Problems

In this question, you may ignore questions of convergence. Let \(y= \dfrac {\arcsin x}{\sqrt{1-x^2}}\,\). Show that \[ (1-x^2)\frac {\d y}{\d x} -xy -1 =0 \] and prove that, for any positive integer \(n\), \[ (1-x^2) \frac{\d^{n+2}y}{\d x^{n+2}} - (2n+3)x \frac{\d ^{n+1}y}{\d x ^{n+1}} -(n+1)^2 \frac{\d^ny}{\d x^n}=0\, . \] Hence obtain the Maclaurin series for \( \dfrac {\arcsin x}{\sqrt{1-x^2}}\,\), giving the general term for odd and for even powers of \(x\). Evaluate the infinite sum \[ 1 + \frac 1 {3!} + \frac{2^2}{5!} + \frac {2^2\times 3^2}{7!}+\cdots + \frac {2^2\times 3^2\times \cdots \times n^2}{(2n+1)!} + \cdots\,. \]

In this question, you may use without proof the following result: \[ \int \sqrt{4-x^2}\, \d x = 2 \arcsin (\tfrac12 x ) + \tfrac 12 x \sqrt{4-x^2} +c\,. \] A random variable \(X\) has probability density function \(\f\) given by \[ \f(x) = \begin{cases} 2k & -a\le x <0 \\[3mm] k\sqrt{4-x^2} & \phantom{-} 0\le x \le 2 \\[3mm] 0 & \phantom{-}\text{otherwise}, \end{cases} \] where \(k\) and \(a\) are positive constants.

1. Find, in terms of \(a\), the mean of \(X\).
2. Let \(d\) be the value of \(X\) such that \(\P(X> d)=\frac1 {10}\,\). Show that \(d < 0\) if \(2a> 9\pi\) and find an expression for \(d\) in terms of \(a\) in this case.
3. Given that \(d=\sqrt 2\), find \(a\).

In this question, you may assume without proof that any function \(\f\) for which \(\f'(x)\ge 0\) is increasing; that is, \(\f(x_2)\ge \f(x_1)\) if \(x_2\ge x_1\,\).

1. 1. Let \(\f(x) =\sin x -x\cos x\). Show that \(\f(x)\) is increasing for \(0\le x \le \frac12\pi\,\) and deduce that \(\f(x)\ge 0\,\) for \(0\le x \le \frac12\pi\,\).
   2. Given that \(\dfrac{\d}{\d x} (\arcsin x) \ge1\) for \(0\le x< 1\), show that \[ \arcsin x\ge x \quad (0\le x < 1). \]
   3. Let \(\g(x)= x\cosec x\, \text{ for }0< x < \frac12\pi\). Show that \(\g\) is increasing and deduce that \[ ({\arcsin x})\, x^{-1} \ge x\,{\cosec x} \quad (0 < x < 1). \]
2. Given that $\dfrac{\d}{\d x} (\arctan x)\le 1\text{ for }x\ge 0$, show by considering the function \(x^{-1} \tan x\) that \[ (\tan x)( \arctan x) \ge x^2 \quad (0< x < \tfrac12\pi). \]

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 \(t\ge1\), \[ \arcsin (t^{-1}) \ge t - \sqrt{t^2-1}\,. \]

A particle of mass \(m\) is initially at rest on a rough horizontal surface. The particle experiences a force \(mg\sin \pi t\), where \(t\) is time, acting in a fixed horizontal direction. The coefficient of friction between the particle and the surface is \(\mu\). Given that the particle starts to move first at \(t=T_0\), state the relation between \(T_0\) and \(\mu\).

1. For \(\mu = \mu_0\), the particle comes to rest for the first time at \(t=1\). Sketch the acceleration-time graph for \(0\le t \le 1\). Show that \[ 1+\left(1-\mu_0^2\right)^{\frac12} -\mu_0\pi +\mu_0 \arcsin \mu_0 =0\,. \]
2. For \(\mu=\mu_0\) sketch the acceleration-time graph for \(0\le t\le 3\). Describe the motion of the particle in this case and in the case \(\mu=0\).

The points \(A\) and \(B\) are \(180\) metres apart and lie on horizontal ground. A missile is launched from \(A\) at speed of \(100\,\)m\,s\(^{-1}\) and at an acute angle of elevation to the line \(AB\) of \(\arcsin \frac35\). A time \(T\) seconds later, an anti-missile missile is launched from \(B\), at speed of \(200\,\)m\,s\(^{-1}\) and at an acute angle of elevation to the line \(BA\) of \(\arcsin \frac45\). The motion of both missiles takes place in the vertical plane containing \(A\) and \(B\), and the missiles collide. Taking \(g =10\,\)m\,s\(^{-2}\) and ignoring air resistance, find \(T\). \noindent [Note that \(\arcsin \frac35\) is another notation for \(\sin^{-1} \frac35\,\).]

Write down a value of \(\theta\,\) in the interval \(\frac{1}{4}\pi< \theta <\frac{1}{2}\pi\) that satisfies the equation \[ 4\cos\theta+ 2\sqrt3\, \sin\theta = 5 \;. \] Hence, or otherwise, show that \[ \pi=3\arccos(5/\sqrt{28}) + 3\arctan(\sqrt3/2)\;. \] Show that \[ \pi=4\arcsin(7\sqrt2/10) - 4\arctan(3/4)\;. \]

Given that \(x+a>0\) and \(x+b>0\,\), and that \(b>a\,\), show that \[ \frac{\mathrm{d} \ }{\mathrm{d} x} \arcsin \left ( \frac{x + a }{ \ x + b} \right) = \frac{ \sqrt{\;b - a\;}} {( x + b ) \sqrt{ a + b + 2x} \ \ } \] and find $\displaystyle \frac{\mathrm{d} \ }{ \mathrm{d} x} \; \mathrm{arcosh} \left ( \frac{x + b }{ \ x + a} \right)$. Hence, or otherwise, integrate, for \(x > -1\,\),

1. \[ \int \frac{1}{ ( x + 1) \sqrt{x + 3} } \mathrm{d} x \]
2. \[ \int \frac{1} {( x + 3 ) \sqrt{x + 1} } \mathrm{d} x \] .

In this question, the function \(\sin^{-1}\) is defined to have domain \( -1\le x \le 1\) and range \linebreak \( - \frac{1}{2}\pi \le x \le \frac{1}{2}\pi\) and the function \(\tan^{-1}\) is defined to have the real numbers as its domain and range \( - \frac{1}{2}\pi < x < \frac{1}{2}\pi\).

1. Let $$ \g(x) = \displaystyle {2x \over 1 + x^2}\;, \ \ \ \ \ \ \ \ \ \ -\infty
2. Let \[ \displaystyle \f \l x \r = \sin^{-1} \l {2x \over 1 + x^2} \r \;,\ \ \ \ \ \ \ \ \ -\infty < x < \infty\;. \] Show that $ \f(x ) = 2 \tan^{-1} x\( for \) -1 \le x \le 1\,\( and \)\f(x) = \pi - 2 \tan^{-1} x \( for \)x\ge1\,$. Sketch the graph of \(\f(x)\).

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 \(n > 0\).

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 \(t\).

Show Solution

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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