Problems

2017 Paper 3 Q1

D: 1700.0 B: 1516.0

Prove that, for any positive integers $n$ and $r$, \[ \frac{1}{^{n+r}\C_{r+1}} =\frac{r+1}{r} \left(\frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}}\right). \] Hence determine \[ \sum_{n=1}^{\infty}{\frac{1}{^{n+r}\C_{r+1}}} \,, \] and deduce that \ $\displaystyle \sum_{n=2}^\infty \frac 1 {^{n+2}\C_3} = \frac12\,$.
Show that, for $n \ge 3\,$, \[ \frac{3!}{n^3} < \frac{1}{^{n+1}\C_{3}} \ \ \ \ \ \text{and} \ \ \ \ \ \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} < \frac{5!}{n^3} \,. \] By summing these inequalities for $n \ge 3\,$, show that \[ \frac{115}{96} < \sum_{n=1}^{\infty}{\frac{1}{n^3}} < \frac{116}{96} \, . \]

{\bf Note: } $^n\C_r$ is another notation for $\displaystyle \binom n r $.

Solution: \begin{align*} \frac{r+1}{r} \left(\frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}}\right) &= \frac{r+1}{r} \l \frac{r!(n-1)!}{(n+r-1)!} - \frac{r!n!}{(n+r)!} \r \\ &= \frac{(r+1)!(n-1)!}{r(n+r-1)!} \l 1 - \frac{n}{n+r} \r \\ &= \frac{(r+1)!(n-1)!}{r(n+r-1)!} \frac{r}{n+r} \\ &= \frac{(r+1)!n!}{(n+r)!} \\ &= \frac{1}{^{n+r}\C_{r+1}} \end{align*} \begin{align*} \sum_{n=1}^{\infty}{\frac{1}{^{n+r}\C_{r+1}}} &= \sum_{n=1}^{\infty} \l \frac{r+1}{r} \left(\frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}}\right) \r \\ &= \frac{r+1}{r} \sum_{n=1}^{\infty} \l \frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}} \r \\ &= \frac{r+1}{r} \lim_{N \to \infty} \sum_{n=1}^{N} \l \frac{1}{^{n+r-1}\C_{r}}-\frac{1}{^{n+r}\C_{r}} \r \\ &= \frac{r+1}{r} \lim_{N \to \infty} \l \frac{1}{^{1+r-1}\C_{r}} - \frac{1}{^{N+r}\C_{r}}\r \\ &= \frac{r+1}{r} \frac{1}{^{1+r-1}\C_{r}} \tag{since $\frac{1}{^{N+r}\C_{r}} \to 0$} \\ &= \frac{r+1}{r} \end{align*} When $r = 2$, we have: \begin{align*} && \frac{3}{2} &= \sum_{n=1}^{\infty}{\frac{1}{^{n+2}\C_{3}}} \\ && &=\frac{1}{^{1+2}\C_{3}} + \sum_{n=2}^{\infty}{\frac{1}{^{n+2}\C_{3}}} \\ && &= 1 + \sum_{n=2}^{\infty}{\frac{1}{^{n+2}\C_{3}}} \\ \Rightarrow && \sum_{n=2}^{\infty}{\frac{1}{^{n+2}\C_{3}}} &= \frac12 \end{align*} \begin{align*} \frac{1}{^{n+1}\C_{3}} &= \frac{3!}{(n+1)n(n-1)} \\ &= \frac{3!}{n^3-n} \\ &> \frac{3!}{n^3} \end{align*} \begin{align*} \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} &= \frac{5!}{(n+1)n(n-1)} - \frac{5!}{(n+2)(n+1)n(n-1)(n-2)} \\ &= \frac{5!}{n^3} \frac{n^2}{n^2-1}\l 1- \frac{1}{n^2-4} \r \\ &= \frac{5!}{n^3} \frac{n^2}{n^2-1}\l \frac{n^2-5}{n^2-4} \r \\ &= \frac{5!}{n^3} \frac{n^2(n^2-5)}{(n^2-1)(n^2-4)} \\ &< \frac{5!}{n^3} \end{align*} Since $k(k-5) < (k-1)(k-4) \Leftrightarrow 0 < 4$, this only makes sense if $n \geq 3$ \begin{align*} &&\frac{3!}{n^3} &< \frac{1}{^{n+1}\C_{3}} \tag{if $n \geq 3$} \\ \Rightarrow &&\sum_{n=3}^\infty \frac{3!}{n^3} &< \sum_{n=3}^\infty \frac{1}{^{n+1}\C_{3}} \\ \Rightarrow && \frac{6}{1^3} + \frac{6}{2^3} + \sum_{n=3}^\infty \frac{3!}{n^3} &< \frac{6}{1^3} + \frac{6}{2^3} + \sum_{n=3}^\infty \frac{1}{^{n+1}\C_{3}} \\ \Rightarrow && \sum_{n=1}^\infty \frac{3!}{n^3} &< 6 + \frac{3}{4} + \sum_{n=2}^\infty \frac{1}{^{n+2}\C_{2+1}} \\ \Rightarrow && \sum_{n=1}^\infty \frac{3!}{n^3} &< 6 + \frac{3}{4} + \frac{1}{2} = \frac{29}{4} \\ \Rightarrow && \sum_{n=1}^\infty \frac{1}{n^3} &< \frac{29}{24} = \frac{116}{96} \\ \end{align*} \begin{align*} && \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} &< \frac{5!}{n^3} \\ \Rightarrow && \sum_{n=3}^\infty \l \frac{20}{^{n+1}\C_3} - \frac{1}{^{n+2}\C_{5}} \r &< \sum_{n=3}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=3}^\infty \frac{20}{^{n+1}\C_3} - \sum_{n=3}^\infty \frac{1}{^{n+2}\C_{5}} &< \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=3}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=2}^\infty \frac{20}{^{n+2}\C_{2+1}} - \sum_{n=1}^\infty \frac{1}{^{n+4}\C_{4+1}} &< \frac{120}{1^3} + \frac{120}{2^3} + \sum_{n=3}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{120}{1^3} + \frac{120}{2^3} + \frac{20}{2} - \frac{4+1}{4} &< \sum_{n=1}^\infty \frac{5!}{n^3} \\ \Rightarrow && \frac{115}{96} &< \sum_{n=1}^\infty \frac{1}{n^3} \\ \end{align*}

