# JEE-Advanced Maths Contest '16 (Continued)

Hello, guys!

This note is the continuation of the JEE-Advanced Maths Contest '16 because of a large number of comments on the previous note. This continued contest note is starting from the 51st problem of the contest. Each Brilliant user is invited to participate in the contest under the following rules:

1. I will start by posting the first problem. If there is a user solves it, then (s)he must post a new one.
2. You may only post a solution of the problem below the thread of problem and post your proposed problem in a new thread. Put them separately.
3. Please make a substantial comment.
4. Make sure you know how to solve your problem before posting it. In case there is no one can answer it within 24 hours, then you must post the solution, and you have a right to post another problem.
5. If the one who solves the last problem does not post his/her problem after solving it within a day, then the one who has a right to post a problem is the last solver before him/her.
6. The scope of questions is only what is covered under JEE-Advanced syllabus. If you are unaware of the syllabus cover, You can check out the information brochure.
7. You can use tricks/short methods/specific cases to post the solution of a problem only if they are logically valid.
8. In the case of any argument or disagreement among the solvers, the decision/judgement is passed to me.
9. DO NOT ask the answer to the problem. Just post your detailed solution solution along with the answer. It will be highly helpful to gain confidence in your problem solving and rock in JEE.
10. Proof problems are not allowed.
11. You can post a problem only from Maths section.

• Please write the detailed solutions to the problems.

Format your post is as follows:

SOLUTION OF PROBLEM xxx (number of problem) :

PROBLEM xxx (number of problem) :

Note by Sandeep Bhardwaj
4 years, 5 months ago

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Problem 51

Let $f$ be a continuous real function defined on positive reals ($\mathbb{R}^+$) satisfying the following conditions: \begin{aligned} f(xy)&=f(x)+f(y) \quad \forall x, y \in \mathbb{R}^+ \\f(e)&=1 \end{aligned}

Find the total number of such functions $f$.

Note: Here $e$ is the base of natural logarithm.

Deeparaj Bhat has provided the correct solution. Feel free to add new approaches. Thanks!

- 4 years, 5 months ago

Solution to problem 51

Define $g(x)=f(e^x) \quad \forall x \in \mathbb{R}$

Then,

$g(x+y)=g(x)+g(y)$ By induction, the following are true: $g\left(\sum_{i=1}^{n} x_i \right)=\sum_{i=1}^{n} g(x_i) \quad x_i \in \mathbb{R}\\ g(x)=cx \quad \forall x \in \mathbb{Q} \text{ where c is a constant }$ The proofs are similar to the ones done by Aakash.

Now, define a sequence $a_n$ which satisfies the following properties: $a_n \in \mathbb{Q} \quad \forall n \in \mathbb{N}\\ a_n\neq \alpha \quad \forall n \in \mathbb{N}\\ \lim_{n \to \infty}a_n = \alpha \quad where \, \alpha \,is \, an \, arbitrary \,real$

So, $g(\alpha)= \lim_{n \to \infty}g(a_n)=\left(\lim_{n \to \infty} ca_n \right)=c\alpha$

So, $g(x)=cx \quad \forall x \in \mathbb{R}$

So, $f(x)=c\ln(x) \quad \forall x \in \mathbb{R}^+$

As $f(e)=1$, only one such function exists.

Note: You might want to check out Cauchy's functional equation.

