Problems

The continuous random variable \(X\) is uniformly distributed on \([a,b]\) where \(0 < a < b\).

1. Let \(\mathrm{f}\) be a function defined for all \(x \in [a,b]\)
   • with \(\mathrm{f}(a) = b\) and \(\mathrm{f}(b) = a\),
   • which is strictly decreasing on \([a,b]\),
   • for which \(\mathrm{f}(x) = \mathrm{f}^{-1}(x)\) for all \(x \in [a,b]\).
   The random variable \(Y\) is defined by \(Y = \mathrm{f}(X)\). Show that \[ \mathrm{P}(Y \leqslant y) = \frac{b - \mathrm{f}(y)}{b - a} \quad \text{for } y \in [a,b]. \] Find the probability density function for \(Y\) and hence show that \[ \mathrm{E}(Y^2) = -ab + \int_a^b \frac{2x\,\mathrm{f}(x)}{b-a} \; \mathrm{d}x. \]
2. The random variable \(Z\) is defined by \(\dfrac{1}{Z} + \dfrac{1}{X} = \dfrac{1}{c}\) where \(\dfrac{1}{c} = \dfrac{1}{a} + \dfrac{1}{b}\). By finding the variance of \(Z\), show that \[ \ln\left(\frac{b-c}{a-c}\right) < \frac{b-a}{c}. \]

The function f satisfies \(f(0) = 0\) and \(f'(t) > 0\) for \(t > 0\). Show by means of a sketch that, for \(x > 0\), $$\int_0^x f(t) \, dt + \int_0^{f(x)} f^{-1}(y) \, dy = xf(x).$$

1. The (real) function g is defined, for all \(t\), by $$(g(t))^3 + g(t) = t.$$ Prove that \(g(0) = 0\), and that \(g'(t) > 0\) for all \(t\). Evaluate \(\int_0^2 g(t) \, dt\).
2. The (real) function h is defined, for all \(t\), by $$(h(t))^3 + h(t) = t + 2.$$ Evaluate \(\int_0^8 h(t) \, dt\).

Show Solution

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Solution:

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Notice the total area is \(xf(x)\) and it is made up of the sum of the two integrals.

1. Suppose \((g(t))^3 + g(t) = t\). Notice that \((g(0))^3 + g(0) =0 \Rightarrow g(0)((g(0))^2 + 1) = 0 \Rightarrow g(0) = 0\). \begin{align*} && t &= (g(t))^3 + g(t) \\ \Rightarrow && 1 &= 3(g(t))^2 g'(t) + g'(t) \\ \Rightarrow && g'(t) &= \frac{1}{1 + 3(g(t))^2} > 0 \end{align*}
   

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   From our sketch, we can see we are interested in: \begin{align*} && \int_0^2 g(t) \d t &= 2 - \int_0^1 (x^3 + x) \d x \\ &&&= 2 - \frac14 - \frac12 = \frac54 \end{align*}
2. \(\,\)
   

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   From our second sketch, we can see that: \begin{align*} && \int_0^8 h(t) \d t &= 16 - \int_1^2 (x^3+x-2) \d x \\ &&&= 16 - \left ( \frac{8}{4} + \frac{2^2}{2} - 2 \cdot 2 \right)+ \left ( \frac{1}{4} + \frac{1}{2} - 2 \right) \\ &&&= \frac{59}{4} \end{align*}

The functions \(\s\) and \(\c\) satisfy \(\s(0)= 0\,\), \(\c(0)=1\,\) and \[ \s'(x) = \c(x)^2 ,\] \[ \c'(x)=-\s(x)^2. \] You may assume that \(\s\) and \(\c\) are uniquely defined by these conditions.

1. Show that \(\s(x)^3+\c(x)^3\) is constant, and deduce that \(\s(x)^3+\c(x)^3=1\,\).
2. Show that \[ \frac{\d }{\d x} \, \Big( \s(x) \c(x) \Big) = 2\c(x)^3-1 \] and find (and simplify) an expression in terms of \(\c(x)\) for $\dfrac{\d }{\d x} \left( \dfrac{\s(x)}{\c(x)} \right) $.
3. Find the integrals \[ \int \s(x)^2 \, \d x \ \ \ \ \ \ \text{and} \ \ \ \ \ \ \int \s(x)^5 \, \d x \,. \]
4. Given that \(\s\) has an inverse function, \(\s^{-1}\), use the substitution \(u = \s(x)\) to show that \[ \int \frac{1}{(1-u^3)^{\frac{2}{3}}} \, \d u = \s^{-1}(u) \, + \text{constant}. \]
5. Find, in terms of \(u\), the integrals \[ \int \frac{1}{{(1-u^3)}^{\frac{4}{3}}} \, \d u \ \ \ \ \ \ \text{and} \ \ \ \ \ \ \int {(1-u^3)}^{\frac{1}{3}} \, \d u \,. \]

In this question, you are not permitted to use any properties of trigonometric functions or inverse trigonometric functions. The function \(\T\) is defined for \(x>0\) by \[ \T(x) = \int_0^x \! \frac 1 {1+u^2} \, \d u\,, \] and $\displaystyle T_\infty = \int_0^\infty \!\! \frac 1 {1+u^2} \, \d u\,$ (which has a finite value).

