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2016 Paper 1 Q2
D: 1516.0 B: 1516.0
differentiation integration product rule chain rule logarithmic function inverse hyperbolic functions simplification integration by parts

Differentiate, with respect to \(x\), \[ (ax^2+bx+c)\,\ln \big( x+\sqrt{1+x^2}\big) +\big(dx+e\big)\sqrt{1+x^2} \,, \] where \(a\), \(b\), \(c\), \(d\) and \(e\) are constants. You should simplify your answer as far as possible. Hence integrate:

  1. \( \ln \big( x+\sqrt{1+x^2}\,\big) \,;\)
  2. \(\sqrt{1+x^2} \,; \)
  3. \( x\ln \big( x+\sqrt{1+x^2}\,\big) \,.\)

Solution: \begin{align*} && y &= (ax^2+bx+c)\,\ln \big( x+\sqrt{1+x^2}\big) +\big(dx+e\big)\sqrt{1+x^2} \\ && y' &= (2ax+b)\,\ln \big( x+\sqrt{1+x^2}\big) + (ax^2+bx+c) \frac{1}{x + \sqrt{1+x^2}} \cdot \left(1 + \frac{x}{\sqrt{1+x^2}} \right) + d\sqrt{1+x^2} + \frac{x(dx+e)}{\sqrt{1+x^2}} \\ &&&= (2ax+b)\,\ln \big( x+\sqrt{1+x^2}\big) + \frac{1}{\sqrt{1+x^2}} \left ( (ax^2+bx+c) + d(1+x^2) + x(dx+e) \right) \\ &&&= (2ax+b)\,\ln \big( x+\sqrt{1+x^2}\big) + \frac{1}{\sqrt{1+x^2}} \left ( (a+2d)x^2+(b+e)x+(d+c) \right) \\ \end{align*}

  1. We want \(a = 0, b = 1, d = 0, e = -1, c =0\), so \begin{align*} I &= \int \ln \big( x+\sqrt{1+x^2}\,\big) \,\d x \\ &= x\ln(x+\sqrt{1+x^2})-\sqrt{1+x^2}+C \end{align*}
  2. We want \(a = b =0, e = 0, d = \frac12, c = \frac12\), so \begin{align*} I &= \int \sqrt{1+x^2}\, \d x \\ &= \frac12\ln(x+\sqrt{1+x^2}) + \frac12x\sqrt{1+x^2}+C \end{align*}
  3. We want \(a = \frac12, b = 0, d = -\frac14, e = 0, c = \frac14\), so \begin{align*} I &= \int x \ln (x+\sqrt{1+x^2}) \, \d x \\ &= \left (\frac12 x^2+\frac14 \right)\ln(x+\sqrt{1+x^2}) -\frac14x\sqrt{1+x^2}+C \end{align*}

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2013 Paper 2 Q1
D: 1600.0 B: 1484.0
differentiation tangent logarithmic function curve sketching inequality concavity optimisation geometric proof

  1. Find the value of \(m\) for which the line \(y = mx\) touches the curve \(y = \ln x\,\). If instead the line intersects the curve when \(x = a\) and \(x = b\), where \(a < b\), show that \(a^b = b^a\). Show by means of a sketch that \(a < \e < b\).
  2. The line \(y=mx+c\), where \(c>0\), intersects the curve \(y=\ln x\) when \(x=p\) and \(x=q\), where \(p < q\). Show by means of a sketch, or otherwise, that \(p^q > q^p\).
  3. Show by means of a sketch that the straight line through the points \((p, \ln p)\) and \((q, \ln q)\), where \(\e\le p < q\,\), intersects the \(y\)-axis at a positive value of \(y\). Which is greater, \(\pi^\e\) or \(\e^\pi\)?
  4. Show, using a sketch or otherwise, that if \(0 < p < q\) and \(\dfrac{\ln q - \ln p}{q-p} = \e^{-1}\), then \(q^p > p^q\).