- 4 years, 5 months ago

$SOLUTION\quad OF\quad PROBLEM\quad 51\\ First\quad we\quad notice\quad that\quad f(1)=0\\ since\quad f(e)=f(e\cdot 1)=f(e)+f(1)\\ Now\quad it\quad is\quad given\quad that\quad f\left( xy \right) =f\left( x \right) +f\left( y \right) \quad \forall x,y\in { \Re }^{ + }\\ \frac { \partial f\left( xy \right) }{ \partial y } =\frac { \partial f\left( x \right) }{ \partial y } +\frac { \partial f\left( y \right) }{ \partial y } \\ xf^{ \prime }\left( xy \right) =f^{ \prime }\left( y \right) \\ inserting\quad y=1\quad we\quad get\quad the\quad diffrential\quad equation,\\ xf^{ \prime }\left( x \right) =f^{ \prime }\left( 1 \right) \\ \frac { dy }{ dx } =\frac { f^{ \prime }\left( 1 \right) }{ x } \\ \int { dy } =f^{ \prime }\left( 1 \right) \int { \frac { dx }{ x } } \\ y=f\left( x \right) =f^{ \prime }\left( 1 \right) \ln { \left| x \right| } +C\quad \{ C\quad is\quad the\quad constant\quad of\quad integration\} \\ as\quad x>0\\ we\quad can\quad write\quad \\ f\left( x \right) =f^{ \prime }\left( 1 \right) \ln { x } +C\\ now\quad using\quad f\left( 1 \right) =0\quad and\quad f\left( e \right) =1\\ we\quad get\quad f^{ \prime }\left( 1 \right) =1\quad and\quad C=0\\ hence\quad ,\quad f\left( x \right) =\ln { x } \\ \therefore \quad Only\quad one\quad such\quad function\quad is\quad possible\quad which\quad satisfies\quad the\quad given\quad conditions$

- 4 years, 5 months ago

Its not given that $f$ is differentiable.

- 4 years, 5 months ago

It isn't given that $f$ is differentiable. So, the solution is wrong, though the anvswer is right.

- 4 years, 5 months ago

Then who has the right to post the next question? P.S:- I didn't read the the question carfeully

- 4 years, 5 months ago

I don't really know since even Aakash's proof is incomplete as it stands.

Could you please decide who'll post the next problem? @Sandeep Bhardwaj

- 4 years, 5 months ago

@Rishi Sharma You can post the next problem.

@Deeparaj Bhat Can you please post the detailed correct solution to your problem?

- 4 years, 5 months ago

IOnly one . $ln(x)=f(x)$.

Proof: Put $x=y=e$.

Then $f(e^{2})=2$.

Similarily $(f(e^n)= n$. $\forall x \in N$.

Now putting $x=y=\sqrt{e}$

We have $f(\sqrt{e})= 1/2$.

$f(e^{2/3}= 2f(e^{1/3})$.

Also $f(e^{4/3}=2f(e^{2/3})= 4f(e^{1/3})= 1+f(e^{1/3})$.

Thus $f(e^{1/3})=1/3$ . . . So on..

$f(e^{1/n})=1/n$.

Hence prooved.

- 4 years, 5 months ago

That seems right for rational values of $x$. Only remains to say that since values at rational arguments are the same, so are limits of them, and hence so are values at irrational arguments. (Probably you took that to be too obvious to say.)

- 4 years, 5 months ago

This proves that $f(x)=ln(x)$ for all rational $x$. Use still need to prove that it is unique even if the whole set of real numbers is taken.

However, if the situation as described by Mark C is right (the statement in the paranthesis in his comment), please add that to your proof.

- 4 years, 5 months ago

1

- 2 years, 3 months ago

Problems 1 to 50 of this contest are covered under JEE-Advanced Maths Contest '16.

Thanks to everyone for the amazing problems and solutions. Keep it up!

- 4 years, 5 months ago

Problem 59:

We have bundle of $n$ (for $n$ even) ropes, each having a top end and a bottom end. We randomly tie ends together, top-to-top and bottom-to-bottom, until every end is tied to one other end.

What is the limit, for large $n$, of the probability that we have created just one loop composed of all of the ropes?

- 4 years, 4 months ago

The top ends will in fact be tied somehow or other; so we can focus on whether, given how the tops will in fact be tied, we will create a loop with the knots on the bottom ends. For convenience, then, imagine that the top ends of our $n$ ropes are already tied and we are starting to knot the bottom ends.

The first knot we tie can defeat us if we tie the bottom ends of ropes that are knotted up top. For any rope I grab, there are $n-2$ of the $n-1$ other ropes that will avoid failure (failure is creating a mini-loop). So the probability of avoiding failure with my first knot is $\frac{n-2}{n-1}$.

