(a) The integrand is \(\frac12e^{2x}+\frac14e^x\). Therefore \[\left[\frac14e^{2x}+\frac14e^x\right]_a^{2a}=50,\] \[\frac14(e^{4a}+e^{2a}-e^{2a}-e^a)=50.\] Thus \(e^{4a}=200+e^a\). Taking logarithms gives \[\boxed{a=\frac14\ln(200+e^a)},\] as required.

(b) Let \(f(a)=a-\frac14\ln(200+e^a)\). Then \[f(1)=-0.327954\ldots<0,\qquad f(1.5)=0.169880\ldots>0.\] Continuity gives a root in \(\boxed{1<a<1.5}\).

(c) Use \(a_{n+1}=\frac14\ln(200+e^{a_n})\), starting at \(a_0=1\), retaining full precision internally: \[\begin{array}{c|rrrr}n&1&2&3&4\\\hline a_n&1.32795&1.32925&1.32926&1.32926\end{array}\] The stable 4-significant-figure answer is \(\boxed{a=1.329}\).

(a) Polynomial division gives \[\begin{array}{r|l}\text{quotient term}&\text{remaining polynomial}\\\hline x^3&4x^2-8x\\2x&-14x\\-7&21\end{array}\] Hence \[p(x)=(2x+3)(x^3+2x-7)+21.\] The quotient is \(\boxed{x^3+2x-7}\) and the remainder is \(\boxed{21}\).

(b) \(p(x)-21=(2x+3)(x^3+2x-7)\). The negative crossing is \(\boxed{\alpha=-3/2}\). The cubic has positive derivative \(3x^2+2\), so its unique root is the positive crossing \(\beta\). It satisfies \[x^3+2x-7=0\quad\Rightarrow\quad\boxed{x=\sqrt[3]{7-2x}}.\]

(c) For \(f(x)=x^3+2x-7\), \[f(1.5)=-0.625<0,\qquad f(1.6)=0.296>0.\] Continuity and this sign change show that \(\boxed{1.5<\beta<1.6}\).

(d) Use \(x_{n+1}=\sqrt[3]{7-2x_n}\), starting with \(x_0=1.5\). Keep full precision internally and display 6 significant figures:

\[\begin{array}{c|rrrrrrrrr}n&1&2&3&4&5&6&7&8&9\\\hline x_n&1.58740&1.56393&1.57030&1.56858&1.56905&1.56892&1.56895&1.56894&1.56895\end{array}\]

The iterations settle to the same 4-significant-figure value: \(\boxed{\beta=1.569}\).

(a) By the factor theorem, \(f(-2)=0\): \[16a-32-12a+42-18=0\ \Longrightarrow\ 4a-8=0\ \Longrightarrow\ \boxed{a=2}.\]

(b) With \(a=2\), divide by \(x+2\): \[f(x)=(x+2)(2x^3-6x-9).\]

The positive root is not \(-2\), so \(2x^3-6x-9=0\). Thus \(x^3=3x+4.5\), giving \(x=\sqrt[3]{3x+4.5}\), as required.

(c) Use \(x_{n+1}=\sqrt[3]{3x_n+4.5}\), starting with \(x_0=2\). Retain full calculator precision between iterations and display each result to 6 significant figures.

The later iterates agree to the required precision. Therefore \(\boxed{x=2.238}\) to exactly 4 significant figures.

(a) Integrating and applying the limits, \[5=\left[\frac14e^{2x}-\frac14e^{-x}\right]_{-2a}^{a}=\frac12e^{2a}-\frac14e^{-a}-\frac14e^{-4a}.\] Hence \[e^{2a}=10+\frac12e^{-a}+\frac12e^{-4a}.\] Taking logarithms and dividing by 2 gives the required equation.

(b) Define \[F(a)=a-\frac12\ln\left(10+\frac12e^{-a}+\frac12e^{-4a}\right).\] Then \(F(1)\approx-0.160855\) and \(F(1.2)\approx0.0410311\). Continuity gives a root between 1 and 1.2. Since the right side of the fixed point equation decreases as \(a\) increases, \(F\) is strictly increasing, so this is the required root.

(c) Start at \(a_0=1.10000\) and retain full accuracy internally:

The iterates stabilise, giving \(a=1.159\) to 4 significant figures.

(a) The exponential decreases from \((0,3)\) and remains positive. The secant curve increases from \((0,1)\), with vertical asymptote \(x=\pi/2\). There is one intersection:

(b) At an intersection, \[3e^{-2x}=\frac1{\cos x}\Rightarrow e^{2x}=3\cos x\Rightarrow x=\frac12\ln(3\cos x).\]

(c) Use \(x_{n+1}=\frac12\ln(3\cos x_n)\) in radian mode, starting at \(x_0=0.5\). Retain full accuracy between iterations:

Thus \(x=0.487\) to 3 decimal places.

