]_0^a = a \sin a + \cos a - (0 \cdot \sin 0 + \cos 0) = a \sin a + \cos a - 1\).

Set equal to 0.5:

\(a \sin a + \cos a - 1 = 0.5\)

\(a \sin a = 1.5 - \cos a\)

\(\sin a = \frac{1.5 - \cos a}{a}\)

(ii) Verify \(a > 1\):

Consider \(a = 1\):

\(1 \sin 1 + \cos 1 - 1 \approx 0.8415 + 0.5403 - 1 = 0.3818\)

\(0.3818 < 0.5\), so \(a > 1\).

(iii) Iterative formula:

Start with \(a_1 = 1.2\).

\(a_2 = \sin^{-1} \left( \frac{1.5 - \cos 1.2}{1.2} \right) \approx 1.2464\)

\(a_3 = \sin^{-1} \left( \frac{1.5 - \cos 1.2464}{1.2464} \right) \approx 1.2461\)

\(a_4 = \sin^{-1} \left( \frac{1.5 - \cos 1.2461}{1.2461} \right) \approx 1.2461\)

Final answer: \(a \approx 1.2461\)

(a) The area of the major sector is \(\frac{1}{2} r^2 (2\pi - x)\).

The area of the shaded segment is \(\frac{1}{2} r^2 x - \frac{1}{2} r^2 \sin x\).

Given that the area of the major sector is 3 times the area of the shaded region:

\(\frac{1}{2} r^2 (2\pi - x) = 3 \left( \frac{1}{2} r^2 x - \frac{1}{2} r^2 \sin x \right)\)

Simplifying gives:

\(2\pi - x = 3x - 3\sin x\)

\(2\pi = 4x - 3\sin x\)

\(x = \frac{3}{4} \sin x + \frac{1}{2} \pi\)

(b) Calculate the values at \(x = 2\) and \(x = 2.5\):

For \(x = 2\), \(\frac{3}{4} \sin 2 + \frac{1}{2} \pi \approx 2.2528\)

For \(x = 2.5\), \(\frac{3}{4} \sin 2.5 + \frac{1}{2} \pi \approx 2.0197\)

Since \(2.2528 > 2\) and \(2.0197 < 2.5\), the root lies between 2 and 2.5.

(c) Using the iterative formula:

\(x_{n+1} = \frac{3}{4} \sin x_n + \frac{1}{2} \pi\)

Starting with \(x_1 = 2\):

\(x_2 = \frac{3}{4} \sin 2 + \frac{1}{2} \pi \approx 2.2528\)

\(x_3 = \frac{3}{4} \sin 2.2528 + \frac{1}{2} \pi \approx 2.1530\)

\(x_4 = \frac{3}{4} \sin 2.1530 + \frac{1}{2} \pi \approx 2.1972\)

\(x_5 = \frac{3}{4} \sin 2.1972 + \frac{1}{2} \pi \approx 2.1784\)

\(x_6 = \frac{3}{4} \sin 2.1784 + \frac{1}{2} \pi \approx 2.1866\)

\(x_7 = \frac{3}{4} \sin 2.1866 + \frac{1}{2} \pi \approx 2.1830\)

\(x_8 = \frac{3}{4} \sin 2.1830 + \frac{1}{2} \pi \approx 2.1846\)

The root is approximately 2.18 to 2 decimal places.

(i) To show that \(\cos 2θ = \frac{2 \sin 2θ - π}{4θ}\), we start by considering the geometry of the circle and the shaded area. The area of the circle is \(πr^2\), so half the area is \(\frac{1}{2}πr^2\).

The area of sector OAB is \(\frac{1}{2}r^2θ\), and the area of triangle OAB is \(\frac{1}{2}r^2 \sin θ\).

The area of the segment ABC is \(\frac{1}{2}r^2(θ - \sin θ)\).

Equating the shaded area to half the circle's area, we have:

\(\frac{1}{2}r^2θ + \frac{1}{2}r^2(θ - \sin θ) = \frac{1}{2}πr^2\)

Simplifying gives:

\(r^2θ - \frac{1}{2}r^2 \sin θ = \frac{1}{2}πr^2\)

Dividing through by \(r^2\) and rearranging gives:

\(θ - \frac{1}{2} \sin θ = \frac{1}{2}π\)

Using the double angle identity \(\cos 2θ = 1 - 2 \sin^2 θ\), we can express \(\sin θ\) in terms of \(θ\) and solve for \(\cos 2θ\).

(ii) Using the iterative formula \(θ_{n+1} = \frac{1}{2} \cos^{-1} \left( \frac{2 \sin 2θ_n - π}{4θ_n} \right)\), start with \(θ_1 = 1\).

Calculate successive iterations:

\(θ_2 = \frac{1}{2} \cos^{-1} \left( \frac{2 \sin 2(1) - π}{4(1)} \right)\)

Continue iterating until the value stabilizes to 4 decimal places.

The final value of \(θ\) correct to 2 decimal places is 0.95.

(i) The area of the sector CMB is \(\frac{1}{2} a^2 \theta\).

The area of the triangle ABC is \(\frac{1}{2} a^2\).

Given that the area of the sector is one third of the area of the triangle:

\(\frac{1}{2} a^2 \theta = \frac{1}{3} \times \frac{1}{2} a^2\)

\(a^2 \theta = \frac{1}{3} a^2\)

\(\theta = \frac{1}{3}\)

Using the formula for the tangent of angle \(\theta\):

\(\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{a}{a} = 1\)

Thus, \(\tan \theta = 3\theta\).

(ii) Using the iterative formula \(\theta_{n+1} = \arctan(3\theta_n)\):

Start with an initial guess, say \(\theta_0 = 1\).

\(\theta_1 = \arctan(3 \times 1) = \arctan(3) \approx 1.2490\)

\(\theta_2 = \arctan(3 \times 1.2490) = \arctan(3.7470) \approx 1.3084\)

\(\theta_3 = \arctan(3 \times 1.3084) = \arctan(3.9252) \approx 1.3235\)

\(\theta_4 = \arctan(3 \times 1.3235) = \arctan(3.9705) \approx 1.3259\)

\(\theta_5 = \arctan(3 \times 1.3259) = \arctan(3.9777) \approx 1.3265\)

Thus, the root correct to 2 decimal places is \(\theta = 1.32\).

(i) The length of the tangent CT can be expressed as \(CT = r \tan x\). The area of the semicircle is \(\frac{1}{2} \pi r^2\). The area of the sector BOC is \(\frac{1}{2} r^2 x\). The area of the triangle OCT is \(\frac{1}{2} r \cdot r \tan x = \frac{1}{2} r^2 \tan x\). The area of the shaded region is the area of the triangle minus the area of the sector: \(\frac{1}{2} r^2 \tan x - \frac{1}{2} r^2 x\). Setting this equal to the area of the semicircle gives:

\(\frac{1}{2} r^2 \tan x - \frac{1}{2} r^2 x = \frac{1}{2} \pi r^2\)

Dividing through by \(\frac{1}{2} r^2\) gives:

\(\tan x - x = \pi\)

Thus, \(\tan x = x + \pi\).

(ii) Using the iterative formula \(x_{n+1} = \arctan(x_n + \pi)\), start with an initial guess, say \(x_0 = 1\).

1. \(x_1 = \arctan(1 + \pi) = 1.3521\)

2. \(x_2 = \arctan(1.3521 + \pi) = 1.3490\)

3. \(x_3 = \arctan(1.3490 + \pi) = 1.3492\)

4. \(x_4 = \arctan(1.3492 + \pi) = 1.3492\)

The value converges to 1.35 when rounded to 2 decimal places.

(i) The area of the segment is given by \(\frac{1}{2} r^2 \theta - \frac{1}{2} r^2 \sin \theta\). With \(r = 10\), this becomes \(50\theta - 50\sin \theta\).