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2017 Paper 3 Q2

D: 1700.0 B: 1500.1

The transformation $R$ in the complex plane is a rotation (anticlockwise) by an angle $\theta$ about the point represented by the complex number $a$. The transformation $S$ in the complex plane is a rotation (anticlockwise) by an angle $\phi$ about the point represented by the complex number $b$.

The point $P$ is represented by the complex number~$z$. Show that the image of $P$ under $R$ is represented by \[ \e^{{\mathrm i} \theta}z + a(1-\e^{{\rm i} \theta})\,. \]
Show that the transformation $SR$ (equivalent to $R$ followed by $S$) is a rotation about the point represented by $c$, where \[ %\textstyle c\,\sin \tfrac12 (\theta+\phi) = a\,\e^{ {\mathrm i}\phi/2}\sin \tfrac12\theta + b\,\e^{-{\mathrm i} \theta/2}\sin \tfrac12 \phi \,, \] provided $\theta+\phi \ne 2n\pi$ for any integer $n$. What is the transformation $SR$ if $\theta +\phi = 2\pi$?
Under what circumstances is $RS =SR$?

Solution:

We can map $a \mapsto 0$, rotate the whole plane, then shift the plane back to $a$, so $z \mapsto (z-a) \mapsto e^{i \theta}(z-a) \mapsto a + e^{i \theta}(z-a) = e^{i \theta}z + a(1 - e^{i\theta})$
$z \mapsto e^{i \theta}z + a(1 - e^{i\theta}) \mapsto e^{i \phi} \l e^{i \theta}z + a(1 - e^{i\theta}) \r + b(1 - e^{i \phi})$ \begin{align*} e^{i \phi} \l e^{i \theta}z + a(1 - e^{i\theta}) \r + b(1 - e^{i \phi}) &= e^{i(\phi + \theta)}z + ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi}) \\ \end{align*} Therefore this is rotation by angle $\phi + \theta$ and about \begin{align*} \frac{ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi})}{1 - e^{i(\phi + \theta)}} &= \frac{e^{-i\frac{(\phi + \theta)}{2}} \l ae^{i\phi} - ae^{i (\theta + \phi)} + b(1 - e^{i \phi}) \r}{e^{-i\frac{(\phi + \theta)}{2}} - e^{i\frac{(\phi + \theta)}2}} \\ &= \frac{\l ae^{i\frac{\phi-\theta}{2}} - ae^{i \frac{(\theta + \phi)}{2}} + b(e^{-i\frac{(\phi + \theta)}{2}} -e^{i\frac{(\phi - \theta)}{2}}) \r}{e^{-i\frac{(\phi + \theta)}{2}} - e^{i\frac{(\phi + \theta)}2}} \\ &= \frac{ae^{i\frac{\phi}{2}} 2i\sin(\frac{\theta}{2}) + be^{-i\frac{\theta}{2}}2i\sin\frac{\phi}{2} }{2i \sin(\frac{\phi + \theta}2)} \\ \end{align*} as required. If $\phi + \theta = 2\pi$, then $z \mapsto z + (b-a)(1 - e^{i\phi})$ which is a translation.