Solution: \begin{questionparts} \item The tangent to \(y = \ln x\) is \begin{align*} && \frac{y - \ln x_1}{x - x_1} &= \frac{1}{x_1} \\ \Rightarrow && \frac{x_1y -x_1 \ln x_1}{ x- x_1} &= 1 \\ \Rightarrow && x_1 y - x_1 \ln x_1 &= x - x_1 \end{align*} So to run through the origin, we need \(\ln x_1 = 1 \Rightarrow x_1 = e\) so the line will be \(y = \frac1{e} x\) If \(ma = \ln a \Rightarrow m = \frac{\ln a}{a} = \frac{\ln b}{b} \Rightarrow b \ln a = a \ln b \Rightarrow a^b = b^a\). \item

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2006 Paper 2 Q2
D: 1600.0 B: 1500.0
exponential series inequality proof by induction factorial curve sketching stationary points logarithmic function differentiation bounding estimates

Using the series \[ \e^x = 1 + x +\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}+\cdots\,, \] show that \(\e>\frac83\). Show that \(n!>2^n\) for \(n\ge4\) and hence show that \(\e<\frac {67}{24}\). Show that the curve with equation \[ y= 3\e^{2x} +14 \ln (\tfrac43-x)\,, \qquad {x<\tfrac43} \] has a minimum turning point between \(x=\frac12\) and \(x=1\) and give a sketch to show the shape of the curve.

Solution: \begin{align*} && e &= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots \\ &&&> 1 + 1+ \frac12 + \frac16 \\ &&&= \frac{12+3+1}{6} = \frac83 \end{align*} \(4! = 24 > 16 = 2^4\), notice that \(n! = \underbrace{n \cdot (n-1) \cdots 5}_{>2^{n-4}} \cdot \underbrace{4!}_{>2^4} >2^n\). \begin{align*} && e &= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots \\ &&&< \frac83 + \frac{1}{2^4} + \frac{1}{2^5} + \cdots \\ &&&= \frac83 + \frac{1}{2^4} \frac{1}{1-\tfrac12} \\ &&&= \frac83 + \frac1{8} \\ &&&= \frac{67}{24} \end{align*} \begin{align*} && y &= 3e^{2x} +14 \ln(\tfrac43-x) \\ && y' &= 6e^{2x} - \frac{14}{\tfrac43-x} \\ && y'(\tfrac12) &= 6e - \frac{14}{\tfrac43-\tfrac12} \\ &&&= 6e -\tfrac{84}{5} = 6(e-\tfrac{14}5) < 0 \\ && y'(1) &= 6e^2 - \frac{14}{\tfrac43-1} \\ &&&= 6e^2 - 42 = 6(e^2-7) \\ &&&> 6(\tfrac{64}{9} - 7) > 0 \end{align*} Therefore \(y'\) changes from negative (decreasing) to positive (increasing) in our range, and therefore there is a minima in this range.

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2006 Paper 2 Q14
D: 1600.0 B: 1500.0
continuous probability distributions probability density function integration curve sketching logarithmic function substitution normalisation asymptotes

Sketch the graph of \[ y= \dfrac1 { x \ln x} \text{ for \(x>0\), \(x\ne1\)}.\] You may assume that \(x\ln x \to 0\) as \(x\to 0\). The continuous random variable \(X\) has probability density function \[ \f(x) = \begin{cases} \dfrac \lambda {x\ln x}& \text{for \(a\le x \le b\)}\;, \\[3mm] \ \ \ 0 & \text{otherwise }, \end{cases} \] where \(a\), \(b\) and \(\lambda\) are suitably chosen constants.

  1. In the case \(a=1/4\) and \(b=1/2\), find \(\lambda\,\).
  2. In the case \(\lambda=1\) and \(a>1\), show that \(b=a^\e\).
  3. In the case \(\lambda =1\) and \(a=\e\), show that \(\P(\e^{3/2}\le X \le \e^2)\approx \frac {31}{108}\,\).
  4. In the case \(\lambda =1\) and \(a=\e^{1/2}\), find \(\P(\e^{3/2}\le X \le \e^2)\;\).

