# Math Challenge - June 2019

• Challenge
• Featured
7. Five vessels contain ##100## balls each. Some vessels contain only balls of ##10\, g## mass, while the other vessels contain only balls of ##11\, g## mass. How can we determine with a single weighing which results in a mass, which vessels contain balls of ##10\, g## and which contain balls of ##11\, g##? (It is allowed to remove balls from the vessels.)
From 1 to 31, all natural numbers can be written in one unique way in the following form $$a\cdot 1 + b\cdot 2 + c\cdot 2^2 +d\cdot 2^3 + e \cdot 2^4 = a\cdot 1 +b\cdot 2 +c\cdot 4 +d\cdot 8+ e\cdot 16$$ wherein a, b, c, d, e are each 0 or 1.

In the first vessel, we leave one weight, in the second vessel 2, 4 in the third, 8 in the fourth and 16 in the fifth. We weigh all of these weights. Since each weight is either 10 or 11 g, if we subtract 310 g, then we get exactly the weight that the 11 g weights contributed to the total weight, with their weight above 10 g. The difference is expressed in the above form, and a, b, c, d, e show which vessels contributed to the total weight by weight over 10 g.

Let's look at an example: First vessel: 10 g; second 11 g; third 11 g; fourth 10 g; fifth 11g. Using the above method, the total weight ## 1 \cdot 10 + 2 \cdot 11 +4 \cdot 11 +8 \cdot 10 +16 \cdot 11 = 332##. Furthermore ##332-310 =22 =2+4+16.##

• timetraveller123 and scottdave
scottdave
Homework Helper
#7
Remove 1 ball from the first vessel, two balls from the second vessel, three from the third, four from the fourth and five from the fifth. Weigh all fifteen balls. If they weigh 150 grams, all vessels contain 10 g balls. If they weigh 152 grams, all vessels but #2 weigh 10 grams and vessel #2 weighs 11 grams. If they weigh 155 g, only #2 and #3 have 11 gram balls.
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Good. I had originally considered the prime quantities, but soon realized there would be duplicates, Then the Binary number idea came to me.

.
Good. I had originally considered the prime quantities, but soon realized there would be duplicates, Then the Binary number idea came to me.
I also thought of prime numbers for the first time, but here we are talking about sums. Then I thought of binary writing of the numbers. It was confusing that 32 and 64 were not needed, only 16.

• Luci
This way you cannot tell which vessel has 10g balls and which 11g balls, only how many of each.
I WILL JUST PUT SOME LABELS ON EACH MEASURED BALL :)

Calculate: $$\lim_{{x}\to{0}}\frac{{(\cos(x))^2-1}}{(\sinh(x))^2}$$ and $$\lim_{{x}\to{0}}\frac{e^{x}+e^{-x}-2-x^2}{(\cos(x)-1)^2}$$
Part (a)
\begin{align*}
&\lim_{{x}\to{0}}\frac{{(\cos(x))^2-1}}{(\sinh(x))^2}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
&\lim_{{x}\to{0}}\frac{({(\cos(x))^2-1}}{((\sinh(x))^2)'}=\lim_{{x}\to{0}}\frac{-2\sin(x)\cos(x)}{(2\sinh(x)\cosh(x))}=\lim_{{x}\to{0}}\frac{-\sin(2x)}{\sinh(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
&\lim_{{x}\to{0}}\frac{-2\cos(2x)}{2\cosh(2x)}=-1
\end{align*}

Part(b)
\begin{align*}
\lim_{{x}\to{0}}\frac{e^{x}+e^{-x}-2-x^2}{(\cos(x)-1)^2}=\lim_{{x}\to{0}}\frac{2\cosh(x)-2-x^2}{(\cos(x)-1)^2}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{(2\cosh(x)-2-x^2)'}{((\cos(x)-1)^2)'}=\lim_{{x}\to{0}}\frac{2\sinh(x)-2x}{((2\sin(x)-\sin(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{2\cosh(x)-2}{2\cos(x)-2\cos(2x)}=\lim_{{x}\to{0}}\frac{\cosh(x)-1}{\cos(x)-\cos(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{\sinh(x)}{-\sin(x)+2\sin(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{\cosh(x)}{-\cos(x)+4\cos(2x)}=\frac{1}{3}
\end{align*}