If $\phi + \theta \neq 2 \pi$ then $RS = ST$ if \begin{align*} && a\,\e^{ {\mathrm i}\phi/2}\sin \tfrac12\theta + b\,\e^{-{\mathrm i} \theta/2}\sin \tfrac12 \phi &= b\,\e^{ {\mathrm i}\theta/2}\sin \tfrac12\phi + a\,\e^{-{\mathrm i} \phi/2}\sin \tfrac12 \theta \\ && a\,(\e^{ {\mathrm i}\phi/2}-\e^{-{\mathrm i}\phi/2})\sin \tfrac12\theta + b\,(\e^{-{\mathrm i} \theta/2}-\e^{+{\mathrm i} \theta/2})\sin \tfrac12 \phi &= 0 \\ && a \sin \frac{\phi}{2} \sin \frac{\theta}{2}-b \sin \frac{\theta}{2} \sin \frac{\phi}{2} &= 0 \\ \Leftrightarrow && a = b \text{ or } \sin \frac{\theta}{2} = 0 \text{ or } \sin \frac{\phi}{2} = 0 \\ \Leftrightarrow && a = b \text{ or } \theta = 0 \text{ or } \phi = 0 \\ \end{align*} If $\phi + \theta \neq 2 \pi$ then $RS = ST$ if $b = a$ or $e^{i\phi} = e^{i \theta}$ ie rotation through the same angle.

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2017 Paper 3 Q3

D: 1700.0 B: 1500.0

Let $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the quartic equation \[ x^4 +px^3 +qx^2 +r x +s =0 \,. \] You are given that, for any such equation, $\,\alpha \beta + \gamma\delta\,$, $\alpha\gamma+\beta\delta\,$ and $\,\alpha \delta + \beta\gamma\,$ satisfy a cubic equation of the form \[ y^3+Ay^2+ (pr-4s)y+ (4qs-p^2s -r^2) =0 \,. \] Determine $A$. Now consider the quartic equation given by $p=0\,$, $q= 3\,$, $r=-6\,$ and $s=10\,$.

Find the value of $\alpha\beta + \gamma \delta$, given that it is the largest root of the corresponding cubic equation.
Hence, using the values of $q$ and $s$, find the value of $(\alpha +\beta)(\gamma+\delta)\,$ and the value of $\alpha\beta$ given that $\alpha\beta >\gamma\delta\,$.
Using these results, and the values of $p$ and $r$, solve the quartic equation.

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2017 Paper 3 Q4

D: 1700.0 B: 1484.0

For any function $\f$ satisfying $\f(x) > 0$, we define the {\em geometric mean}, F, by \[ \F(y) \; = \mbox{ \fontsize{12}{15.6}\selectfont $\e$} \mbox{ \fontsize{14}{15.6}\selectfont $ ^{\! \raisemath {3pt} {\frac{1}{y} \! \int_{\raisemath{-1pt}{0}}^{\raisemath{1pt}{y}} \ln \f(x) \, \d x} } $ } \ \ \ \ \ \ (y>0)\,. \]

The function f satisfies $\f(x) > 0$ and $a$ is a positive number with $a\ne1$. Prove that \[ \F(y) = \mbox{ \fontsize{12}{15.6}\selectfont $a$} \mbox{ \fontsize{14}{15.6}\selectfont $ ^{ \! \raisemath {3pt} {\frac{1}{y} \! \int_{\raisemath{-1pt}{0}}^{\raisemath{1pt}{y}} \log_a \f(x) \, \d x} } $ } . \]
The functions f and g satisfy $\f(x) > 0$ and $\g(x) > 0$, and the function $\h$ is defined by $\h(x) = \f(x)\g(x)$. Their geometric means are F, G and H, respectively. Show that $\H(y)= \F(y) \G(y)\,$.
Prove that, for any positive number $b$, the geometric mean of $b^x$ is $\sqrt{b^y}\,$.
Prove that, if $\f(x)>0$ and the geometric mean of $\f(x)$ is $\sqrt{\f(y)}\,$, then $\f(x) = b^x$ for some positive number $b$.