Solution:

  1. \begin{align*} 1 &= \int_{1/4}^{1/2} \frac{\lambda}{x\ln x} \, dx \\ &= \lambda\left [ \ln |\ln x| \right ]_{1/4}^{1/2} \\ &= \lambda \l \ln |-\ln 2| - \ln |-2 \ln 2| \r \\ &= \lambda (-\ln 2) \end{align*} So \(\lambda = -\frac{1}{\ln 2} = \frac{1}{\ln \frac12}\)
  2. \begin{align*} 1 &= \int_{a}^{b} \frac{1}{x\ln x} \, dx \\ &= \left [ \ln |\ln x| \right ]_{a}^{b} \\ &= \l \ln \ln b - \ln \ln a \r \\ &= \ln \l \frac{\ln b}{\ln a} \r \\ \end{align*} So \(b = e^{a}\)
  3. If \(\lambda = 1, a = e, b = e^e\), then \begin{align*} \P(\e^{3/2}\le X \le \e^2) &= \int_{e^{3/2}}^{e^2} \frac{1}{x \ln x} \, dx \\ &= \left [ \ln \ln x \right]_{e^{3/2}}^{e^2} \\ &= \ln 2 - \ln \frac{3}{2} \\ &= \ln \frac{4}{3} \\ &= \ln \l 1 + \frac{1}{3} \r \\ &\approx \frac{1}{3} - \frac{1}{2 \cdot 3^2} + \frac{1}{3 \cdot 3^3} - \frac{1}{4 \cdot 3^4} \\ &= \frac{31}{108} \end{align*}
  4. Note that \(2 > e^{\frac12} > 1\) so \(a = e^{\frac12}, b = e^{\frac{e}2}\). Since \(3 > e \Rightarrow e^{3/2} > e^{\frac{e}{2}}\) this probability is out of range, therefore \(\P(\e^{3/2}\le X \le \e^2) = 0\)

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2002 Paper 3 Q1
D: 1700.0 B: 1500.0
integration integration by parts volume of revolution logarithmic function improper integral area under curve limits at infinity convergence

Find the area of the region between the curve \(\displaystyle y = {\ln x \over x}\,\) and the \(x\)-axis, for \(1 \le x \le a\). What happens to this area as \(a\) tends to infinity? Find the volume of the solid obtained when the region between the curve \(\displaystyle y = {\ln x \over x}\,\) and the \(x\)-axis, for \(1 \le x\le a\), is rotated through \(2 \pi\) radians about the \(x\)-axis. What happens to this volume as \(a\) tends to infinity?

Solution: \begin{align*} && \int_1^a \frac{\ln x}{x} \d x &= \left [ \ln x \cdot \ln x\right ]_1^a - \int_1^a \frac{\ln x}{x} \d x \\ \Rightarrow && \int_1^a \frac{\ln x}{x} \d x &= \frac12 \left ( \ln a \right) ^2 \\ && \int_1^\infty \frac{\ln x}{x} \d x &= \lim_{a \to \infty} \frac12 (\ln a)^2 \\ &&&= \infty \end{align*} \begin{align*} && \pi \int_1^a \left ( \frac{\ln x}{x} \right)^2 \d x &= \pi \int_{u=0}^{u=\ln a} \left ( \frac{u}{e^u} \right)^2 e^u \d u \\ &&&= \pi \int_0^{\ln a} u^2 e^{-u} \d u \\ &&&= \pi \left [-u^2e^{-u} \right]_0^{\ln a} +\pi \int_0^{\ln a} 2u e^{-u} \d u \\ &&&= -\frac{\pi}{a} (\ln a)^2 + \pi \left [-2u e^{-u} \right]_0^{\ln a} + \pi \int_0^{\ln a} e^{-u} \d u \\ &&&= -\frac{\pi}{a} (\ln a)^2- \frac{2 \pi}{a}\ln a+\pi \left (1 - \frac{1}{a} \right) \\ \\ && \pi \int_1^{\infty} \left ( \frac{\ln x}{x} \right)^2 \d x &= \lim_{a \to \infty} \left ( -\frac{\pi}{a} (\ln a)^2- \frac{2 \pi}{a}\ln a+\pi \left (1 - \frac{1}{a} \right) \right) \\ &&&= \pi \end{align*}