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• fresh_42
Mentor
Calculate: $$\lim_{{x}\to{0}}\frac{{(\cos(x))^2-1}}{(\sinh(x))^2}$$ and $$\lim_{{x}\to{0}}\frac{e^{x}+e^{-x}-2-x^2}{(\cos(x)-1)^2}$$
Part (a)
\begin{align*}
&\lim_{{x}\to{0}}\frac{{(\cos(x))^2-1}}{(\sinh(x))^2}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
&\lim_{{x}\to{0}}\frac{({(\cos(x))^2-1}}{((\sinh(x))^2)'}=\lim_{{x}\to{0}}\frac{-2\sin(x)\cos(x)}{(2\sinh(x)\cosh(x))}=\lim_{{x}\to{0}}\frac{-\sin(2x)}{\sinh(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
&\lim_{{x}\to{0}}\frac{-2\cos(2x)}{2\cosh(2x)}=-1
\end{align*}

Part(b)
\begin{align*}
\lim_{{x}\to{0}}\frac{e^{x}+e^{-x}-2-x^2}{(\cos(x)-1)^2}=\lim_{{x}\to{0}}\frac{2\cosh(x)-2-x^2}{(\cos(x)-1)^2}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{(2\cosh(x)-2-x^2)'}{((\cos(x)-1)^2)'}=\lim_{{x}\to{0}}\frac{(2\sinh(x)-2x)'}{((2\sin(x)-\sin(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{2\cosh(x)-2}{2\cos(x)-2\cos(2x)}=\lim_{{x}\to{0}}\frac{\cosh(x)-1}{\cos(x)+\cos(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{\sinh(x)}{-\sin(x)+2\sin(2x)}=\frac{0}{0} & ( \text{Since the limit is indetermined, we can use L'Hôpital's rule.})\\
\lim_{{x}\to{0}}\frac{\cosh(x)}{-\cos(x)+4\cos(2x)}=\frac{1}{3}
\end{align*}
Modulo a typo or two this is correct. The idea is that de L'Hôpital can be applied several times.

• cbarker1
Solve the following equation: $$(3x^2y^2+x^2)dx+(2x^3y+y^2)dy=0$$ (1).

Solution:
We need to determine if (1) is an exact equation or not.
Let $$M(x,y)=(3x^2y^2+x^2)$$ and $$N(x,y)=(2x^3y+y^2)$$. For testing for exactness, we need to take the partial derivatives with respect to y and x for M and N, respectively.
So, $$M_y(x,y)=6x^2y$$ and $$N_x(x,y)=6x^2y$$.
Since $$M_y(x,y)=N_x(x,y)$$, we know that (1) is an exact equation. So
\begin{align*}
F_x=(3x^2y^2+x^2)\\
&\int F_x \, dx =\int (3x^2y^2+x^2) \, dx\\
&F(x,y)= x^3y^2+\frac{x^3}{3}+g(y) \, & (\text{we need to determine what g(y), so taking the partial derivative of F with respect to y and set equal to N(x,y) and integrate both side by y})\\
&F_y(x,y)=2x^3y+g'(y)\\
2x^3y+y^2=2x^3y+g'(y)\\
g(y)=\frac{y^3}{3}
\end{align*}
The final solution of (1) is $$F(x,y)=x^3y^2+\frac{x^3}{3}+\frac{y^3}{3}=C$$.

• fresh_42
mfb
Mentor
Is "Ernest [is] the only one who didn't win." really needed to make the chess solution unique? Don't worry, I won't post solutions here, but I was curious and I think I didn't use that information.

Mentor
Is "Ernest [is] the only one who didn't win." really needed to make the chess solution unique? Don't worry, I won't post solutions here, but I was curious and I think I didn't use that information.
Probably not, as we always have that the entire sum of points equals ten, which should be an equivalent information, but I haven't checked. I just sounds better in the question to mention all players.