(a) The modulus graph has vertex \((3/2,0)\). The logarithmic graph is defined for \(x>-1\), passes through the origin, and increases with vertical asymptote \(x=-1\). There are two intersections:

(b) The smaller intersection lies on the decreasing branch \(y=3-2x\). Thus \[3-2\alpha=\ln(\alpha+1)\Rightarrow\alpha=1.5-0.5\ln(\alpha+1).\]

(c) Use \(x_{n+1}=1.5-0.5\ln(x_n+1)\), starting at \(x_0=1.1\) and retaining full accuracy internally:

Hence \(\alpha=1.12\) to 3 significant figures.

(d) For the larger intersection use the increasing branch. Let \[F(x)=2x-3-\ln(x+1).\] Then \[F(2.055)\approx-0.00677959,\qquad F(2.065)\approx0.00995243.\] Continuity and the change of sign give \(2.055<\beta<2.065\).

(a) \[7=\left[5\ln(2x+1)\right]_a^{a^3}=5\ln\frac{2a^3+1}{2a+1}.\] Hence \(2a^3+1=e^{1.4}(2a+1)\), so \[a^3=0.5e^{1.4}(2a+1)-0.5,\quad a=\sqrt[3]{0.5e^{1.4}(2a+1)-0.5}.\]

(b) Use \(a_{n+1}=[0.5e^{1.4}(2a_n+1)-0.5]^{1/3}\), with \(a_0=2\). Successive values, displayed to 5 significant figures while retaining calculator precision, are 2.1281, 2.1657, 2.1765, 2.1795, 2.1804, 2.1807, 2.1807, 2.1807, 2.1808, 2.1808. Thus \(a=2.18\) to 3 significant figures.

(a) The exponential curve passes through (0,2) with horizontal asymptote y=1. The modulus graph has vertex (4,0), y-intercept 4, and one intersection P:

(b) At P, \(x<4\), so \(1+e^{2x}=4-x\). Hence \(e^{2x}=3-x\) and \(x=\frac12\ln(3-x)\).

(c) Starting with \(x_0=0.45\), use \(x_{n+1}=\frac12\ln(3-x_n)\). Values to 5 significant figures: 0.46805, 0.4645, 0.4652, 0.46506, 0.46509, 0.46508, 0.46508, 0.46508, 0.46508. Hence \(x=0.465\) to 3 significant figures.

(a) \[\frac{dy}{dx}=\frac{2e^{2x}(1+3x)-3(1+e^{2x})}{(1+3x)^2}.\] At P, \(e^{2x}(6x-1)=3\), so \(x=1/6+\frac12e^{-2x}\).

(b) Let \(g(x)=x-1/6-\frac12e^{-2x}\). Then \(g(0.35)=-0.0649593\), \(g(0.45)=0.0800485\). Continuity and the sign change give a root between these values.

(c) Starting at \(x_0=0.4\), use \(x_{n+1}=1/6+\frac12e^{-2x_n}\). Values to 5 significant figures: 0.39133, 0.39526, 0.39347, 0.39428, 0.39391, 0.39408, 0.39401, 0.39404, 0.39402, 0.39403, 0.39403, 0.39403, 0.39403. Hence \(x=0.394\) to 3 significant figures.

(a) \[\frac{dy}{dx}=\frac{\frac2{2x+1}(x+3)-\ln(2x+1)}{(x+3)^2}.\]

(b) Setting the numerator to zero gives \(2(x+3)=(2x+1)\ln(2x+1)\). Rearranging gives \(x=(x+3)/\ln(2x+1)-1/2\).

(c) Define \(g(x)=x-(x+3)/\ln(2x+1)+1/2\). Then \(g(2.5)=-0.0696084\), \(g(3)=0.416610\). Continuity and the sign change show a root in (2.5,3).

(d) Starting with \(x_0=2.5\), use \(x_{n+1}=(x_n+3)/\ln(2x_n+1)-0.5\). Values to 6 significant figures: 2.56961, 2.56917, 2.56917, 2.56917, 2.56917, 2.56917. Hence \(x=2.569\) to 4 significant figures.

(a) \[\frac{dy}{dx}=\frac{3x^2(x+2)-x^3}{(x+2)^2}.\] Equating to 6 gives \(2x^3+6x^2=6x^2+24x+24\), so \(x^3=12x+12\), giving \(x=\sqrt[3]{12x+12}\).