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2001 Paper 3 Q1
D: 1700.0 B: 1500.0
Taylor series Maclaurin series proof by induction implicit differentiation Leibniz rule recurrence relation logarithmic function inverse hyperbolic function

Given that \(y = \ln ( x + \sqrt{x^2 + 1})\), show that \( \displaystyle \frac{\d y}{\d x} = \frac1 {\sqrt{x^2 + 1} }\;\). Prove by induction that, for \(n \ge 0\,\), \[ \l x^2 + 1 \r y^{\l n + 2 \r} + \l 2n + 1 \r x y^{\l n + 1 \r} + n^2 y^{\l n \r} = 0\;, \] where \(\displaystyle y^{(n)} = \frac{\d^n y}{\d x^n}\) and \(y^{(0)} =y\,\). Using this result in the case \(x = 0\,\), or otherwise, show that the Maclaurin series for \(y\) begins \[ x - {x^3 \over 6} +{3 x^5 \over 40} \] and find the next non-zero term.

Solution: \begin{align*} && y & = \ln ( x + \sqrt{x^2+1}) \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{1}{x+\sqrt{x^2+1}} \cdot \frac{\d }{\d x} \left ( x + \sqrt{x^2+1} \right) \\ &&&= \frac{1}{x+\sqrt{x^2+1}} \left (1 + \frac12 \frac{2x}{\sqrt{x^2+1}} \right) \\ &&&= \frac{1}{x+\sqrt{x^2+1}} \left ( \frac{\sqrt{x^2+1} + x}{\sqrt{x^2+1}}\right) \\ &&&= \frac{1}{\sqrt{x^2+1}} \end{align*} Note that \(\displaystyle y^{(2)} = - \frac12 \frac{2x}{(x^2+1)^{3/2}} = - \frac{x}{(x^2+1)^{3/2}}\), and in particular \((x^2+1)y^{(2)} + xy^{(1)} = 0\). Now applying Leibnitz formula: \begin{align*} 0 &= \left ( (x^2+1)y^{(2)} + xy^{(1)} \right )^{(n)} \\ &= \left ( (x^2+1)y^{(2)}\right )^{(n)} + \left (xy^{(1)} \right )^{(n)} \\ &= (x^2+1)y^{(n+2)} +n2xy^{(n+1)} + \binom{n}{2}2y^{(n)} + xy^{(n+1)} + n y^{(n)} \\ &= (x^2+1)y^{(n+2)} + (2n+1)xy^{(n+1)} + (n^2-n+n)y^{(n)} \\ &= (x^2+1)y^{(n+2)} + (2n+1)xy^{(n+1)} + n^2y^{(n)} \end{align*} as required. When \(x = 0\): \begin{align*} && y(0) &= \ln(0 + \sqrt{0^2+1}) \\ &&&= \ln 1 = 0 \\ && y'(0) &= \frac{1}{\sqrt{0^2+1}} = 1 \\ && y^{(n+2)} &= -n^2 y^{(n)} \\ && y^{(2k)} &= 0 \\ && y^{(3)} &= -1 \\ && y^{(5)} &= 3^2 \\ && y^{(7)} &= - 5^2 \cdot 3^2 \\ \end{align*} Therefore the Maclaurin series about \(x = 0\) is \begin{align*} y &= x - \frac{1}{3!} x^3 + \frac{3^2}{5!} x^5 - \frac{3^2 \cdot 5^2}{7!} x^7 + \cdots \\ &= x - \frac{1}{6} x^3 + \frac{3}{1 \cdot 2 \cdot 4 \cdot 5} x^5 - \frac{5}{1 \cdot 2 \cdot 4 \cdot 2 \cdot 7} x^7 + \cdots \\ &= x - \frac{1}{6}x^3 + \frac{3}{40} x^5 - \frac{5}{56} x^7 + \cdots \end{align*}