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2017 Paper 3 Q5

D: 1700.0 B: 1484.0

The point with cartesian coordinates $(x,y)$ lies on a curve with polar equation $r=\f(\theta)\,$. Find an expression for $\dfrac{\d y}{\d x}$ in terms of $\f(\theta)$, $\f'(\theta)$ and $\tan\theta\,$. Two curves, with polar equations $r=\f(\theta)$ and $r=\g(\theta)$, meet at right angles. Show that where they meet \[ \f'(\theta) \g'(\theta) +\f(\theta)\g(\theta) = 0 \,. \] The curve $C$ has polar equation $r=\f(\theta)$ and passes through the point given by $r=4$, $\theta = - \frac12\pi$. For each positive value of $a$, the curve with polar equation $r= a(1+\sin\theta)$ meets~$C$ at right angles. Find $\f(\theta)\,$. Sketch on a single diagram the three curves with polar equations $r= 1+\sin\theta\,$, \ $r= 4(1+\sin\theta)$ and $r=\f(\theta)\,$.

Solution: $(x, y) = (f(\theta)\cos(\theta), f(\theta)\sin(\theta))$ so \begin{align*} \frac{dy}{d\theta} &= -f(\theta)\sin(\theta) + f'(\theta)\cos(\theta) \\ \frac{dx}{d\theta} &= f(\theta)\cos(\theta) + f'(\theta)\sin(\theta) \\ \frac{dy}{dx} &= \frac{-f(\theta)\sin(\theta) + f'(\theta)\cos(\theta)}{f(\theta)\cos(\theta) + f'(\theta)\sin(\theta) } \\ &= \frac{-f(\theta)\tan(\theta) + f'(\theta)}{f(\theta) + f'(\theta)\tan(\theta) } \end{align*} If the curves meet at right angles then the product of their gradients is $-1$, ie \begin{align*} \frac{-f(\theta)\tan(\theta) + f'(\theta)}{f(\theta) + f'(\theta)\tan(\theta) } \cdot \frac{-g(\theta)\tan(\theta) + g'(\theta)}{g(\theta) + g'(\theta)\tan(\theta) } &= -1 \\ f(\theta)g(\theta)\tan^2 \theta - f(\theta)g'(\theta)\tan \theta - f'(\theta)g(\theta)\tan \theta + f'(\theta)g'(\theta) &= \\ \quad - \l f(\theta)g(\theta) + f(\theta)g'(\theta)\tan(\theta) + f'(\theta)g(\theta)\tan(\theta) + f'(\theta)g'(\theta)\tan^2 \theta \r \\ \tan^2\theta \l f(\theta)g(\theta) + f'(\theta)g'(\theta) \r + f'(\theta)g'(\theta) + f(\theta)g(\theta) &= 0 \\ (\tan^2\theta + 1) \l f(\theta)g(\theta) + f'(\theta)g'(\theta) \r &= 0 \\ f(\theta)g(\theta) + f'(\theta)g'(\theta) &= 0 \end{align*} $g(\theta) = a(1+\sin\theta), g'(\theta) = a\cos\theta$ Therefore $f'(\theta)a\cos \theta+f(\theta)a(1+\sin(\theta)) = 0$ \begin{align*} && \frac{f'(\theta)}{f(\theta)} &= -\sec(\theta) - \tan(\theta) \\ \Rightarrow && \ln(f(\theta)) &= -\ln |\tan(\theta) + \sec(\theta)| + \ln |\cos(\theta)| + C \\ \Rightarrow && f(\theta) &= A \frac{\cos \theta}{\tan \theta + \sec \theta} \\ &&&= A \frac{\cos^2 \theta}{\sin \theta + 1} \\ &&&= A \frac{1-\sin^2 \theta}{\sin \theta + 1} \\ &&&= A (1-\sin \theta) \end{align*} When $\theta = -\frac12 \pi, r = 4$, so $A = 2$.

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2017 Paper 3 Q6

D: 1700.0 B: 1500.0

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).