Denote equivalence classes of $R/I$ as $[r] := r + I, r\in R$. If $R$ is commutative, then of course, $R/I$ is as well. The symbol $\subset$ will denote proper containment.

We show $R/I$ is an integral domain iff $I$ is a prime ideal. Recall that an ideal $I$ of a commutative ring is prime if it's a proper subset satisfying
$$rs \in I \Rightarrow r\in I \lor s\in I.$$
Suppose $I$ is a prime ideal in $R$. Let $[rs] = $. Then $rs\in I$ and by assumption $r\in I$ i.e $[r] = $. Therefore, $R/I$ is an integral domain.

Conversely, let $R/I$ be an integral domain and let $rs\in I$. Then $[rs] = $ and by assumption at least one of the terms $r,s$ belongs to $I$. Hence, $I$ is a prime ideal.

We show $R/I$ is a field iff $I$ is maximal. We know the following.
Let $I\subseteq J\subseteq R$ be ideals of $R$. Then $J/I$ is an ideal of $R/I$. In fact, any ideal of $R/I$ is of the form $J/I$ where $I\subseteq J$ and $J$ is an ideal of $R$.

Let $R/I$ be a field, which is a simple ring. Suppose $I\subset J$ for some ideal $J$ of $R$. Then $J/I$ is a nonzero ideal and due to simpleness $J=R$, thus $I$ is maximal.

Conversely, suppose $I$ is maximal. Suffices to show $R/I$ doesn't contain proper nonzero ideals. Indeed, for if a ring $S$ is not a field, then for a non-invertible $s\neq 0$ we have the ideal $0\neq sS \subset S$, because $1\notin sS$.

Per initial remark, let $J/I$ be a nonzero ideal of $R/I$. Then $I\subset J$, thus the only nonzero ideal is $R/I$ itself.

Modulo a few pedantic comments, this should do the trick.

I like #15, makes use of identities and integration tricks.

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There are definitely more elegant ways to do this, but oh well.

Let’s establish an interval to work with. Disregarding the information given about Bernie and Ernest we know that the max points in the chess tournament is 4 and min is 0. Now we’ll use process of elimination.

We know that Alan lost a match since Bernie is the only one who didn’t lose. This puts Alan at a max of 3 points (1 loss, 3 wins) and Bernie at a min of 2.5 (1 win, 3 ties), but because of unambiguity (Alan’s score is greater than Bernie’s) Bernie’s final score must then be 2.5 which means that Alan must’ve had 3 points (3 wins, 1 loss). Alan didn’t tie a match, so he must’ve lost to Bernie who tied all matches except 1. Let’s move on to Ernest. We know that Ernest lost at least one match (since Bernie was the only who didn’t lose) and that he couldn’t have gotten 1.5 points or above because of unambiguity (David and Chuck both have a greater score than Ernest and Chuck has a greater score than David but less points than Bernie). We also know that Ernest didn’t win a single match. This means Ernest had to have lost at least two matches (three ties would put him at 1.5 which isn’t possible). Since Alan didn’t play any ties and Ernest didn’t win a match, Alan must have beaten Ernest. Following the logic Bernie tied with Ernest (putting Ernest at a minimum of 0.5 points). Since Ernest couldn’t have gotten 1.5 points he had to have lost at least one match against either Chuck or David. We know that Chuck lost against Alan and tied against Bernie, and that he won a match (Ernest was the only one who didn’t win a match) leaving him at 1.5 points so far. Likewise David won at least a match, tied against Bernie and lost against Alan leaving him at a minimum of 1.5 points as well. Since Bernie had 2.5 points and since David and Chuck couldn’t have ended up with the same points (ambiguity), Chuck must’ve had 2 points and David 1.5 points. This means that Chuck won against Ernest, tied with David and Bernie and lost to Alan. David lost to Alan and tied all his other matches. Ernest then lost two matches (against Chuck and Alan) and tied against Bernie and David leaving him at 1 point. As mentioned earlier, Bernie won against Alan and tied against the rest and Alan lost against Bernie and won against the rest.

Mentor
There are definitely more elegant ways to do this, but oh well.