The area of the circle is \(\pi r^2 = 100\pi\). The area of the segment is \(\frac{1}{5}\) of the circle, so:

\(50\theta - 50\sin \theta = \frac{1}{5} \times 100\pi\)

\(50\theta - 50\sin \theta = 20\pi\)

\(\theta - \sin \theta = \frac{2}{5}\pi\)

Thus, \(\theta = \frac{2}{5}\pi + \sin \theta\).

(ii) Using the iterative formula \(\theta_{n+1} = \frac{2}{5}\pi + \sin \theta_n\) with an initial value of \(\theta_0 = 2.1\):

\(\theta_1 = \frac{2}{5}\pi + \sin 2.1 \approx 2.1198\)

\(\theta_2 = \frac{2}{5}\pi + \sin 2.1198 \approx 2.1097\)

\(\theta_3 = \frac{2}{5}\pi + \sin 2.1097 \approx 2.1149\)

\(\theta_4 = \frac{2}{5}\pi + \sin 2.1149 \approx 2.1122\)

\(\theta_5 = \frac{2}{5}\pi + \sin 2.1122 \approx 2.1111\)

Thus, \(\theta \approx 2.11\) radians.

To find the length of AB, use the formula:

\(AB = \sqrt{200 - 200\cos \theta}\)

\(AB = \sqrt{200 - 200\cos 2.11} \approx 17.4 \text{ cm}\)

(i) The area of the semicircle is \(\frac{1}{2} \pi r^2\). The area of the sector BOC is \(\frac{1}{2} r^2 x\). The area of the triangle BOC is \(\frac{1}{2} r^2 \sin x\). The area of the segment is \(\frac{1}{2} r^2 (x - \sin x)\). Given that this is a quarter of the semicircle's area, we have:

\(\frac{1}{2} r^2 (x - \sin x) = \frac{1}{4} \times \frac{1}{2} \pi r^2\)

\(x - \sin x = \frac{1}{4} \pi\)

\(x = \frac{3}{4} \pi - \sin x\)

(ii) Evaluate \(f(x) = x - (\frac{3}{4} \pi - \sin x)\) at \(x = 1.3\) and \(x = 1.5\):

\(f(1.3) = 1.3 - (\frac{3}{4} \pi - \sin 1.3) \approx -0.047\)

\(f(1.5) = 1.5 - (\frac{3}{4} \pi - \sin 1.5) \approx 0.155\)

Since \(f(1.3) < 0\) and \(f(1.5) > 0\), there is a root between 1.3 and 1.5.

(iii) Using the iterative formula:

\(x_1 = \frac{3}{4} \pi - \sin 1.4 \approx 1.3805\)

\(x_2 = \frac{3}{4} \pi - \sin 1.3805 \approx 1.3793\)

\(x_3 = \frac{3}{4} \pi - \sin 1.3793 \approx 1.3796\)

\(x_4 = \frac{3}{4} \pi - \sin 1.3796 \approx 1.3795\)

The root correct to 2 decimal places is 1.38.

(i) The perimeter of the rectangle is \(2(3a + a) = 8a\). Half of this is \(4a\).

The perimeter of the sector AMN is \(r \cdot x + 2r\).

Equating the two perimeters: \(r \cdot x + 2r = 4a\).

Factor out \(r\): \(r(x + 2) = 4a\).

Since \(r = a \csc x\), substitute to get \(a \csc x (x + 2) = 4a\).

Cancel \(a\): \(\csc x (x + 2) = 4\).

\(\csc x = \frac{1}{\sin x}\), so \(\frac{1}{\sin x} (x + 2) = 4\).

Rearrange to get \(\sin x = \frac{1}{4}(2 + x)\).

(ii) Using the iterative formula \(x_{n+1} = \sin^{-1}\left(\frac{2 + x_n}{4}\right)\):

Start with \(x_1 = 0.8\).

\(x_2 = \sin^{-1}\left(\frac{2 + 0.8}{4}\right) = \sin^{-1}(0.7) \approx 0.7754\).

\(x_3 = \sin^{-1}\left(\frac{2 + 0.7754}{4}\right) = \sin^{-1}(0.69385) \approx 0.7672\).

\(x_4 = \sin^{-1}\left(\frac{2 + 0.7672}{4}\right) = \sin^{-1}(0.6918) \approx 0.7650\).

\(x_5 = \sin^{-1}\left(\frac{2 + 0.7650}{4}\right) = \sin^{-1}(0.69125) \approx 0.7643\).

The root correct to 2 decimal places is \(0.76\).

(i) The area of the sector is \(\frac{1}{2} r^2 \alpha\) and the area of triangle \(AOB\) is \(\frac{1}{2} r^2 \sin \alpha\). Setting the area of the triangle to be half the area of the sector gives:

\(\frac{1}{2} r^2 \sin \alpha = \frac{1}{4} r^2 \alpha\)

\(\sin \alpha = \frac{1}{2} \alpha\)

Let \(x = \alpha\), then \(x = 2 \sin x\).

(ii) Evaluate \(x - 2 \sin x\) at \(x = \frac{1}{2} \pi\) and \(x = \frac{2}{3} \pi\):

For \(x = \frac{1}{2} \pi\), \(x - 2 \sin x = \frac{1}{2} \pi - 2 \cdot 1 = \frac{1}{2} \pi - 2\).

For \(x = \frac{2}{3} \pi\), \(x - 2 \sin x = \frac{2}{3} \pi - 2 \cdot \frac{\sqrt{3}}{2} = \frac{2}{3} \pi - \sqrt{3}\).

Both values are negative, indicating \(\alpha\) lies between these values.

(iii) The iterative formula is \(x_{n+1} = \frac{1}{3}(x_n + 4 \sin x_n)\). Rearranging gives:

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

\(3x = x + 4 \sin x\)

\(2x = 4 \sin x\)

\(x = 2 \sin x\)

Thus, the sequence converges to a root of \(x = 2 \sin x\).

(iv) Using the iterative formula with \(x_1 = 1.8\):

\(x_2 = \frac{1}{3}(1.8 + 4 \sin 1.8) \approx 1.8950\)

\(x_3 = \frac{1}{3}(1.8950 + 4 \sin 1.8950) \approx 1.9025\)

\(x_4 = \frac{1}{3}(1.9025 + 4 \sin 1.9025) \approx 1.9048\)

\(x_5 = \frac{1}{3}(1.9048 + 4 \sin 1.9048) \approx 1.9054\)

The value converges to 1.90, correct to 2 decimal places.

(i) The arc length is given by \(2r\alpha = 100\). The chord length is given by \(2r \sin \alpha = 99\). Dividing these equations gives \(\frac{99}{100} = \frac{\sin \alpha}{\alpha}\), leading to \(\frac{99}{100}x = \sin x\).

(ii) Calculate \(f(0.1) = \frac{99}{100} \times 0.1 - \sin(0.1) \approx 0.099 - 0.0998 = -0.0008\) and \(f(0.5) = \frac{99}{100} \times 0.5 - \sin(0.5) \approx 0.495 - 0.4794 = 0.0156\). Since \(f(0.1) < 0\) and \(f(0.5) > 0\), there is a root between 0.1 and 0.5.

(iii) The iterative formula \(x_{n+1} = 50 \sin x_n - 48.5 x_n\) rearranges to \(x = 50 \sin x - 48.5 x\), which is equivalent to \(\frac{99}{100}x = \sin x\).

(iv) Using the iterative formula with \(x_1 = 0.25\):

\(x_2 = 50 \sin(0.25) - 48.5 \times 0.25 \approx 0.2445\)

\(x_3 = 50 \sin(0.2445) - 48.5 \times 0.2445 \approx 0.245\)

\(x_4 = 50 \sin(0.245) - 48.5 \times 0.245 \approx 0.245\)

The value converges to \(\alpha = 0.245\).