Having tied the first knot, if I have avoided failure I am in essentially the situation I would be if I started with $n-2$ ropes: there are $n-2$ untied bottom ends of ropes, each of which is paired with another by being tied to it either directly up top or through a chain of knots. So the chance of avoiding failure this time is $\frac{n-4}{n-3}$.

Thus the chance of my succeeding is the product of the chances of my avoiding failure at every turn:

$P_n = \frac{n-2}{n-1}\cdot\frac{n-4}{n-3}\cdots\frac{4}{5}\cdot\frac{2}{3}$

$\lim_{n\to\infty}(P_n) = \frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdots = \prod_{i=1}^\infty (1-\frac{1}{2i+1})$ $= \exp(\ln(\prod_{i=1}^{\infty}(1-\frac{1}{2i+1}))) = \exp (\sum_{i=1}^\infty \ln(1- \frac{1}{2i+1}))$ ($\exp(x)$ means $e^x$. Since for $x>0$, $\ln(x) \leq x - 1$:) $\leq \exp (-\sum_{i=1}^\infty(\frac{1}{2i+1})) \leq \exp (-\sum_{i=1}^\infty(\frac{1}{i})) = e^{-\infty} = \boxed{0}.$

- 4 years, 4 months ago

$PROBLEM\quad NO\quad 52\\ Consider\quad two\quad complex\quad no's\quad { z }_{ 1 }\& { z }_{ 2 }\quad satisfying\\ S:\left| z-5i \right| =4\\ such\quad that\quad arg({ z }_{ 1 })=arg({ z }_{ 2 })=\frac { 3\pi }{ 10 } \\ and\quad \left| { z }_{ 1 } \right| <\left| { z }_{ 2 } \right| \\ If\quad { z }_{ 3 }\quad is\quad the\quad point\quad of\quad intersection\quad of\quad tangent\quad \\ drawn\quad at\quad { z }_{ 2 }\quad on\quad S\quad with\quad real\quad axis.\quad Find\\ arg(\frac { 5i-{ z }_{ 3 } }{ { z }_{ 2 }-{ z }_{ 3 } } )\\ BONUS:\quad Can\quad you\quad do\quad it\quad without\quad co-ordinate\quad bashing.\\ \\ \quad \\ \quad$

Deeparaj Bhat has solved this problem along with the solution. Feel free to add new approaches. Thanks!

- 4 years, 5 months ago

Solution to problem 52:

@Rishi Sharma No coordinate bash!

- 4 years, 5 months ago

Don't enclose the text into latex brackets. You can ping me if you need any type of help regarding latex. Thanks!

- 4 years, 5 months ago

Problem 53:

Let $\Gamma_1$ be the circumcircle of $\Delta ABC$ and $\Gamma_2$ be the circumcircle of $\Delta DEF$ where $D, E, F$ are the foot of the altitudes of $\Delta ABC$.

Find the value of the following expression when the length of common chord of $\Gamma_1$ and $\Gamma_2$ is maximum for a given circumradius:

$\sum_{cyc} \cos 2A$

Deeparaj Bhat has provided the solution himself. Feel free to add new approaches. Thanks!

- 4 years, 5 months ago

Since nobody has answered this question in 24 hours, I'll post my solution.

Solution to problem 53:

First of all, note that $\Gamma_2$ (call it's centre $O_9$) is the nine point circle of $\Delta ABC$.

Call the centre of $\Gamma_1$ as $O$ and it's radius $R$. It now follows that the radius of $\Gamma_2$ is $\dfrac{R}{2}$. As the radius of $\Gamma_2$ is less than that of $\Gamma_1$, the length of the common chord is maximum if and only if it (the chord) is the diameter of $\Gamma_2$.

So, when maximum occurs, using Pythagoras' theorem, $OO_9=\frac{\sqrt3}{2}R$

Now, we use the following equations which hold in general:

$2 \cdot OO_9= OH\\ OH^2=9R^2-(a^2+b^2+c^2)\\ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

Where $H$ denotes the orthocentre of $\Delta ABC$.