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2001 Paper 3 Q7
D: 1700.0 B: 1516.0
differential equation separable equation orthogonal trajectories curve sketching parabola family of curves logarithmic function

Sketch the graph of the function \(\ln x - {1 \over 2} x^2\). Show that the differential equation \[ {\mathrm{d} y \over \mathrm{d} x} = {2xy \over x^2 - 1} \] describes a family of parabolas each of which passes through the points \((1,0)\) and \((-1,0)\) and has its vertex on the \(y\)-axis. Hence find the equation of the curve that passes through the point \((1,1)\) and intersects each of the above parabolas orthogonally. Sketch this curve. [Two curves intersect orthogonally if their tangents at the point of intersection are perpendicular.]

Solution:

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\begin{align*} && y' &= \frac{2xy}{x^2-1} \\ \Rightarrow && \int \frac{1}{y} \d y &= \int \frac{2x}{x^2-1} \d x \\ \Rightarrow && \ln |y| &= \ln |x^2-1| + C \\ \Rightarrow && y &= A(x^2-1) \end{align*} which is a family of parabolas each passing through \((\pm1, 0)\) and with a vertex on the \(y\)-axis. The curve we seek must satisfy \begin{align*} && y' &= \frac{1-x^2}{2xy} \\ \Rightarrow && \int2 y \d y &= \int \left ( \frac{1}{x} - x \right) \d x \\ \Rightarrow && y^2 &= \ln x - \tfrac12 x^2 + C \\ (1,1): && 1 &= -\tfrac12+C \\ \Rightarrow && C &= \frac32 \\ \Rightarrow && y^2 &= \tfrac32 + \ln x - \tfrac12 x^2 \end{align*}
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1994 Paper 3 Q5
D: 1700.0 B: 1486.8
Taylor series Maclaurin series inverse trigonometric functions arcsin Leibniz rule differential equation logarithmic function inequality coefficient comparison proof by induction

The function \(\mathrm{f}\) is given by \(\mathrm{f}(x)=\sin^{-1}x\) for \(-1 < x < 1.\) Prove that \[ (1-x^{2})\mathrm{f}''(x)-x\mathrm{f}'(x)=0. \] Prove also that \[ (1-x^{2})\mathrm{f}^{(n+2)}(x)-(2n+1)x\mathrm{f}^{(n+1)}(x)-n^{2}\mathrm{f}^{(n)}(x)=0, \] for all \(n>0\), where \(\mathrm{f}^{(n)}\) denotes the \(n\)th derivative of \(\mathrm{f}\). Hence express \(\mathrm{f}(x)\) as a Maclaurin series. The function \(\mathrm{g}\) is given by \[ \mathrm{g}(x)=\ln\sqrt{\frac{1+x}{1-x}}, \] for \(-1 < x < 1.\) Write down a power series expression for \(\mathrm{g}(x),\) and show that the coefficient of \(x^{2n+1}\) is greater than that in the expansion of \(\mathrm{f},\) for each \(n > 0\).