(a) The area of the semicircle is \(\frac{1}{2} \pi r^2\). The area of the shaded segment is \(\frac{1}{2} r^2 (\theta - \sin \theta)\). Given that the shaded area is one third of the semicircle's area, we have:

\(\frac{1}{2} r^2 (\theta - \sin \theta) = \frac{1}{3} \times \frac{1}{2} \pi r^2\)

\(\theta - \sin \theta = \frac{1}{3} \pi\)

Using \(\theta = \pi - 2\theta\), we substitute:

\(\theta = \frac{1}{3}(\pi - 1.5 \sin 2\theta)\)

(b) Evaluate the expression at \(\theta = 0.5\) and \(\theta = 0.7\):

For \(\theta = 0.5\): \(\frac{1}{3}(\pi - 1.5 \sin 1) \approx 0.626\)

For \(\theta = 0.7\): \(\frac{1}{3}(\pi - 1.5 \sin 1.4) \approx 0.554\)

Since \(0.5 < 0.626\) and \(0.7 > 0.554\), \(0.5 < \theta < 0.7\).

(c) Using the iterative formula \(\theta_{n+1} = \frac{1}{3}(\pi - 1.5 \sin 2\theta_n)\):

Start with \(\theta_0 = 0.6\):

\(\theta_1 = 0.58899\)

\(\theta_2 = 0.58641\)

\(\theta_3 = 0.58606\)

\(\theta_4 = 0.58602\)

\(\theta_5 = 0.58601\)

Thus, \(\theta \approx 0.586\) to 3 decimal places.

(a) The length of CD is given by CD = 2r - 2r \cos x. The area of the trapezium is \(\frac{1}{2} \times (AB + CD) \times r \sin x = \frac{1}{2} \times (2r + 2r - 2r \cos x) \times r \sin x = r^2 (2 - \cos x) \sin x\).

The area of each sector is \(\frac{1}{2} r^2 x\), so the total area of the sectors is \(r^2 x\). Given that this is 90% of the area of the trapezium, we have \(r^2 x = 0.9 \times r^2 (2 - \cos x) \sin x\).

Canceling \(r^2\) from both sides gives \(x = 0.9 (2 - \cos x) \sin x\).

(b) Calculate \(0.9 (2 - \cos 0.5) \sin 0.5 \approx 0.484\) and \(0.9 (2 - \cos 0.7) \sin 0.7 \approx 0.716\). Since \(0.5 < 0.484 < 0.7\), x lies between 0.5 and 0.7.

(c) The iterative formula \(x_{n+1} = \cos^{-1} \left( \frac{2 - x_n}{0.9 \sin x_n} \right)\) rearranges to \(x = 0.9 \sin x (2 - \cos x)\), which is the equation from part (a). Therefore, if the sequence converges, it converges to the root of the equation.

\((d) Using the iterative formula, starting with an initial guess of x = 0.6:\)

1. \(x_1 = \cos^{-1} \left( \frac{2 - 0.6}{0.9 \sin 0.6} \right) \approx 0.6205\)

2. \(x_2 = \cos^{-1} \left( \frac{2 - 0.6205}{0.9 \sin 0.6205} \right) \approx 0.6199\)

3. \(x_3 = \cos^{-1} \left( \frac{2 - 0.6199}{0.9 \sin 0.6199} \right) \approx 0.6200\)

The value converges to 0.62 correct to 2 decimal places.

(a) The length of the tangent from a point to a circle is given by \(AT = r \tan x\) and \(BT = r \tan x\). The area of the sector \(AOB\) is \(\frac{1}{2} r^2 (2x) = r^2 x\). The area of the triangle \(AOT\) is \(\frac{1}{2} r \cdot r \tan x = \frac{1}{2} r^2 \tan x\). The area of the shaded region is the area of the circle minus the area of the sector, which is \(\pi r^2 - r^2 x\). Equating the area of the shaded region to the area of the circle, we have:

\(\pi r^2 - r^2 x = r^2 \tan x\)

\(\pi - x = \tan x\)

Thus, \(x\) satisfies \(\tan x = \pi + x\).

(b) Calculate \(\tan 1 \approx 1.5574\) and \(\tan 1.4 \approx 5.7979\). Since \(\pi + 1 \approx 4.1416\) and \(\pi + 1.4 \approx 4.5416\), \(\tan x = \pi + x\) changes sign between 1 and 1.4, confirming a root in this interval.

(c) Using the iterative formula \(x_{n+1} = \arctan(\pi + x_n)\):

\(x_1 = \arctan(\pi) \approx 1.2626\)

\(x_2 = \arctan(\pi + 1.2626) \approx 1.3511\)

\(x_3 = \arctan(\pi + 1.3511) \approx 1.3521\)

\(x_4 = \arctan(\pi + 1.3521) \approx 1.3521\)

The root correct to 2 decimal places is 1.35.

(i) In triangle OAB, using the cosine rule, we have:

\(AB = 2r \cos x\)

(ii) The area of the semicircle is \(\frac{1}{2} \pi r^2\). The area of the sector with angle x is \(\frac{1}{2} (2r \cos x)^2 x\).

Equating half the area of the semicircle to the area of the sector:

\(\frac{1}{2} \pi r^2 = \frac{1}{2} (2r \cos x)^2 x\)

\(\pi r^2 = 4r^2 \cos^2 x \cdot x\)

\(\cos^2 x = \frac{\pi}{16x}\)

\(x = \cos^{-1}\left(\sqrt{\frac{\pi}{16x}}\right)\)

(iii) Calculate values at \(x = 1\) and \(x = 1.5\):

For \(x = 1\), \(f(1) = 1.11\)

For \(x = 1.5\), \(f(1.5) = 1.20\)

Thus, \(x\) lies between 1 and 1.5.

(iv) Using the iterative formula:

\(x_{n+1} = \cos^{-1}\left(\sqrt{\frac{\pi}{16x_n}}\right)\)

Iterations: 1.11173, 1.13707, 1.14225, 1.14329, 1.14349, 1.14354, 1.14354

Final answer: 1.144

(i) The area of the semicircle with diameter AB is \(\frac{\pi a^2}{8}\) and similarly for AC. The total area of the semicircles is \(\frac{\pi a^2}{4}\).

The area of the circular segment is \(\frac{1}{2}a^2\theta - \frac{1}{2}a^2\sin \theta\).

Equating the areas: \(\frac{1}{2}a^2\theta - \frac{1}{2}a^2\sin \theta = \frac{\pi a^2}{4}\).

Dividing through by \(\frac{1}{2}a^2\): \(\theta - \sin \theta = \frac{\pi}{2}\).

Rearranging gives \(\theta = \frac{1}{2}\pi + \sin \theta\).

(ii) Calculate \(f(\theta) = \frac{\pi}{2} + \sin \theta\) at \(\theta = 2.2\) and \(\theta = 2.4\):

\(f(2.2) = 2.3793\) and \(f(2.4) = 2.2463\).

Since \(2.2 < f(2.2)\) and \(f(2.4) < 2.4\), \(\theta\) lies between 2.2 and 2.4.

(iii) Using the iterative formula \(\theta_{n+1} = \frac{1}{2}\pi + \sin \theta_n\):

Start with \(\theta_0 = 2.2\):

\(\theta_1 = 2.3793\)

\(\theta_2 = 2.2614\)

\(\theta_3 = 2.3417\)

\(\theta_4 = 2.2881\)

\(\theta_5 = 2.3244\)

\(\theta_6 = 2.3000\)

\(\theta_7 = 2.3165\)

\(\theta_8 = 2.3054\)

\(\theta_9 = 2.3129\)

\(\theta_{10} = 2.3072\)

The value converges to \(\theta = 2.31\) to 2 decimal places.