(b) Let \(g(x)=x-\sqrt[3]{12x+12}\). Then \(g(3.8)=-0.06195754\) and \(g(4.0)=0.08513236\). The continuous function changes sign, so its root lies between 3.8 and 4.0.

(c) Starting at \(x_0=3.9\), use \(x_{n+1}=(12x_n+12)^{1/3}\). Values to 5 significant figures: 3.8886, 3.8856, 3.8848, 3.8846, 3.8845, 3.8845, 3.8845, 3.8845, 3.8845, 3.8845, 3.8845. Hence \(x=3.88\) to 3 significant figures.

(a) Implicit differentiation gives \[2e^{2x}-18+(3y^2+1)\frac{dy}{dx}=0.\] At a stationary point \(dy/dx=0\), so \(e^{2p}=9\) and \(p=\ln3\).

(b) Substitute \(p=\ln3\) into the curve: \[9-18\ln3+q^3+q=11,\qquad q^3=2+18\ln3-q,\] yielding the required cube-root equation.

(c) Let \(F(q)=q^3+q-2-18\ln3\). Then \(F(2.5)\approx-3.65002\) and \(F(3)\approx8.22498\). As \(F\) is continuous and changes sign, a root lies between 2.5 and 3.0.

(d) Use \(q_{n+1}=(2+18\ln3-q_n)^{1/3}\), starting with \(q_0=2.5\).

The values settle to \(q=2.673\), correct to 4 significant figures.

(a) \[\frac{dx}{dt}=\frac6{2t-3},\qquad\frac{dy}{dt}=4\ln t+4,\qquad\frac{dy}{dx}=\frac{(4\ln t+4)(2t-3)}6.\] At \(A\), \(x=0\), so \(2t-3=1\) and \(t=2\). The gradient is \(\frac23(\ln2+1)\).

(b) At \(B\), \[(4\ln t+4)(2t-3)=72,\qquad(1+\ln t)(2t-3)=18.\] Rearranging gives \(t=9/(1+\ln t)+3/2\).

(c) Use \(t_{n+1}=9/(1+\ln t_n)+3/2\), with \(t_0=5\).

The result is \(t=4.96\), correct to 3 significant figures.

(a) \[\left[\frac32e^{2x}-x\right]_0^a=12,\qquad\frac32e^{2a}-a-\frac32=12.\] Thus \(e^{2a}=9+2a/3\), and \(a=\frac12\ln(9+2a/3)\).

(b) Use \(a_{n+1}=\frac12\ln(9+2a_n/3)\), with \(a_0=1\).

The result is \(a=1.139\), correct to 4 significant figures.

(a) The modulus graph has vertex \((4/5,0)\) and slopes \(-5,5\). The sketch shows two intersections.

For \(x\le4/5\), the difference is \(5x-1-e^{-x/2}\), which is strictly increasing, is negative as \(x\to-\infty\), and is positive at \(x=4/5\). For \(x\ge4/5\), the difference is \(7-5x-e^{-x/2}\); its derivative \(-5+\frac12e^{-x/2}\) is negative there, and it changes from positive to negative. Thus each branch has exactly one intersection.

(b) Let \(F(x)=5x-1-e^{-x/2}\). Then \(F(0.36)=-0.0352702\ldots\) and \(F(0.37)=0.0188957\ldots\). Continuity and the sign change give \(0.36<\alpha<0.37\).

(c) Starting with \(x_0=1\):

The result is \(\beta=1.295\), correct to 4 significant figures.

(a) \[\left[2\ln(1+2x)+3\ln x\right]_1^a=\ln10,\] so \(2\ln((1+2a)/3)+3\ln a=\ln10\). Since \(a>1\), all logarithms are defined, and exponentiating gives \[a^3(1+2a)^2=90,\qquad a=\sqrt[3]{90(1+2a)^{-2}}.\]

(b) Use \(a_{n+1}=\sqrt[3]{90/(1+2a_n)^2}\), starting at \(a_0=1.7\).

The result is \(a=1.68\), correct to 3 significant figures.

(a) Sketch \(y=e^{-x/2}\), which is positive and strictly decreasing through \((0,1)\), and \(y=x^5\), which is strictly increasing through the origin.

For \(x\le0\), \(x^5\le0<e^{-x/2}\), so there is no intersection. At 0, the exponential is above \(x^5\); at 1 it is below. The two opposite monotonic directions give exactly one intersection in \((0,1)\).

(b) The iteration is equivalently \(x_{n+1}=e^{-x_n/10}\). Starting with \(x_0=1\), the values below are shown to 6 significant figures; unrounded values are used in each next step.