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1992 Paper 2 Q6
D: 1600.0 B: 1485.5
curve sketching solving equations numerically sec logarithmic function tangent line Newton-Raphson trigonometric graphical methods optimisation

Sketch the graphs of \(y=\sec x\) and \(y=\ln(2\sec x)\) for \(0\leqslant x\leqslant\frac{1}{2}\pi\). Show graphically that the equation \[ kx=\ln(2\sec x) \] has no solution with \(0\leqslant x<\frac{1}{2}\pi\) if \(k\) is a small positive number but two solutions if \(k\) is large. Explain why there is a number \(k_{0}\) such that \[ k_{0}x=\ln(2\sec x) \] has exactly one solution with \(0\leqslant x<\frac{1}{2}\pi\). Let \(x_{0}\) be this solution, so that \(0\leqslant x_{0}<\frac{1}{2}\pi\) and \(k_{0}x_{0}=\ln(2\sec x_0)\). Show that \[ x_{0}=\cot x_{0}\ln(2\sec x_{0}). \] Use any appropriate method to find \(x_{0}\) correct to two decimal places. Hence find an approximate value for \(k_{0}\).

Solution:

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The red line is \(y = \ln (2 \sec x)\), blue is \(y = \sec x\). We can see that if the gradient is too small it never touches the red line. If it is large it will cross the red line twice in that interval. For some value it will be perfectly tangent. Since the line is tangent we must have \begin{align*} && y &= \ln (2 \sec x) \\ \Rightarrow && \frac{\d y}{ \d x} &= \frac{1}{2 \sec x} \cdot 2\sec x \tan x \\ &&&= \tan x \\ \Rightarrow && k_0 &=\tan x_0 \\ \Rightarrow && k_0 x_0 &= \ln(2 \sec x_0 ) \\ \Rightarrow && x_0 &= \cot x_0 \ln (2 \sec x_0) \end{align*} If \(f(x) =x- \cot x \ln (2 \sec x)\), then \(f'(x) =1 - 1+\ln(2\sec x) \cosec^2x = \ln(2 \sec x)\cosec^2x \) so we should look at \begin{align*} x_{n+1} &= x_n - \frac{f(x_n)}{f'(x_n)} \\ &= x_n - \frac{x_n- \cot x_n \ln (2 \sec x_n)}{\ln(2 \sec x_n)\cosec^2x_n } \\ &= x_n \left (1 - \frac{\sin^2 x_n}{\ln (2 \sec x_n)}\right) +\sin x_n \cos x_n \end{align*} \begin{array}{c|c} n & x_n \\ \hline 1 & \frac{\pi}{4} \\ 2 & 0.907701\ldots \\ 3 & 0.91439340\ldots \\ 4 & 0.914403867\ldots \\ 5 & 0.91440386\ldots \\ 6 & 0.91440386\ldots \\ \end{array} The sign change test shows that \(x_0 \approx 0.91\) and \(k_0 = \tan(x_0) \approx 1.30\)

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1988 Paper 1 Q9
D: 1500.0 B: 1516.7
integration integration by parts substitution logarithmic function trigonometric definite integral algebraic manipulation partial fractions

Find the following integrals:

  1. \(\ {\displaystyle \int_{1}^{\mathrm{e}}\frac{\ln x}{x^{2}}\,\mathrm{d}x}\,,\)
  2. \(\ {\displaystyle \int\frac{\cos x}{\sin x\sqrt{1+\sin x}}\,\mathrm{d}x.}\)

Solution:

  1. \begin{align*} \int_{1}^{\mathrm{e}}\frac{\ln x}{x^{2}}\,\mathrm{d}x &= \left [-\frac{\ln x}{x} \right]_1^e + \int_1^e \frac{1}{x^2} \, \d x \\ &= -\frac{1}{e} + \left [ -\frac{1}{x} \right]_1^e \\ &= 1 - \frac{2}{e} \end{align*}
  2. \begin{align*} \int\frac{\cos x}{\sin x\sqrt{1+\sin x}}\,\mathrm{d}x &= \int \frac{2u}{(u^2-1)u} \d u \tag{\(u^2 = 1+\sin x\)} \\ &= \int \frac{1}{u-1} - \frac{1}{u+1} \d u \\ &= \ln(u-1) - \ln (u+1) + C \\ &= \ln \l \frac{u-1}{u+1} \r + C \\ &= \ln \l \frac{\sqrt{\sin x + 1} + 1}{\sqrt{\sin x + 1} -1} \r + C \end{align*}

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