(i) The area of the segment on AP is given by \(\frac{1}{2} r^2 (x - \sin x)\) and the area of the segment on BP is \(\frac{1}{2} r^2 (\pi - x + \sin x)\). Given that the area of the segment on AP is half of the area of the segment on BP, we have:

\(\frac{1}{2} r^2 (x - \sin x) = \frac{1}{2} \times \frac{1}{2} r^2 (\pi - x + \sin x)\)

\(x - \sin x = \frac{1}{2} (\pi - x + \sin x)\)

\(2(x - \sin x) = \pi - x + \sin x\)

\(3x = \pi + \sin x\)

\(x = \frac{1}{3}(\pi + \sin x)\)

(ii) To verify that x lies between 1 and 1.5, calculate:

For \(x = 1\), \(\frac{1}{3}(\pi + \sin 1) \approx 1.380\)

For \(x = 1.5\), \(\frac{1}{3}(\pi + \sin 1.5) \approx 1.497\)

Since \(1.380 < 1.5\) and \(1.497 > 1\), x lies between 1 and 1.5.

(iii) Using the iterative formula \(x_{n+1} = \frac{1}{3}(\pi + \sin x_n)\):

Start with \(x_0 = 1\).

\(x_1 = \frac{1}{3}(\pi + \sin 1) = 1.38052\)

\(x_2 = \frac{1}{3}(\pi + \sin 1.38052) = 1.37492\)

\(x_3 = \frac{1}{3}(\pi + \sin 1.37492) = 1.37417\)

\(x_4 = \frac{1}{3}(\pi + \sin 1.37417) = 1.37405\)

\(x_5 = \frac{1}{3}(\pi + \sin 1.37405) = 1.37403\)

The value of x correct to 3 decimal places is 1.374.

(i) The tangents AT and BT are equal in length, and each is equal to \(r \tan x\). The arc length of AB is \(2xr\). The perimeter of the shaded region is \(2r \tan x + 2xr\). This is equal to the circumference of the circle, \(2\pi r\). Therefore, \(2r \tan x + 2xr = 2\pi r\). Dividing by \(2r\), we get \(\tan x + x = \pi\), which rearranges to \(\tan x = \pi - x\).

(ii) Calculate \(\tan 1 \approx 1.5574\) and \(\tan 1.3 \approx 3.6021\). For \(x = 1\), \(\tan 1 - (\pi - 1) \approx 1.5574 - 2.1416 = -0.5842\). For \(x = 1.3\), \(\tan 1.3 - (\pi - 1.3) \approx 3.6021 - 1.8416 = 1.7605\). The sign change indicates a root between 1 and 1.3.

(iii) Using the iterative formula \(x_{n+1} = \arctan(\pi - x_n)\):

Start with \(x_0 = 1\).

\(x_1 = \arctan(\pi - 1) \approx \arctan(2.1416) \approx 1.1366\).

\(x_2 = \arctan(\pi - 1.1366) \approx \arctan(2.0044) \approx 1.1071\).

\(x_3 = \arctan(\pi - 1.1071) \approx \arctan(2.0345) \approx 1.1102\).

\(x_4 = \arctan(\pi - 1.1102) \approx \arctan(2.0314) \approx 1.1098\).

The root correct to 2 decimal places is 1.11.

(i) The perimeter of the shaded region is given by the arc length \(r \cdot x\) plus the length of the arc \(2r \cdot \theta\), where \(\theta = \frac{\pi}{2} - x\). The total perimeter is \(r \cdot x + 2r \cdot \left( \frac{\pi}{2} - x \right) = \frac{1}{2} \cdot 2\pi r\). Simplifying gives \(x = \cos^{-1} \left( \frac{\pi}{4 + 4x} \right)\).

(ii) Calculate \(\cos^{-1} \left( \frac{\pi}{4 + 4 \cdot 1} \right) \approx 1.318\) and \(\cos^{-1} \left( \frac{\pi}{4 + 4 \cdot 1.5} \right) \approx 1.047\). Since \(1.047 < x < 1.318\), x lies between 1 and 1.5.

(iii) Using the iterative formula:

\(x_1 = \cos^{-1} \left( \frac{\pi}{4 + 4 \cdot 1} \right) = 1.3183\)

\(x_2 = \cos^{-1} \left( \frac{\pi}{4 + 4 \cdot 1.3183} \right) = 1.2146\)

\(x_3 = \cos^{-1} \left( \frac{\pi}{4 + 4 \cdot 1.2146} \right) = 1.2099\)

\(x_4 = \cos^{-1} \left( \frac{\pi}{4 + 4 \cdot 1.2099} \right) = 1.2109\)

\(x_5 = \cos^{-1} \left( \frac{\pi}{4 + 4 \cdot 1.2109} \right) = 1.2106\)

The value of x correct to 2 decimal places is 1.21.

(a) Sketch the graphs of \(y = \cot x\) and \(y = 2 - \cos x\) for \(0 < x \leq \frac{1}{2}\pi\). The intersection point of these graphs indicates the root of the equation \(\cot x = 2 - \cos x\). The graph shows one intersection in the given interval.

(b) Calculate \(\cot(0.6)\) and \(\cot(0.8)\) and compare with \(2 - \cos(0.6)\) and \(2 - \cos(0.8)\). For \(x = 0.6\), \(\cot(0.6) \approx 1.46\) and \(2 - \cos(0.6) \approx 1.34\). For \(x = 0.8\), \(\cot(0.8) \approx 1.03\) and \(2 - \cos(0.8) \approx 1.29\). The change in sign between these values indicates a root between 0.6 and 0.8.

(c) Use the iterative formula \(x_{n+1} = \arctan\left( \frac{1}{2 - \cos x_n} \right)\) starting with an initial guess, such as \(x_0 = 0.7\). Perform iterations:

\(x_1 = \arctan\left( \frac{1}{2 - \cos(0.7)} \right) \approx 0.6806\)

\(x_2 = \arctan\left( \frac{1}{2 - \cos(0.6806)} \right) \approx 0.6855\)

\(x_3 = \arctan\left( \frac{1}{2 - \cos(0.6855)} \right) \approx 0.6843\)

\(x_4 = \arctan\left( \frac{1}{2 - \cos(0.6843)} \right) \approx 0.6846\)

The iterations converge to approximately 0.68, confirming the root to 2 decimal places.

(i) Sketch the graphs of \(y = x^3\) and \(y = 3 - x\). The intersection of these graphs gives the solution to \(x^3 = 3 - x\). The graph of \(y = x^3\) is a cubic curve passing through the origin, and \(y = 3 - x\) is a straight line with a negative slope. They intersect at exactly one point, indicating one real root.

(ii) Given the iterative formula \(x_{n+1} = \frac{2x_n^3 + 3}{3x_n^2 + 1}\), if it converges, then \(x_{n+1} = x_n = x\). Substituting into the formula gives \(x = \frac{2x^3 + 3}{3x^2 + 1}\). Rearranging, we get \(x(3x^2 + 1) = 2x^3 + 3\), which simplifies to \(x^3 = 3 - x\), showing convergence to the root found in part (i).

(iii) Using the iterative formula, start with an initial guess, say \(x_0 = 1\). Calculate successive iterations to 5 decimal places:

\(x_1 = \frac{2(1)^3 + 3}{3(1)^2 + 1} = 1.25\)

\(x_2 = \frac{2(1.25)^3 + 3}{3(1.25)^2 + 1} = 1.21622\)

\(x_3 = \frac{2(1.21622)^3 + 3}{3(1.21622)^2 + 1} = 1.21305\)

\(x_4 = \frac{2(1.21305)^3 + 3}{3(1.21305)^2 + 1} = 1.21300\)

\(x_5 = \frac{2(1.21300)^3 + 3}{3(1.21300)^2 + 1} = 1.21300\)

The root correct to 3 decimal places is 1.213.