The values stabilise at approximately 0.912765. Thus the root to 4 significant figures is \[x=0.9128.\]

(a) Differentiating by the product rule, \[\frac{dy}{dx}=\ln(4x+1)+\frac{4x}{4x+1}-3.\] At P this is zero. Since \(4x/(4x+1)=1-1/(4x+1)\), \[(4x+1)\ln(4x+1)=8x+3.\] Dividing by \(4\ln(4x+1)\) gives \[x+\frac14=\frac{2x+3/4}{\ln(4x+1)},\] which is the required equation.

(b) Let \[F(x)=x-\frac{2x+0.75}{\ln(4x+1)}+0.25.\] It is continuous on \([1.8,1.9]\), and \[F(1.8)\approx-0.01736,\qquad F(1.9)\approx0.03545.\] The sign change proves that the stationary x-coordinate lies between 1.8 and 1.9.

(c) Use \[x_{n+1}=\frac{2x_n+0.75}{\ln(4x_n+1)}-0.25,\qquad x_0=1.85.\] The table gives each iteration to 5 significant figures; unrounded values are carried into the next step.

The sequence converges to approximately 1.832843. Therefore \[x=1.83\quad\text{(3 significant figures)}.\]

(a) Sketch the V-shaped graph with vertex \((2.5,0)\) and the increasing logarithmic graph through \((1,0)\).

For \(0<x\le1\), the logarithm is non-positive while the modulus is positive. On \((1,2.5)\), the modulus decreases and the logarithm increases, giving exactly one intersection. For \(x>2.5\), the difference \(2x-5-3\ln x\) is strictly increasing because its derivative is \(2-3/x>0\); it starts negative and becomes positive, giving exactly one further intersection. Thus there are exactly two roots.

(b) At the larger root, \(x>2.5\), so \[2x-5=3\ln x\quad\Longrightarrow\quad x=2.5+1.5\ln x.\]

(c) Let \(F(x)=x-2.5-1.5\ln x\). Then \[F(4.5)\approx-0.25612,\qquad F(5.0)\approx0.08584.\] Continuity and the sign change show that the larger root lies between 4.5 and 5.0.

(d) Use \(x_{n+1}=2.5+1.5\ln x_n\), starting with \(x_0=4.75\). The values are displayed to 5 significant figures, with unrounded values used in each next step.

The values converge to approximately 4.876705, so the larger root is \[x=4.88\quad\text{(3 significant figures)}.\]

(a) \[120=\int_a^{a+1}3e^{2x-1}\,dx=\left[\frac32e^{2x-1}\right]_a^{a+1}=\frac32\left(e^{2a+1}-e^{2a-1}\right).\] Therefore \[e^{2a+1}=80+e^{2a-1}.\] Taking logarithms and rearranging gives \[2a+1=\ln\bigl(80+e^{2a-1}\bigr),\qquad a=\frac12\ln\bigl(80+e^{2a-1}\bigr)-\frac12.\]

(b) Start with \(a_0=1.7\), and use \[a_{n+1}=\frac12\ln\bigl(80+e^{2a_n-1}\bigr)-\frac12.\] Each result is displayed to 5 significant figures; unrounded values are used in the next step.

The sequence converges to approximately 1.76372. Hence \[\boxed{a=1.76}\quad\text{(3 significant figures)}.\]

(a) Differentiating implicitly gives \[2e^{2x}y+e^{2x}\frac{dy}{dx}-e^y\frac{dy}{dx}=0.\] Hence \[(e^y-e^{2x})\frac{dy}{dx}=2e^{2x}y,\] which gives the required derivative.

(b) A stationary point would require \(2e^{2x}y=0\). Since \(e^{2x}>0\), this would mean \(y=0\). Substituting into the original curve gives \(-1=100\), which is impossible. Therefore there are no stationary points.

(c) At a vertical tangent, \(e^y-e^{2x}=0\), so \(y=2x\). Substituting into the curve gives \[e^{2x}(2x-1)=100.\] This also implies \(2x-1>0\). Taking logarithms, \[2x+\ln(2x-1)=\ln100=2\ln10,\] so \[x=\ln10-\frac12\ln(2x-1).\] The derivative numerator is non-zero there, as \(y=2x>1\), confirming a vertical tangent.

(d) Use \[x_{n+1}=\ln10-\frac12\ln(2x_n-1),\qquad x_0=2.\] The table gives each iteration to 5 significant figures; unrounded values are used in each next step.