1. By making an appropriate substitution in the integral for \(\T(x)\), show that \[\T(x) = \T_\infty - \T(x^{-1})\,.\]
2. Let \(v= \dfrac{u+a}{1-au}\), where \(a\) is a constant. Verify that, for \(u\ne a^{-1}\), \[ \frac{\d v}{\d u} = \frac{1+v^2}{1+u^2} \,. \] Hence show that, for \(a>0\) and \(x< \dfrac1a\,\), \[ \T(x) = \T\left(\frac{x+a}{1-ax}\right) -\T(a) \,. \] Deduce that \[ \T(x^{-1}) = 2\T_\infty -\T\left(\frac{x+a}{1-ax}\right) -\T(a^{-1}) \] and hence that, for \(b>0\) and \(y>\dfrac1b\,\), \[ \T(y) =2\T_\infty - \T\left(\frac{y+b}{by-1}\right) - \T(b) \,. \]
3. Use the above results to show that \(\T(\sqrt3)= \tfrac23 \T_\infty \,\) and \(\T(\sqrt2 -1)= \frac14 \T_\infty\,\).

The following result applies to any function \(\f\) which is continuous, has positive gradient and satisfies \(\f(0)=0\,\): \[ ab\le \int_0^a\f(x)\, \d x + \int_0^b \f^{-1}(y)\, \d y\,, \tag{\(*\)}\] where \(\f^{-1}\) denotes the inverse function of \(\f\), and \(a\ge 0\) and \(b\ge 0\).

1. By considering the graph of \(y=\f(x)\), explain briefly why the inequality \((*)\) holds. In the case \(a>0\) and \(b>0\), state a condition on \(a\) and \(b\) under which equality holds.
2. By taking \(\f(x) = x^{p-1}\) in \((*)\), where \(p>1\), show that if \(\displaystyle \frac 1p + \frac 1q =1\) then \[ ab \le \frac{a^p}p + \frac{b^q}q\,. \] Verify that equality holds under the condition you stated above.
3. Show that, for \(0\le a \le \frac12 \pi\) and \(0\le b \le 1\), \[ ab \le b\arcsin b + \sqrt{1-b^2} \, - \cos a\,. \] Deduce that, for \(t\ge1\), \[ \arcsin (t^{-1}) \ge t - \sqrt{t^2-1}\,. \]

The function \(\f\) is defined by \[ \f(x) = \frac{\e^x-1}{\e-1}, \ \ \ \ \ x\ge0, \] and the function \(\g\) is the inverse function to \(\f\), so that \(\g(\f(x))=x\). Sketch \(\f(x)\) and \(\g(x)\) on the same axes. Verify, by evaluating each integral, that \[ \int_0^\frac12 \f(x) \,\d x + \int_0^k \g(x) \,\d x = \frac1 {2(\sqrt \e +1)}\,, \] where \(\displaystyle k= \frac 1{\sqrt\e+1}\), and explain this result by means of a diagram.

A transformation \(T\) of the real numbers is defined by \[ y=T(x)=\frac{ax-b}{cx-d}\,, \] where \(a,b,c\), \(d\) are real numbers such that \(ad\neq bc\). Find all numbers \(x\) such that \(T(x)=x.\) Show that the inverse operation, \(x=T^{-1}(y)\) expressing \(x\) in terms of \(y\) is of the same form as \(T\) and find corresponding numbers \(a',b',c'\),\(d'\). Let \(S_{r}\) denote the set of all real numbers excluding \(r\). Show that, if \(c\neq0,\) there is a value of \(r\) such that \(T\) is defined for all \(x\in S_{r}\) and find the image \(T(S_{r}).\) What is the corresponding result if \(c=0\)? If \(T_{1},\) given by numbers \(a_{1},b_{1},c_{1},d_{1},\) and \(T_{2},\) given by numbers \(a_{2},b_{2},c_{2},d_{2}\) are two such transformations, show that their composition \(T_{3},\) defined by \(T_{3}(x)=T_{2}(T_{1}(x)),\) is of the same form. Find necessary and sufficient conditions on the numbers \(a,b,c,d\) for \(T^{2}\), the composition of \(T\) with itself, to be the identity. Hence, or otherwise, find transformations \(T_{1},T_{2}\) and their composition \(T_{3}\) such that \(T_{1}^{2}\) and \(T_{2}^{2}\) are each the identity but \(T_{3}^{2}\) is not.