By making an appropriate substitution in the integral for $\T(x)$, show that \[\T(x) = \T_\infty - \T(x^{-1})\,.\]
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 $y>\dfrac1b\,$, \[ \T(y) =2\T_\infty - \T\left(\frac{y+b}{by-1}\right) - \T(b) \,. \]
Use the above results to show that $\T(\sqrt3)= \tfrac23 \T_\infty \,$ and $\T(\sqrt2 -1)= \frac14 \T_\infty\,$.

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2017 Paper 3 Q7

D: 1700.0 B: 1500.0

Show that the point $T$ with coordinates \[ \left( \frac{a(1-t^2)}{1+t^2} \; , \; \frac{2bt}{1+t^2}\right) \tag{$*$} \] (where $a$ and $b$ are non-zero) lies on the ellipse \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} =1 \,. \]

The line $L$ is the tangent to the ellipse at $T$. The point $(X,Y)$ lies on $L$, and $X^2\ne a^2$. Show that \[ (a+X)bt^2 -2aYt +b(a-X) =0 \,. \] Deduce that if $a^2Y^2>(a^2-X^2)b^2$, then there are two distinct lines through $(X,Y)$ that are tangents to the ellipse. Interpret this result geometrically. Show, by means of a sketch, that the result holds also if $X^2=a^2\,$.
The distinct points $P$ and $Q$ are given by $(*)$, with $t=p$ and $t=q$, respectively. The tangents to the ellipse at $P$ and $Q$ meet at the point with coordinates $(X,Y)$, where $X^2\ne a^2\,$. Show that \[ (a+X)pq = a-X\] and find an expression for $p+q$ in terms of $a$, $b$, $X$ and $Y$. Given that the tangents meet the $y$-axis at points $(0,y_1)$ and $(0,y_2)$, where $y_1+y_2 = 2b\,$, show that \[ \frac{X^2}{a^2} +\frac{Y}{b}= 1 \,. \]

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2017 Paper 3 Q8

D: 1700.0 B: 1500.0

Prove that, for any numbers $a_1$, $a_2$, $\ldots$\,, and $b_1$, $b_2$, $\ldots$\,, and for $n\ge1$, \[ \sum_{m=1}^n a_m(b_{m+1} -b_m) = a_{n+1}b_{n+1} -a_1b_1 -\sum_{m=1}^n b_{m+1}(a_{m+1} -a_m) \,. \]

By setting $b_m = \sin mx$, show that \[ \sum_{m=1}^n \cos (m+\tfrac12)x = \tfrac12 \big(\sin (n+1)x - \sin x \big) \cosec \tfrac12 x \,. \] {\bf Note:} $\sin A - \sin B = \displaystyle 2 \cos \big( \tfrac{{\displaystyle A+B\vphantom{_1}}} {\displaystyle 2\vphantom{^1}} \big)\, \sin\big( \tfrac{{\displaystyle A-B\vphantom{_1}}}{\displaystyle 2\vphantom{^1}} \big)\, $.
Show that \[ \sum_{m=1}^n m\sin mx = \big (p \sin(n+1)x +q \sin nx\big) \cosec^2 \tfrac12 x \,, \] where $p$ and $q$ are to be determined in terms of $n$. \vspace{3mm} {\bf Note:} $2\sin A \sin B = \cos (A-B) - \cos (A+B)\,$; \\[2mm] \phantom {\bf Note:} $2\cos A \sin B = \sin (A+B) - \sin (A-B)\,$.

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2017 Paper 3 Q9

D: 1700.0 B: 1500.9

Two particles $A$ and $B$ of masses $m$ and $2 m$, respectively, are connected by a light spring of natural length $a$ and modulus of elasticity $\lambda$. They are placed on a smooth horizontal table with $AB$ perpendicular to the edge of the table, and $A$ is held on the edge of the table. Initially the spring is at its natural length. Particle $A$ is released. At a time $t$ later, particle $A$ has dropped a distance $y$ and particle $ B$ has moved a distance $x$ from its initial position (where $x < a$). Show that $ y + 2x= \frac12 gt^2$. The value of $\lambda$ is such that particle $B$ reaches the edge of the table at a time $T$ given by $T= \sqrt{6a/g\,}\,$. By considering the total energy of the system (without solving any differential equations), show that the speed of particle $B$ at this time is $\sqrt{2ag/3\,}\,$.