Let’s establish an interval to work with. Disregarding the information given about Bernie and Ernest we know that the max points in the chess tournament is 4 and min is 0. Now we’ll use process of elimination.

We know that Alan lost a match since Bernie is the only one who didn’t lose. This puts Alan at a max of 3 points (1 loss, 3 wins) and Bernie at a min of 2.5 (1 win, 3 ties), but because of unambiguity (Alan’s score is greater than Bernie’s) Bernie’s final score must then be 2.5 which means that Alan must’ve had 3 points (3 wins, 1 loss). Alan didn’t tie a match, so he must’ve lost to Bernie ...
\begin{array}{|c|c|c|c|c|c|}
\hline & A & B & C & D & E \\
\hline A & - &0 &1 &1 &1 \\
\hline B & 1& - &1/2 &1/2 & 1/2\\
\hline C & & & - & & \\
\hline D & & & & - & \\
\hline E & & & & & - \\
\hline
\end{array}
... who tied all matches except 1. Let’s move on to Ernest. We know that Ernest lost at least one match (since Bernie was the only who didn’t lose) and that he couldn’t have gotten 1.5 points or above because of unambiguity (David and Chuck both have a greater score than Ernest and Chuck has a greater score than David but less points than Bernie). We also know that Ernest didn’t win a single match. This means Ernest had to have lost at least two matches (three ties would put him at 1.5 which isn’t possible). Since Alan didn’t play any ties and Ernest didn’t win a match, Alan must have beaten Ernest. Following the logic Bernie tied with Ernest (putting Ernest at a minimum of 0.5 points). Since Ernest couldn’t have gotten 1.5 points he had to have lost at least one match against either Chuck or David. We know that Chuck lost against Alan and tied against Bernie, and that he won a match (Ernest was the only one who didn’t win a match) leaving him at 1.5 points so far. Likewise David won at least a match, tied against Bernie and lost against Alan leaving him at a minimum of 1.5 points as well. Since Bernie had 2.5 points and since David and Chuck couldn’t have ended up with the same points (ambiguity), Chuck must’ve had 2 points and David 1.5 points.
...
\begin{array}{|c|c|c|c|c|c|}
\hline & A & B & C & D & E \\
\hline A & - &0 &1 &1 &1 \\
\hline B & 1& - &1/2 &1/2 & 1/2\\
\hline C &0 &1/2 & - & 1/2&1 \\
\hline D &0 &1/2 & 1/2& - & 1/2 \\
\hline E & 0& 1/2&0 & 1/2 & - \\
\hline
\end{array}
...
This means that Chuck won against Ernest, tied with David and Bernie and lost to Alan. David lost to Alan and tied all his other matches. Ernest then lost two matches (against Chuck and Alan) and tied against Bernie and David leaving him at 1 point. As mentioned earlier, Bernie won against Alan and tied against the rest and Alan lost against Bernie and won against the rest.
Was hard to figure out what was thoughts and what result in this wall-o-text but I think I protocolled it according to what you wrote.

It is almost correct (Alan and Bernie), but you have David and Ernest how didn't win while Ernest was the only one!

...

It is almost correct (Alan and Bernie), but you have David and Ernest how didn't win while Ernest was the only one!
My bad. David won against Ernest, lost to Chuck (which means he has 1.5 points). Chuck then won against David and tied against Ernest leaving him at 2 points. This means that Ernest tied with Chuck and Bernie and lost his matches against David and Alan.

\begin{array}{|c|c|c|c|c|c|}
\hline & A & B & C & D & E \\
\hline A & - &0 &1 &1 &1 \\
\hline B & 1& - &1/2 &1/2 & 1/2\\
\hline C &0 &1/2 & - & 1&1/2 \\
\hline D &0 &1/2 & 0& - & 1 \\
\hline E & 0& 1/2&1/2 & 0 & - \\
\hline
\end{array}

which is correct.

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• fresh_42
Let's call the two bands band 1 and band 2.
Light band 1 on fire in both ends of the band and at the same time light band 2 on fire in one end. When band 1 is burnt through 30 mins will have passed. Now light the other end of band 2 on fire. This way it'll be 15 minutes between band 1 being burnt through and band 2 being burnt all the way through resulting in 45 minutes.