(i) Differentiate \(y = \frac{\ln x}{3 + x}\) using the quotient rule:

\(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\)

Let \(u = \ln x\) and \(v = 3 + x\). Then \(\frac{du}{dx} = \frac{1}{x}\) and \(\frac{dv}{dx} = 1\).

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

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

\(3 + x - x \ln x = 0\)

\(x \ln x = 3 + x\)

\(\ln x = 1 + \frac{3}{x}\)

(ii) Sketch the graphs of \(y = \ln x\) and \(y = 1 + \frac{3}{x}\). The intersection point represents the root. Since \(\ln x\) is a monotonically increasing function and \(1 + \frac{3}{x}\) is decreasing for \(x > 0\), they intersect at only one point.

(iii) Use the iterative formula \(x_{n+1} = \frac{3 + x_n}{\ln x_n}\):

Start with an initial guess, say \(x_1 = 4.5\).

Calculate successive iterations:

\(x_2 = \frac{3 + 4.5}{\ln 4.5} \approx 4.9686\)

\(x_3 = \frac{3 + 4.9686}{\ln 4.9686} \approx 4.9701\)

\(x_4 = \frac{3 + 4.9701}{\ln 4.9701} \approx 4.9700\)

Continue until the value stabilizes to 4 decimal places. The final answer is 4.97.

(i) Sketch the graphs of \(y = e^{2x}\) and \(y = 6 + e^{-x}\). The intersection of these graphs represents the solution to the equation \(e^{2x} = 6 + e^{-x}\). The graphs intersect at exactly one point, indicating one real root.

(ii) Calculate \(e^{2(0.5)} = e \approx 2.718\) and \(6 + e^{-0.5} \approx 6.606\). Since \(e^{2(0.5)} < 6 + e^{-0.5}\), check \(x = 1\): \(e^{2(1)} = e^2 \approx 7.389\) and \(6 + e^{-1} \approx 6.368\). Since \(e^{2(1)} > 6 + e^{-1}\), the root lies between 0.5 and 1.

(iii) The iterative formula \(x_{n+1} = \frac{1}{3} \ln(1 + 6e^{x_n})\) is derived from rearranging the original equation. If the sequence converges, it converges to the root of \(e^{2x} = 6 + e^{-x}\).

(iv) Using the iterative formula, start with an initial guess, say \(x_0 = 0.5\):

\(x_1 = \frac{1}{3} \ln(1 + 6e^{0.5}) \approx 0.91894\)

\(x_2 = \frac{1}{3} \ln(1 + 6e^{0.91894}) \approx 0.92754\)

\(x_3 = \frac{1}{3} \ln(1 + 6e^{0.92754}) \approx 0.92823\)

\(x_4 = \frac{1}{3} \ln(1 + 6e^{0.92823}) \approx 0.92828\)

\(x_5 = \frac{1}{3} \ln(1 + 6e^{0.92828}) \approx 0.92828\)

The root correct to 3 decimal places is 0.928.

(i) Sketch the graphs of \(y = e^{-\frac{1}{2}x}\) and \(y = 4 - x^2\). The intersection points of these graphs represent the roots of the equation. The graph \(y = e^{-\frac{1}{2}x}\) is an exponential decay curve, while \(y = 4 - x^2\) is a downward-opening parabola. The graphs intersect at one positive and one negative point, indicating one positive root and one negative root.

(ii) Calculate the values at \(x = -1\) and \(x = -1.5\):

For \(x = -1\):

\(e^{-\frac{1}{2}(-1)} = e^{0.5} \approx 1.6487\)

\(4 - (-1)^2 = 3\)

Since \(1.6487 > 3\), the function changes sign between \(-1\) and \(-1.5\).

For \(x = -1.5\):

\(e^{-\frac{1}{2}(-1.5)} = e^{0.75} \approx 2.1170\)

\(4 - (-1.5)^2 = 1.75\)

Since \(2.1170 > 1.75\), the function changes sign between \(-1\) and \(-1.5\).

(iii) Use the iterative formula \(x_{n+1} = -\sqrt{4 - e^{-\frac{1}{2}x_n}}\):

Start with an initial guess, say \(x_0 = -1\).

\(x_1 = -\sqrt{4 - e^{-\frac{1}{2}(-1)}} \approx -1.4142\)

\(x_2 = -\sqrt{4 - e^{-\frac{1}{2}(-1.4142)}} \approx -1.4106\)

\(x_3 = -\sqrt{4 - e^{-\frac{1}{2}(-1.4106)}} \approx -1.4100\)

\(x_4 = -\sqrt{4 - e^{-\frac{1}{2}(-1.4100)}} \approx -1.4100\)

The iterations converge to \(-1.41\) to 2 decimal places.

(i) Sketch the graphs of \(y = \csc \frac{1}{2}x\) and \(y = \frac{1}{3}x + 1\). The intersection of these graphs in the interval \(0 < x \leq \pi\) indicates a root.

(ii) Calculate the values of the expressions at \(x = 1.4\) and \(x = 1.6\). If there is a sign change, the root lies between these values.

(iii) Rearrange the iterative formula \(x_{n+1} = 2 \sin^{-1} \left( \frac{3}{x_n + 3} \right)\) to show it is equivalent to \(\csc \frac{1}{2}x = \frac{1}{3}x + 1\). This shows convergence to the root.

(iv) Use the iterative formula starting with an initial guess. Calculate successive values to 5 decimal places until they stabilize to 3 decimal places. The final answer is 1.471.

(i) To show that the equation \(5e^{-x} = \sqrt{x}\) has one root, sketch the graphs of \(y = 5e^{-x}\) and \(y = \sqrt{x}\). The intersection point of these graphs represents the root. The exponential function \(5e^{-x}\) decreases rapidly, while \(\sqrt{x}\) increases slowly. They intersect at one point, indicating one root.

(ii) The iterative formula is \(x_{n+1} = \frac{1}{2} \ln\left(\frac{25}{x_n}\right)\). Rearrange this to show it is equivalent to \(5e^{-x} = \sqrt{x}\). If the sequence converges, it must satisfy the original equation, thus converging to the root.

(iii) Using the iterative formula with \(x_1 = 1\):

\(x_2 = \frac{1}{2} \ln\left(\frac{25}{1}\right) = 1.6094\)

\(x_3 = \frac{1}{2} \ln\left(\frac{25}{1.6094}\right) = 1.4332\)

\(x_4 = \frac{1}{2} \ln\left(\frac{25}{1.4332}\right) = 1.4292\)

\(x_5 = \frac{1}{2} \ln\left(\frac{25}{1.4292}\right) = 1.4286\)

The sequence converges to 1.43, correct to 2 decimal places.

(i) Sketch the curve \(y = \ln(x + 1)\), which is an increasing curve passing through the origin for \(x \geq 0\). The second curve is \(y = 40 - x^3\), which is a decreasing curve for \(x > 0\). The intersection of these curves indicates the root of the equation \(x^3 + \ln(x + 1) = 40\).

(ii) Calculate \(x^3 + \ln(x + 1) - 40\) at \(x = 3\) and \(x = 4\):

For \(x = 3\), \(3^3 + \ln(4) - 40 = 27 + 1.386 - 40 = -11.614\).

For \(x = 4\), \(4^3 + \ln(5) - 40 = 64 + 1.609 - 40 = 25.609\).

Since the sign changes between \(x = 3\) and \(x = 4\), the root lies between these values.

(iii) Using the iterative formula \(x_{n+1} = \sqrt[3]{40 - \ln(x_n + 1)}\):

Start with \(x_1 = 3.5\).

\(x_2 = \sqrt[3]{40 - \ln(3.5 + 1)} = 3.377\).

Continue iterations until the value stabilizes to 3 decimal places: \(x \approx 3.377\).

(iv) For \((e^y - 1)^3 + y = 40\), solve for \(y\):

\(y = \ln(x + 1) \approx 1.48\).