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2017 Paper 3 Q10

D: 1700.0 B: 1484.0

A uniform rod $PQ$ of mass $m$ and length $3a$ is freely hinged at $P$. The rod is held horizontally and a particle of mass $m$ is placed on top of the rod at a distance~$\ell$ from $P$, where $\ell <2a$. The coefficient of friction between the rod and the particle is $\mu$. The rod is then released. Show that, while the particle does not slip along the rod, \[ (3a^2+\ell^2)\dot \theta^2 = g(3a+2\ell)\sin\theta \,, \] where $\theta$ is the angle through which the rod has turned, and the dot denotes the time derivative. Hence, or otherwise, find an expression for $\ddot \theta$ and show that the normal reaction of the rod on the particle is non-zero when~$\theta$ is acute. Show further that, when the particle is on the point of slipping, \[ \tan\theta = \frac{\mu a (2a-\ell)}{2(\ell^2 + a\ell +a^2)} \,. \] What happens at the moment the rod is released if, instead, $\ell>2a$?

Solution:

By energy considerations, the initial energy is $0$.

\begin{tabular}{l|c|c} & Inital & \@ $\theta$ \\ \hline Rotational KE of rod & $0$ & $\frac{1}{2}I\dot{\theta}^2 = \frac{1}{2} \frac{1}{3} m (3a)^2 \dot{\theta}^2 = \frac32 m a^2 \dot{\theta}^2$\\ KE of particle & $0$ & $\frac12 m \ell^2\dot{\theta}^2$\\ GPE of rod & $0$ & $-\frac{3}{2}mga \sin \theta$\\ GPE of particle & $0$ & $-mg \ell \sin \theta$ \\ \hline Total & $0$ & $\frac12m \l \l 3a^2 + \ell^2\r \dot{\theta}^2 - \l 3a + 2\ell \r g \sin \theta \r$ \end{tabular}

Therefore: \begin{align*} && \l 3a^2 + \ell^2\r \dot{\theta}^2 &= \l 3a + 2\ell \r g \sin \theta \\ \Rightarrow && \l 3a^2 + \ell^2\r 2\dot{\theta} \ddot{\theta} &= \l 3a + 2\ell \r g \cos\theta \dot{\theta} \tag{$\frac{\d}{\d t}$} \\ \Rightarrow && 2\l 3a^2 + \ell^2\r \ddot{\theta} &= \l 3a + 2\ell \r g \cos\theta \\ \Rightarrow && \ddot{\theta} &= \boxed{\frac{3a + 2\ell }{2(3a^2 + \ell^2)}g \cos\theta} \\ \end{align*} \begin{align*} \text{N}2(\perp PQ): && mg \cos \theta - R &= m \ell \ddot{\theta} \\ && R &= mg \cos \theta - m \ell \l \frac{3a + 2\ell }{2(3a^2 + \ell^2)}g \cos\theta \r \\ && &= mg\cos \theta \l 1 - \ell \frac{3a + 2\ell }{2(3a^2 + \ell^2)} \r \\ && &= mg \cos \theta \l \frac{6a^2 + 2\ell^2 - 3a\ell - 2\ell^2}{2(3a^2 + \ell^2)} \r \\ && &= mg \cos \theta \l \frac{3a(2a - \ell)}{2(3a^2 + \ell^2)} \r > 0 \tag{since $2a > \ell$} \end{align*} At limiting equilibrium, $F = \mu R$. \begin{align*} \text{N}2(\parallel PQ): && \mu R - mg \sin \theta &= m \ell \dot{\theta}^2 \\ \Rightarrow && \mu mg \cos \theta \l \frac{3a(2a - \ell)}{2(3a^2 + \ell^2)} \r - mg \sin \theta &= m \ell \frac{(3a+2\ell)}{(3a^2+\ell^2)} g \sin \theta \\ \Rightarrow && \mu \l 3a(2a - \ell) \r - \l 2(3a^2 + \ell^2) \r \tan \theta &= 2\ell (3a+2\ell) \tan \theta \\ \Rightarrow && \mu \l 3a(2a - \ell) \r &= \l 6a\ell + 6a^2 + 6\ell^2 \r \tan \theta \\ \Rightarrow && \tan \theta &= \boxed{\frac{\mu a(2a-\ell)}{2(a^2 + a\ell + \ell^2)}} \end{align*} If $\ell > 2a$, then the initial reaction force will be $0$, ie the particle will have no contact with the rod. In other words, the rod will rotate faster than the particle will free-fall and the particle immediately loses contact with the rod.

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