• fresh_42
lekh2003
Gold Member
I solved HS Question 15, I guess? But its just simple integration and then I graphed it and found an arbitrary value for L for which it functions. Is there a more elegant solution?

Mentor
It is a simple integration practicing trig functions. But L is not arbitrary.

Let's setup three equations:

$$x \cdot y = 1 \cdot y + 1$$
$$-x \cdot z = 1 \cdot z - 1$$
$$z+y=1$$
x is the speed of the dog, y is the time the dog uses to run in front of the flock and z is the time the dog uses to run back to the end of the flock. The sheeps speed is set to 1 km/h, but that doesn't matter since it is the ratio between the sheeps and dogs speeds we are interested in. The two times z+y need to be 1 since the sheep have to move 1 km with a speed of 1 km/h. $$x \cdot y = 1 \cdot y +1$$ since the dog needs to move 1 km farther in order to get to the front of the sheep flock. $$-x \cdot z = 1 \cdot z - 1$$ since the dog has to move 1 km to get to the back of the flock. When these three equations are solved we see that the dog needs to move 2.41 the speed of the sheep resulting in the times 0.707 and 0.293 hours for y and z respectively. Using these times we get the following distances.
Distance covered when dog goes to front of flock:
$$2.41 km/h \cdot 0.707 h = 1.7 km$$
$$1 km/h \cdot 0.707 h = 0.707 km$$
Distance covered when dog moves to back of flock:

$$2.41 km/h \cdot 0.293 h = 0.707 km$$
$$1 km/h \cdot 0.293 h = 0.293 km$$.

We know add the distances for the dog and get that the dog ran $$1.7 km + 0.707 km = 2.41 km$$

Mentor
Let's setup three equations:

$$x \cdot y = 1 \cdot y + 1$$
$$-x \cdot z = 1 \cdot z - 1$$
$$z+y=1$$
x is the speed of the dog, y is the time the dog uses to run in front of the flock and z is the time the dog uses to run back to the end of the flock. The sheeps speed is set to 1 km/h, but that doesn't matter since it is the ratio between the sheeps and dogs speeds we are interested in. The two times z+y need to be 1 since the sheep have to move 1 km with a speed of 1 km/h. $$x \cdot y = 1 \cdot y +1$$ since the dog needs to move 1 km farther in order to get to the front of the sheep flock. $$-x \cdot z = 1 \cdot z - 1$$ since the dog has to move 1 km to get to the back of the flock. When these three equations are solved we see that the dog needs to move 2.41 the speed of the sheep resulting in the times 0.707 and 0.293 hours for y and z respectively. Using these times we get the following distances.
Distance covered when dog goes to front of flock:
$$2.41 km/h \cdot 0.707 h = 1.7 km$$
$$1 km/h \cdot 0.707 h = 0.707 km$$
Distance covered when dog moves to back of flock:

$$2.41 km/h \cdot 0.293 h = 0.707 km$$
$$1 km/h \cdot 0.293 h = 0.293 km$$.

We know add the distances for the dog and get that the dog ran $$1.7 km + 0.707 km = 2.41 km$$
Correct, well done. Just allow me some remarks.

In physics speed is called velocity, as it is in general not only the speed, but the direction, too. Hence it is abbreviated by ##v##. If we want to indicate whose velocity it is, we use indices, in this case ##v_B## for Boy and ##v_F## for the flock. Time is accordingly abbreviated by ##t##. Such conventions make reading a lot easier, since one hasn't to learn beforehand what ##x,y,z## are.
Another hint is: use exact numbers as long as possible, in our case ## 1+\sqrt{2}\, , \,\sqrt{2}\, , \,\frac{1}{\sqrt{2}}##. Only the final answer should be an approximation. The reason is, that this reduces approximation errors during longer calculations than this one.