The function \(\mathrm{f}\) is defined for \(x<2\) by \[ \mathrm{f}(x)=2| x^{2}-x|+|x^{2}-1|-2|x^{2}+x|. \] Find the maximum and minimum points and the points of inflection of the graph of \(\mathrm{f}\) and sketch this graph. Is \(\mathrm{f}\) continuous everywhere? Is \(\mathrm{f}\) differentiable everywhere? Find the inverse of the function \(\mathrm{f}\), i.e. expressions for \(\mathrm{f}^{-1}(x),\) defined in the various appropriate intervals.

Show Solution

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Solution: \[ f(x) = 2|x(x-1)| + |(x-1)(x+1)|-2|x(x+1)| \] Therefore the absolute value terms will change behaviour at \(x = -1, 0, 1\). Then \begin{align*} f(x) &= \begin{cases} 2(x^2-x)+(x^2-1)-2(x^2+x) & x \leq -1 \\ 2(x^2-x)-(x^2-1)+2(x^2+x) & -1 < x \leq 0 \\ -2(x^2-x)-(x^2-1)-2(x^2+x) & 0 < x \leq 1 \\ 2(x^2-x)+(x^2-1)-2(x^2+x) & 1 < x\end{cases} \\ &= \begin{cases} x^2-4x-1 & x \leq -1 \\ 3x^2+1& -1 < x \leq 0 \\ -5x^2+1& 0 < x \leq 1 \\ x^2-4x-1 & 1 < x\end{cases} \\ \\ f'(x) &= \begin{cases} 2x-4 & x <-1 \\ 6x & -1 < x < 0 \\ -10x & 0 < x < 1 \\ 2x-4 & 1 < x\end{cases} \\ \end{align*} Therefore \(f'(x) = 0 \Rightarrow x = 0, 2\) and so we should check all the turning points. Therefore the minimum is \(x = 2, y = -5\), maximum is \(x = -2, y = 11\) (assuming the range is actually \(|x| < 2\). There is a point of inflection at \(x = 0, y = 1\).

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\(f\) is continuous everywhere as a sum of continuous functions. \(f\) is not differentiable at \(x = -1, 1\) Suppose \begin{align*} &&y &=x^2-4x-1 \\ &&&= (x-2)^2 -5 \\ \Rightarrow &&x &= 2\pm \sqrt{y+5} \\ \\ && y &= 3x^2+1 \\ \Rightarrow && x &= \pm \sqrt{\frac{y-1}{3}} \\ \\ && y &= -5x^2+1 \\ \Rightarrow && x &=\pm \sqrt{\frac{1-y}{5}} \\ \\ \Rightarrow && f^{-1}(y) &= \begin{cases} 2 - \sqrt{y+5} & y > 4 \\ -\sqrt{\frac{y-1}{3}} & 1 < y < 4 \\ \sqrt{\frac{1-y}{5}} & -4 < y < 1 \\ 2 + \sqrt{y+5} & y < -4 \end{cases} \end{align*}

The function \(\mathrm{f}\) satisfies the condition \(\mathrm{f}'(x)>0\) for \(a\leqslant x\leqslant b\), and \(\mathrm{g}\) is the inverse of \(\mathrm{f}.\) By making a suitable change of variable, prove that \[ \int_{a}^{b}\mathrm{f}(x)\,\mathrm{d}x=b\beta-a\alpha-\int_{\alpha}^{\beta}\mathrm{g}(y)\,\mathrm{d}y, \] where \(\alpha=\mathrm{f}(a)\) and \(\beta=\mathrm{f}(b)\). Interpret this formula geometrically, in the case where \(\alpha\) and \(a\) are both positive. Prove similarly and interpret (for \(\alpha>0\) and \(a>0\)) the formula \[ 2\pi\int_{a}^{b}x\mathrm{f}(x)\,\mathrm{d}x=\pi(b^{2}\beta-a^{2}\alpha)-\pi\int_{\alpha}^{\beta}\left[\mathrm{g}(y)\right]^{2}\,\mathrm{d}y. \]

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Solution: Let \(u = f(x)\) then \(\frac{\d u}{\d x} = f'(x)\) and \begin{align*} \int_a^b f(x) \d x &\underbrace{=}_{\text{IBP}} \left [ xf(x) \right]_a^b - \int_a^b x f'(x) \d x \\ &\underbrace{=}_{u = f(x)} b \beta - a \alpha - \int_{u = f(a) = \alpha}^{u = f(b) = \beta} g(u) \d u \\ &= b \beta - a \alpha - \int_{\alpha}^{\beta} g(u) \d u \end{align*}