(i) Sketch the graphs of \(y = \csc x\) and \(y = x(\pi - x)\) for \(0 < x < \pi\). The graph of \(y = \csc x\) has vertical asymptotes at \(x = 0\) and \(x = \pi\), and the graph of \(y = x(\pi - x)\) is a downward-opening parabola with roots at \(x = 0\) and \(x = \pi\). The intersection points of these graphs represent the roots of the equation \(\csc x = x(\pi - x)\). By analyzing the graphs, we find exactly two intersection points in the interval \(0 < x < \pi\).

(ii) Start with \(\csc x = x(\pi - x)\), which is \(\frac{1}{\sin x} = x(\pi - x)\). Rearrange to get \(x = \frac{1 + x^2 \sin x}{\pi \sin x}\).

(iii)(a) Use the iterative formula \(x_{n+1} = \frac{1 + x_n^2 \sin x_n}{\pi \sin x_n}\) starting with an initial guess. Iterate until the value stabilizes to 4 decimal places. The iterations converge to \(\alpha = 0.66\).

(iii)(b) From the mark scheme, \(\beta = 2.48\).

(i) Sketch the graphs of \(y = \sec x\) and \(y = 3 - \frac{1}{2}x^2\) over the interval \(0 < x < \frac{1}{2}\pi\). The intersection of these graphs indicates a root. The sketch should show that the graphs intersect, confirming a root exists in the interval.

(ii) Calculate \(\sec 1 - (3 - \frac{1}{2} \times 1^2)\) and \(\sec 1.4 - (3 - \frac{1}{2} \times 1.4^2)\). If the signs of these values are different, a root exists between 1 and 1.4.

(iii) Rearrange \(\sec x = 3 - \frac{1}{2}x^2\) to \(x = \cos^{-1}\left( \frac{2}{6 - x^2} \right)\) to show equivalence.

(iv) Use the iterative formula \(x_{n+1} = \cos^{-1}\left( \frac{2}{6 - x_n^2} \right)\) starting with an initial guess. Iterate until the value stabilizes to 4 decimal places. The iterations yield \(x \approx 1.13\).

(i) Sketch the graphs of \(y = \cot x\) and \(y = 1 + x^2\) over the interval \(0 < x < \frac{1}{2}\pi\). The graph of \(\cot x\) decreases from infinity to 0, while \(y = 1 + x^2\) is a parabola opening upwards starting at 1. They intersect at only one point in this interval, indicating one root.

(ii) Calculate \(\cot 0.5 - (1 + 0.5^2)\) and \(\cot 0.8 - (1 + 0.8^2)\). The sign changes between these values, confirming a root exists between 0.5 and 0.8.

(iii) Use the iterative formula \(x_{n+1} = \arctan\left( \frac{1}{1 + x_n^2} \right)\) starting with \(x_0 = 0.5\). Iterate until the result stabilizes to 2 decimal places:

1. \(x_1 = \arctan\left( \frac{1}{1 + 0.5^2} \right) = 0.5880\)

2. \(x_2 = \arctan\left( \frac{1}{1 + 0.5880^2} \right) = 0.6176\)

3. \(x_3 = \arctan\left( \frac{1}{1 + 0.6176^2} \right) = 0.6233\)

4. \(x_4 = \arctan\left( \frac{1}{1 + 0.6233^2} \right) = 0.6216\)

5. \(x_5 = \arctan\left( \frac{1}{1 + 0.6216^2} \right) = 0.6219\)

The root correct to 2 decimal places is 0.62.

(a) Sketch the graphs of \(y = e^x - 3\) and \(y = \sqrt{x}\). The graph of \(y = e^x - 3\) cuts the vertical axis at (0, -2) and has an increasing gradient. The graph of \(y = \sqrt{x}\) starts at (0, 0) and has a reducing gradient. The intersection of these graphs indicates the root.

(b) Calculate \(e^1 - 3 = -0.28\) and \(e^2 - 3 = 4.39\). Since \(\sqrt{1} = 1\) and \(\sqrt{2} \approx 1.41\), we have \(1 > -0.28\) and \(1.41 < 4.39\). Therefore, the root lies between 1 and 2.

(c) Rearrange \(\sqrt{x} = e^x - 3\) to \(x = \ln(3 + \sqrt{x})\). The iterative formula \(x_{n+1} = \ln(3 + \sqrt{x_n})\) converges to the root of the equation.

(d) Using the iterative formula, start with an initial guess, e.g., \(x_1 = 1\):

\(x_2 = \ln(3 + \sqrt{1}) = 1.3864\)

\(x_3 = \ln(3 + \sqrt{1.3864}) = 1.4297\)

\(x_4 = \ln(3 + \sqrt{1.4297}) = 1.4341\)

\(x_5 = \ln(3 + \sqrt{1.4341}) = 1.4344\)

\(x_6 = \ln(3 + \sqrt{1.4344}) = 1.4345\)

The root correct to 2 decimal places is 1.43.

(i) To show that \(4x^2 - 1 = \cot x\) has only one root in the interval \(0 < x < \frac{1}{2}\pi\), sketch the graphs of \(y = 4x^2 - 1\) and \(y = \cot x\). The graph of \(y = 4x^2 - 1\) is a parabola opening upwards with vertex at \((0, -1)\). The graph of \(y = \cot x\) has vertical asymptotes at \(x = 0\) and \(x = \frac{1}{2}\pi\), and decreases from positive infinity to negative infinity. The two graphs intersect only once in the interval \(0 < x < \frac{1}{2}\pi\).

(ii) To verify the root lies between 0.6 and 1, calculate \(4x^2 - 1 - \cot x\) at \(x = 0.6\) and \(x = 1\). At \(x = 0.6\), \(4(0.6)^2 - 1 - \cot(0.6)\) is negative, and at \(x = 1\), \(4(1)^2 - 1 - \cot(1)\) is positive. This indicates a sign change, confirming a root exists between 0.6 and 1.

(iii) Using the iterative formula \(x_{n+1} = \frac{1}{2}\sqrt{1 + \cot x_n}\), start with an initial guess, say \(x_0 = 0.7\). Calculate successive iterations:

\(x_1 = \frac{1}{2}\sqrt{1 + \cot(0.7)}\)

\(x_2 = \frac{1}{2}\sqrt{1 + \cot(x_1)}\)

Continue until the value stabilizes to 2 decimal places. The iterations converge to approximately 0.73.

(i) Integrate by parts: let \(u = x\) and \(dv = e^{\frac{1}{2}x} \, dx\). Then \(du = dx\) and \(v = 2e^{\frac{1}{2}x}\).

\(\int xe^{\frac{1}{2}x} \, dx = 2xe^{\frac{1}{2}x} - 2\int e^{\frac{1}{2}x} \, dx\)

\(= 2xe^{\frac{1}{2}x} - 4e^{\frac{1}{2}x} + C\)

Substitute limits and equate to 6: \(2ae^{\frac{1}{2}a} - 4e^{\frac{1}{2}a} = 6\)

Rearrange: \(a = e^{-\frac{1}{2}a} + 2\)

(ii) Sketch the graphs of \(y = x\) and \(y = 2 + e^{-\frac{1}{2}x}\). They intersect at one point, indicating one root.

(iii) Check the sign of \(x - e^{-\frac{1}{2}x} - 2\) at \(x = 2\) and \(x = 2.5\). The sign changes, confirming a root between these values.

(iv) Use the iterative formula \(x_{n+1} = 2 + e^{-\frac{1}{2}x_n}\) with initial guess \(x_0 = 2\).

Calculate iterations: \(x_1 = 2.3033\), \(x_2 = 2.3113\), \(x_3 = 2.3105\), \(x_4 = 2.3107\).

The value of a is 2.31 to 2 decimal places.