• bodycare
Problem 6:
The population equations can be factored as
$$\frac{dx}{dt} = (a - b y) x ,\ \frac{dy}{dt} = (-c + d x) y$$
One can find the interrelationship between x and y by dividing the second equation by the first one. The time drops out and one finds:
$$\frac{dy}{dx} = \frac{(-c + dx)y}{(a - by)x} = \frac{-c/x + d}{a/y - b}$$
This equation is easy to integrate. Multiply both sides by both sides' denominators:
$$\left( \frac{a}{y} - b \right) dy = \left( - \frac{c}{x} + d \right) dx$$
Integrate both sides:
$$a \log y - b y = - c \log x + d x + C$$
or
$$c \log x - d x + a \log y - b y = C$$
where C is the integration constant. Plugging in the numbers gives us
$$7 \log x - x + 10 \log y - 2 y = C$$

Mentor
Problem 6:
The population equations can be factored as
$$\frac{dx}{dt} = (a - b y) x ,\ \frac{dy}{dt} = (-c + d x) y$$
One can find the interrelationship between x and y by dividing the second equation by the first one. The time drops out and one finds:
$$\frac{dy}{dx} = \frac{(-c + dx)y}{(a - by)x} = \frac{-c/x + d}{a/y - b}$$
This equation is easy to integrate. Multiply both sides by both sides' denominators:
$$\left( \frac{a}{y} - b \right) dy = \left( - \frac{c}{x} + d \right) dx$$
Integrate both sides:
$$a \log y - b y = - c \log x + d x + C$$
or
$$c \log x - d x + a \log y - b y = C$$
where C is the integration constant. Plugging in the numbers gives us
$$7 \log x - x + 10 \log y - 2 y = C$$
Correct, well done.
$$y^{10} e^{-2y} = k\,x^{-7}e^{x} \text{ with }k=e^C$$ In Problem HS 13, is multiplication supposed to be commutative and/or associative?

I will take on Problem HS 15.
For this problem, we first do the indefinite integrals, then specialize to the bounds. The bounds are the same, so we can use one expression, by taking (LHS) - (RHS):
$$J(x) = \int {dx} \left( \frac{1}{\sin^2 x} - \frac{1}{1 - \cos x} - 6 \frac{\cot x}{\sin x} \right)$$
Doing the integrals, I find
$$J(x) = - \cot x + \frac{1 + \cos x}{\sin x} + 6 \csc x = 7 \csc x$$
I thus have to find L such that ## J(L) - J(\pi/6) = 0 ## or ## J(L) = J(\pi/6) ##. That involves solving ## 7 \csc L = 7 \csc (\pi/6) ## or ## \sin L = \sin (\pi/6) ##. The sine function rises from 0 at 0 to 1 at ##\pi/2## and then falls to 0 at ##\pi##. It also satisfies ## \sin x = \sin (\pi - x) ##. Thus, the smallest L greater than ##\pi/6## is ##\pi - \pi/6 = 5\pi/6##.

The solution: ## L = 5\pi/6##

Mentor
In Problem HS 13, is multiplication supposed to be commutative and/or associative?
Both, as I mentioned the integers for the sake of simplicity. The question which property of the integers is used is the interesting point, more than the proof.
I will take on Problem HS 15.
For this problem, we first do the indefinite integrals, then specialize to the bounds. The bounds are the same, so we can use one expression, by taking (LHS) - (RHS):
$$J(x) = \int {dx} \left( \frac{1}{\sin^2 x} - \frac{1}{1 - \cos x} - 6 \frac{\cot x}{\sin x} \right)$$
Doing the integrals, I find
$$J(x) = - \cot x + \frac{1 + \cos x}{\sin x} + 6 \csc x = 7 \csc x$$
I thus have to find L such that ## J(L) - J(\pi/6) = 0 ## or ## J(L) = J(\pi/6) ##. That involves solving ## 7 \csc L = 7 \csc (\pi/6) ## or ## \sin L = \sin (\pi/6) ##. The sine function rises from 0 at 0 to 1 at ##\pi/2## and then falls to 0 at ##\pi##. It also satisfies ## \sin x = \sin (\pi - x) ##. Thus, the smallest L greater than ##\pi/6## is ##\pi - \pi/6 = 5\pi/6##.

The solution: ## L = 5\pi/6##
Correct.