The sequence converges to approximately 1.8179665. Hence \[\boxed{x=1.82}\quad\text{(3 significant figures)}.\]

(a) Apply the product and chain rules: \[\frac{dy}{dx}=e^{2x}+2xe^{2x}-5e^{-x}.\] At M the derivative is zero. Multiplying by \(e^x\) gives \[e^{3x}(1+2x)=5.\] This implies \(1+2x>0\), so taking logarithms gives \[3x+\ln(1+2x)=\ln5,\qquad\boxed{x=\frac13\ln5-\frac13\ln(1+2x)}.\]

(b) Use \[x_{n+1}=\frac{\ln5-\ln(1+2x_n)}3,\qquad x_0=0.35.\] Keep full precision for subsequent calculations, displaying each iterate to 5 significant figures:

The values stabilise to 3 significant figures, giving \[\boxed{x_M=0.357}.\] The second derivative \(e^{2x}(4+4x)+5e^{-x}\) is positive at this point, consistent with a minimum.

(a) Sketch \(y=\ln x\) and \(y=2e^{-x}\). For \(x>0\), the logarithm increases and the exponential decreases. For \(0<x\le1\) the logarithm is non-positive and the exponential positive, so they cannot meet there. After x=1, their opposing monotonicity and limiting values force exactly one intersection.

(b) Let \(h(x)=\ln x-2e^{-x}\). Then \[h(1.5)=-0.0407952\ldots,\qquad h(1.6)=0.0662106\ldots.\] The function is continuous, so the sign change confirms a root between 1.5 and 1.6.

(c) If the sequence converges to a limit L, taking limits gives \(L=e^{2e^{-L}}\). Taking natural logarithms yields \(\ln L=2e^{-L}\), the original equation.

(d) Start with \(x_0=1.5\) and calculate from the given recurrence, retaining full precision:

The iterates agree to 3 significant figures. Thus \[\boxed{x=1.54}.\]

(a) At P, \(6p\sin2p=3\). Dividing by \(6\sin2p\) gives \(\boxed{p=1/(2\sin2p)}\).

(b) Define \(h(p)=p-1/(2\sin2p)\). Then \[h(0.5)=-0.0941976\ldots,\qquad h(0.6)=0.0635418\ldots.\] The function is continuous on this interval, so the sign change establishes a root between the endpoints.

(c) Use \(p_{n+1}=1/(2\sin2p_n)\), beginning with \(p_0=0.55\) and retaining full precision internally:

The values stabilise to give \(\boxed{p=0.557}\) to 3 significant figures.

(d) Differentiate the parametric equations: \[\frac{dx}{dt}=4+2e^{2t},\qquad\frac{dy}{dt}=6\sin2t+12t\cos2t.\] Therefore \[\frac{dy}{dx}=\frac{6\sin2t+12t\cos2t}{4+2e^{2t}}.\] Substituting the unrounded parameter \(p=0.557078\ldots\) gives \[\boxed{\left.\frac{dy}{dx}\right|_P=0.826}\quad\text{(3 significant figures)}.\]

(a) By the quotient rule, \[\boxed{\frac{dy}{dx}=\frac{3\ln x-(3x+2)/x}{(\ln x)^2}}.\] At M this equals zero, so \[3\ln x=\frac{3x+2}{x},\qquad\boxed{x=\frac{3x+2}{3\ln x}}.\]

(b) Let \(h(x)=x-(3x+2)/(3\ln x)\). Then \[h(3)=-0.337543\ldots,\qquad h(4)=0.633711\ldots.\] Continuity and the sign change show that the required root lies between 3 and 4.

(c) Use \[x_{n+1}=\frac{3x_n+2}{3\ln x_n},\qquad x_0=3.5.\] Retain full precision and display the iterates to 7 significant figures:

The values agree to 5 significant figures, giving \[\boxed{x_M=3.3223}.\]

(a) The logarithms require \(x>0\). Combine them and exponentiate: \[\ln(x(x+1)^2)=\ln(x+9),\qquad x(x+1)^2=x+9.\] Expand and cancel x: \[x^3+2x^2=9,\qquad x^2(x+2)=9,\qquad x^2=\frac9{x+2}.\] Since x is positive, take the positive square root, giving the required result.

(b) Define \(h(x)=x-\sqrt{9/(x+2)}\). Then \[h(1.5)=-0.103567\ldots,\qquad h(2)=0.5.\] The function is continuous and changes sign, so the root lies between the endpoints.

(c) Use \(x_{n+1}=\sqrt{9/(x_n+2)}\), starting at \(x_0=1.75\). Retain full precision for calculations and display each value to 5 significant figures:

The iterates agree to 3 significant figures, giving \[\boxed{x=1.58}.\]