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\[ \underbrace{\int_{a}^{b}\mathrm{f}(x)\,\mathrm{d}x}_{\text{red area}}=\underbrace{b\beta}_{\text{whole area}}-\underbrace{a\alpha}_{\text{area in green}}-\underbrace{\int_{\alpha}^{\beta}\mathrm{g}(y)\,\mathrm{d}y}_{\text{area in blue}}, \] \begin{align*} 2\pi \int_a^b x f(x) \d x &\underbrace{=}_{\text{IBP}}\pi \left [ x^2 f(x) \right]_a^b - \pi \int_a^b x^2 f'(x) \d x \\ &\underbrace{=}_{x = g(u)} \pi (b^2 \beta - a^2 \alpha) - \pi \int_{u = f(a) = \alpha}^{u = f(b) = \beta} [g(u)]^2 \d u \\ &= \pi(b^2 \beta - a^2 \alpha) - \pi \int_\alpha^\beta [g(u)]^2 \d u \end{align*} This is the volume outside the function in the volume of revolution about the \(y\) axis between \( \alpha\) and \(\beta\).

Let \(y=\mathrm{f}(x)\), \((0\leqslant x\leqslant a)\), be a continuous curve lying in the first quadrant and passing through the origin. Suppose that, for each non-negative value of \(y\) with \(0\leqslant y\leqslant\mathrm{f}(a)\), there is exactly one value of \(x\) such that \(\mathrm{f}(x)=y\); thus we may write \(x=\mathrm{g}(y)\), for a suitable function \(\mathrm{g}.\) For \(0\leqslant s\leqslant a,\) \(0\leqslant t\leqslant \mathrm{f}(a)\), define \[ \mathrm{F}(s)=\int_{0}^{s}\mathrm{f}(x)\,\mathrm{d}x,\qquad\mathrm{G}(t)=\int_{0}^{t}\mathrm{g}(y)\,\mathrm{d}y. \] By a geometrical argument, show that \[ \mathrm{F}(s)+\mathrm{G}(t)\geqslant st.\tag{*} \] When does equality occur in \((*)\)? Suppose that \(y=\sin x\) and that the ranges of \(x,y,s,t\) are restricted to \(0\leqslant x\leqslant s\leqslant\frac{1}{2}\pi,\) \(0\leqslant y\leqslant t\leqslant1\). By considering \(s\) such that the equality holds in \((*)\), show that \[ \int_{0}^{t}\sin^{-1}y\,\mathrm{d}y=t\sin^{-1}t-\left(1-\cos(\sin^{-1}t)\right). \] Check this result by differentiating both sides with respect to \(t\).

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Solution:

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The blue area is \(F(s)\) the red area is \(G(t)\), the dashed rectangle (which is a subset of the red and blue areas) has area \(st\) therefore \(F(s) + G(t) \geq st\). Equality holds if \(f(s) = t\). \begin{align*} && \int_0^t \sin^{-1} y \d y + \int_0^{\sin^{-1} t} \sin x \d x &= t \sin^{-1} t \\ \Rightarrow && \int_0^t \sin^{-1} y \d y &= t \sin^{-1} t - \left [ -\cos (x) \right]_0^{\sin^{-1} t} \\ &&&= t \sin^{-1} t - (1- \cos (\sin^{-1} t)) \end{align*} Let \(y = t \sin^{-1} t - (1- \cos (\sin^{-1} t))\) then, \begin{align*} \frac{\d y}{\d t} &= \sin^{-1} t +t \frac{\d}{\d t} \l \sin^{-1} (t) \r - \sin ( \sin^{-1} t) \frac{\d}{\d t} \l \sin^{-1} (t) \r \\ &= \sin^{-1} t \end{align*} as required

If \(y=\mathrm{f}(x)\), then the inverse of \(\mathrm{f}\) (when it exists) can be obtained from Lagrange's identity. This identity, which you may use without proof, is \[ \mathrm{f}^{-1}(y)=y+\sum_{n=1}^{\infty}\frac{1}{n!}\frac{\mathrm{d}^{n-1}}{\mathrm{d}y^{n-1}}\left[y-\mathrm{f}\left(y\right)\right]^{n}, \] provided the series converges.

1. Verify Lagrange's identity when \(\mathrm{f}(x)=\alpha x\), \((0<\alpha<2)\).
2. Show that one root of the equation \[ \tfrac{1}{2}=x-\tfrac{1}{4}x^{3} \] is \[ x=\sum_{n=0}^{\infty}\frac{\left(3n\right)!}{n!\left(2n+1\right)!2^{4n+1}} \]
3. Find a solution for \(x\), as a series in \(\lambda,\) of the equation \[ x=\mathrm{e}^{\lambda x}. \]