(i) Sketch the graphs of \(y = 2 - x\) and \(y = \ln x\). The intersection point of these graphs represents the root of the equation \(2 - x = \ln x\). Since \(y = 2 - x\) is a straight line with a negative slope and \(y = \ln x\) is a logarithmic curve, they intersect at only one point, indicating one root.

(ii) Evaluate \(2 - x - \ln x\) at \(x = 1.4\) and \(x = 1.7\):

For \(x = 1.4\), \(2 - 1.4 - \ln(1.4) \approx 0.246\).

For \(x = 1.7\), \(2 - 1.7 - \ln(1.7) \approx -0.036\).

Since the sign changes between \(x = 1.4\) and \(x = 1.7\), the root lies in this interval.

(iii) Rearrange the equation \(x = \frac{1}{3}(4 + x - 2 \ln x)\) to \(2 - x = \ln x\), confirming that the root satisfies both equations.

(iv) Use the iterative formula:

\(x_1 = 1.5\)

\(x_2 = \frac{1}{3}(4 + 1.5 - 2 \ln(1.5)) \approx 1.5581\)

\(x_3 = \frac{1}{3}(4 + 1.5581 - 2 \ln(1.5581)) \approx 1.5560\)

\(x_4 = \frac{1}{3}(4 + 1.5560 - 2 \ln(1.5560)) \approx 1.5563\)

\(x_5 = \frac{1}{3}(4 + 1.5563 - 2 \ln(1.5563)) \approx 1.5562\)

The root correct to 2 decimal places is 1.56.

(i) Sketch the graphs of \(y = 2 \cot x\) and \(y = 1 + e^x\). The intersection of these graphs in the interval \(0 < x < \frac{1}{2} \pi\) indicates the root. The behavior of \(2 \cot x\) and \(1 + e^x\) suggests only one intersection point in this interval.

(ii) Evaluate \(2 \cot x - 1 - e^x\) at \(x = 0.5\) and \(x = 1.0\). The sign change between these points confirms a root exists between them.

(iii) Rearrange the original equation to show it is equivalent to \(x = \arctan\left(\frac{2}{1 + e^x}\right)\).

(iv) Use the iterative formula:

\(x_1 = 0.7\)

\(x_2 = \arctan\left(\frac{2}{1 + e^{0.7}}\right) = 0.6104\)

\(x_3 = \arctan\left(\frac{2}{1 + e^{0.6104}}\right) = 0.6100\)

\(x_4 = \arctan\left(\frac{2}{1 + e^{0.6100}}\right) = 0.6100\)

The root correct to 2 decimal places is 0.61.

(i) Sketch the graphs of \(y = \csc x\) and \(y = \frac{1}{2}x + 1\) over the interval \(0 < x < \frac{1}{2}\pi\). The intersection of these graphs indicates a root in this interval.

(ii) Evaluate \(\csc x - \frac{1}{2}x - 1\) at \(x = 0.5\) and \(x = 1\). If the signs are different, a root exists between these values.

(iii) Rearrange \(\csc x = \frac{1}{2}x + 1\) to \(x = \sin^{-1} \left( \frac{2}{x+2} \right)\) to show equivalence.

(iv) Use the iterative formula \(x_{n+1} = \sin^{-1} \left( \frac{2}{x_n+2} \right)\) starting with \(x_1 = 0.75\). Calculate successive iterations until the value stabilizes to 2 decimal places: \(x_2 = 0.8041\), \(x_3 = 0.8002\), \(x_4 = 0.8000\). The root is approximately 0.80.

(i) The area of triangle ONB is \(\frac{1}{2} r^2 \cos \alpha \sin \alpha\).

The area of sector OAB is \(\frac{1}{2} r^2 \alpha\).

Given that the area of triangle ONB is half the area of sector OAB, we have:

\(\frac{1}{2} r^2 \cos \alpha \sin \alpha = \frac{1}{2} \left( \frac{1}{2} r^2 \alpha \right)\)

Simplifying gives \(\sin 2\alpha = \alpha\).

(ii) Sketch the graphs of \(y = \sin 2x\) and \(y = x\) over the interval \(0 < x < \frac{1}{2}\pi\). The graphs intersect exactly once in this interval, indicating one root.

(iii) Using the iterative formula \(x_{n+1} = \sin(2x_n)\) with \(x_1 = 1\):

\(x_2 = \sin(2 \times 1) = \sin(2) \approx 0.909\)

\(x_3 = \sin(2 \times 0.909) \approx 0.973\)

\(x_4 = \sin(2 \times 0.973) \approx 0.929\)

\(x_5 = \sin(2 \times 0.929) \approx 0.964\)

\(x_6 = \sin(2 \times 0.964) \approx 0.945\)

\(x_7 = \sin(2 \times 0.945) \approx 0.955\)

\(x_8 = \sin(2 \times 0.955) \approx 0.949\)

\(x_9 = \sin(2 \times 0.949) \approx 0.953\)

The iterations converge to \(\alpha = 0.95\) correct to 2 decimal places.

(i) Sketch the graphs of \(y = \sec x\) and \(y = 3 - x^2\) for \(0 < x < \frac{1}{2}\pi\). The graph of \(y = \sec x\) has a vertical asymptote at \(x = \frac{1}{2}\pi\) and is positive in the interval. The graph of \(y = 3 - x^2\) is a downward-opening parabola with a positive y-intercept. The intersection of these graphs in the given interval shows exactly one root.

(ii) Assume \(\alpha = \cos^{-1} \left( \frac{1}{3-\alpha^2} \right)\). Rearrange to get \(\cos \alpha = \frac{1}{3-\alpha^2}\), which implies \(\sec \alpha = 3 - \alpha^2\). Thus, if the sequence converges, it converges to a root of the equation \(\sec x = 3 - x^2\).

(iii) Using the iterative formula \(x_{n+1} = \cos^{-1} \left( \frac{1}{3-x_n^2} \right)\) with \(x_1 = 1\):

1. Calculate \(x_2 = \cos^{-1} \left( \frac{1}{3-1^2} \right) \approx 1.047\).

2. Calculate \(x_3 = \cos^{-1} \left( \frac{1}{3-(1.047)^2} \right) \approx 1.032\).

3. Calculate \(x_4 = \cos^{-1} \left( \frac{1}{3-(1.032)^2} \right) \approx 1.030\).

The sequence converges to 1.03, correct to 2 decimal places.

(a) Sketch the graphs of \(y = \ln x\) and \(y = 3x - x^2\). The graph of \(y = \ln x\) passes through (1, 0) and has the y-axis as an asymptote. The graph of \(y = 3x - x^2\) is a parabola opening downwards, symmetric about the line \(x = 1.5\), passing through (0, 0) and (3, 0). The intersection point indicates one real root.

(b) Calculate \(\ln 2 \approx 0.693\) and \(3(2) - 2^2 = 2\). Since \(0.693 < 2\), the function changes sign between 2 and 2.8. Calculate \(\ln 2.8 \approx 1.029\) and \(3(2.8) - 2.8^2 = 1.04\). Since \(1.029 < 1.04\), the root lies between 2 and 2.8.

(c) Use the iterative formula \(x_{n+1} = \sqrt{3x_n - \ln x_n}\) starting with \(x_1 = 2.5\):

\(x_2 = \sqrt{3(2.5) - \ln(2.5)} \approx 2.6178\)

\(x_3 = \sqrt{3(2.6178) - \ln(2.6178)} \approx 2.6311\)

\(x_4 = \sqrt{3(2.6311) - \ln(2.6311)} \approx 2.6300\)

The root is approximately 2.63 to 2 decimal places.

(a) Sketch the graph of \(y = 4 - x^2\) and \(y = \sec \frac{1}{2}x\). The intersection of these graphs in the interval \(0 \leq x < \pi\) shows the root. The graph of \(y = 4 - x^2\) is a downward-opening parabola with vertex at (0, 4) and intersects the x-axis at \(x = 2\) and \(x = -2\). The graph of \(y = \sec \frac{1}{2}x\) has a vertical asymptote at \(x = \pi\) and intersects the y-axis at (0, 1). The graphs intersect exactly once in the interval \(0 \leq x < \pi\).

(b) Calculate \(4 - x^2\) and \(\sec \frac{1}{2}x\) at \(x = 1\) and \(x = 2\):

At \(x = 1\), \(4 - 1^2 = 3\) and \(\sec \frac{1}{2}(1) = \sec 0.5 \approx 1.139\).

At \(x = 2\), \(4 - 2^2 = 0\) and \(\sec \frac{1}{2}(2) = \sec 1 \approx 1.850\).

Since \(3 > 1.139\) and \(0 < 1.850\), the root lies between 1 and 2.

(c) Use the iterative formula \(x_{n+1} = \sqrt{4 - \sec \frac{1}{2}x_n}\):

Start with \(x_0 = 1.5\).

\(x_1 = \sqrt{4 - \sec \frac{1}{2}(1.5)} \approx 1.6161\)

\(x_2 = \sqrt{4 - \sec \frac{1}{2}(1.6161)} \approx 1.6025\)

\(x_3 = \sqrt{4 - \sec \frac{1}{2}(1.6025)} \approx 1.6003\)

\(x_4 = \sqrt{4 - \sec \frac{1}{2}(1.6003)} \approx 1.6000\)

The root correct to 2 decimal places is 1.60.

(a) Sketch the graphs of \(y = \cot \frac{1}{2}x\) and \(y = 1 + e^{-x}\). The intersection of these graphs in the interval \(0 < x \leq \pi\) shows there is exactly one root.

(b) Calculate the values at \(x = 1\) and \(x = 1.5\):

For \(x = 1\), \(\cot \frac{1}{2} \times 1 = \cot 0.5 \approx 1.8305\) and \(1 + e^{-1} \approx 1.3679\).

For \(x = 1.5\), \(\cot \frac{1}{2} \times 1.5 = \cot 0.75 \approx 1.3764\) and \(1 + e^{-1.5} \approx 1.2231\).

The change in sign between these values confirms a root between 1 and 1.5.

(c) Using the iterative formula:

Start with an initial guess, say \(x_0 = 1\).

\(x_1 = 2 \arctan \left( \frac{1}{1 + e^{-1}} \right) \approx 1.3524\)

\(x_2 = 2 \arctan \left( \frac{1}{1 + e^{-1.3524}} \right) \approx 1.3402\)

\(x_3 = 2 \arctan \left( \frac{1}{1 + e^{-1.3402}} \right) \approx 1.3398\)

\(x_4 = 2 \arctan \left( \frac{1}{1 + e^{-1.3398}} \right) \approx 1.3398\)

The root correct to 2 decimal places is 1.34.

(a) To show that the equation \(\csc x = 1 + e^{-\frac{1}{2}x}\) has exactly two roots in the interval \(0 < x < \pi\), sketch the graphs of \(y = \csc x\) and \(y = 1 + e^{-\frac{1}{2}x}\).

The graph of \(y = \csc x\) is U-shaped, roughly symmetrical about \(x = \frac{\pi}{2}\), with \(y\left(\frac{\pi}{2}\right) = 1\), and the domain is at least \(\left(\frac{\pi}{6}, \frac{5\pi}{6}\right)\).

The graph of \(y = 1 + e^{-\frac{1}{2}x}\) is an exponential curve with \(y(0) = 2\), a negative gradient, always increasing, and \(y(\pi) > 1\). Mark the intersections with dots or crosses to show the roots.

(b) Use the iterative formula \(x_{n+1} = \pi - \sin^{-1}\left( \frac{1}{e^{-\frac{1}{2}x_n} + 1} \right)\) with \(x_1 = 2\).

Calculate the iterations:

\(x_2 = 2.3217\)

\(x_3 = 2.2760\)

\(x_4 = 2.2824\)

The sequence converges to 2.28 when rounded to 2 decimal places.

(a) To show that the equation \(x^5 = 2 + x\) has exactly one real root, sketch the graphs of \(y = x^5\) and \(y = x + 2\). The intersection of these graphs represents the solution to the equation. Since \(x^5\) is a strictly increasing function and \(x + 2\) is a linear function, they intersect at exactly one point, confirming one real root.

(b) Consider the iterative formula \(x_{n+1} = \frac{4x_n^5 + 2}{5x_n^4 - 1}\). If this sequence converges, it must satisfy the equation \(x = \frac{4x^5 + 2}{5x^4 - 1}\). Rearranging gives \(x^5 = 2 + x\), which is the equation from part (a). Therefore, if the sequence converges, it converges to the root of the equation.

(c) Using the iterative formula with \(x_1 = 1.5\):

1. Calculate \(x_2 = \frac{4(1.5)^5 + 2}{5(1.5)^4 - 1} = 1.27103\).

2. Calculate \(x_3 = \frac{4(1.27103)^5 + 2}{5(1.27103)^4 - 1} = 1.26713\).

3. Calculate \(x_4 = \frac{4(1.26713)^5 + 2}{5(1.26713)^4 - 1} = 1.26700\).

4. Calculate \(x_5 = \frac{4(1.26700)^5 + 2}{5(1.26700)^4 - 1} = 1.26700\).

The sequence converges to 1.267, correct to 3 decimal places.

(a) Sketch the graph \(y = \sec x\) and the line \(y = 2 - \frac{1}{2}x\). The intersection point in the interval \(0 \leq x < \frac{1}{2}\pi\) indicates the root. The graphs intersect exactly once in this interval.

(b) Calculate \(\sec(0.8) \approx 1.139\) and \(2 - \frac{1}{2}(0.8) = 1.6\). Calculate \(\sec(1) \approx 1.850\) and \(2 - \frac{1}{2}(1) = 1.5\). Since \(\sec(0.8) < 1.6\) and \(\sec(1) > 1.5\), the root lies between 0.8 and 1.

(c) Using the iterative formula:

\(x_1 = \cos^{-1}\left(\frac{2}{4-0.8}\right) \approx 0.9273\)

\(x_2 = \cos^{-1}\left(\frac{2}{4-0.9273}\right) \approx 0.8834\)

\(x_3 = \cos^{-1}\left(\frac{2}{4-0.8834}\right) \approx 0.8800\)

\(x_4 = \cos^{-1}\left(\frac{2}{4-0.8800}\right) \approx 0.8798\)

The root correct to 2 decimal places is 0.88.

(i) Sketch the graphs of \(y = \ln(x+2)\) and \(y = 4e^{-x}\). The graph of \(y = \ln(x+2)\) is a logarithmic curve starting from \(x = -2\) and increasing. The graph of \(y = 4e^{-x}\) is an exponential decay curve starting from \(y = 4\) when \(x = 0\) and decreasing. These graphs intersect at exactly one point, indicating one real root.

(ii) Calculate \(\ln(1+2) = \ln(3) \approx 1.0986\) and \(4e^{-1} \approx 1.4715\). Since \(\ln(3) < 4e^{-1}\), the function \(\ln(x+2) - 4e^{-x}\) changes sign between \(x = 1\) and \(x = 1.5\), confirming a root in this interval.

(iii) Using the iterative formula:

1. Start with \(x_0 = 1\).

2. \(x_1 = \ln\left( \frac{4}{\ln(1 + 2)} \right) \approx \ln\left( \frac{4}{1.0986} \right) \approx 1.2809\).

3. \(x_2 = \ln\left( \frac{4}{\ln(1.2809 + 2)} \right) \approx \ln\left( \frac{4}{1.2809} \right) \approx 1.2345\).

4. \(x_3 = \ln\left( \frac{4}{\ln(1.2345 + 2)} \right) \approx \ln\left( \frac{4}{1.2345} \right) \approx 1.2301\).

The root correct to 2 decimal places is 1.23.