Using the above three equations, we get:

\begin{aligned} \sum_{cyc}\sin^2A&=\frac{3}{2} \\ \implies \sum_{cyc}\cos^2A&=\frac{3}{2}\\ \implies \sum_{cyc}\cos 2A&=\boxed{0}\end{aligned}

- 4 years, 5 months ago

Problem 57 :

Let $f:R\rightarrow R$ be a function satisfying

$\left| f\left( x+y \right) -f\left( x-y \right) -y \right| \le { y }^{ 2 } \quad \forall x,y \in R$.

If $f\left( 0 \right) =\frac { 5 }{ 2 }$ find $f\left( 3 \right)$

Mark C has provided a complete solution to this problem. Thanks!

- 4 years, 4 months ago

Solution of Problem 57

$|f((x+\frac{\Delta x}{2})+\frac{\Delta x}{2}) - f((x+\frac{\Delta x}{2})-\frac{\Delta x}{2}) - \frac{\Delta x}{2}| \leq \frac{\Delta x^2}{4}$

$|f(x+\Delta x)-f(x) - \frac{\Delta x}{2}| = |\Delta f(x) - \frac{\Delta x}{2}| \leq \frac{\Delta x^2}{4}$

As $\Delta x$ approaches zero, $\Delta x^2$ is negligible and so $\Delta f(x)$ approaches $\frac{\Delta x}{2}$. Thus the slope is everywhere $\frac{1}{2}$ and the function is:

$f(x) = \frac{1}{2}x + \frac{5}{2}$

$f(3) = \boxed{\frac{8}{2}}$

- 4 years, 4 months ago

That is correct!!

- 4 years, 4 months ago

Problem 54:

Let $\alpha$ and $\beta$ be two distinct, randomly chosen roots of the following equation: $z^{42}=1$

If the probabilty that $\sqrt{2+\sqrt3}\leq|\alpha+\beta|$ is equal to $\dfrac{a}{b}$ where $a$ and $b$ are coprimes, find $a+b$.

Vighnesh Shenoy has provided the correct solution. Feel free to add new approaches. Thanks!

- 4 years, 5 months ago

$\alpha , \beta$ satisfy $z^{42} = 1$.
Thus, $\alpha, \beta$ are the forty second roots of unity, and lie on a unit circle centered at origin.
Let,
$\alpha =e^{i\frac{2\pi k}{42}} , \beta = e^{i \frac{2\pi m}{42}}$
$| \alpha + \beta | = \left| \left(\cos\left(\dfrac{k\pi}{21}\right) + \cos\left(\dfrac{m\pi}{21}\right) \right)+ i\left(\sin \left(\dfrac{k\pi}{21}\right) + \sin \left(\dfrac{m\pi}{21}\right) \right) \right|$, $0 \le k,m \le 41$
$\therefore | \alpha + \beta | = \sqrt{2 + 2\cos\left(\dfrac{(k-m)\pi}{21}\right) }$
$\sqrt{2 + 2\cos\left(\dfrac{(k-m)\pi}{21}\right) } \ge \sqrt{2+\sqrt{3}}$
$\cos\left(\dfrac{(k-m)\pi}{21}\right) \ge \dfrac{\sqrt{3}}{2}$
$\cos\left(\dfrac{(k-m)\pi}{21}\right) \ge \cos\left(\dfrac{\pi}{6}\right)$
$2n\pi - \dfrac{\pi}{6} \le \dfrac{(k-m)\pi}{21} \le 2n\pi + \dfrac{\pi}{6}$
$42n - \dfrac{7}{2} \le k-m \le 42n + \dfrac{7}{2}$
Also note that,
$-41 \le k - m \le 41$
This restricts us to $n = -1,0,1$
When $n = -1$ ,
$-45.5 \le k-m \le -38.5$
$-41 \le k-m \le -39$ which gives us 6 cases.
When $n = 0$,
$-3.5 \le k-m \le 3.5$
$-3 \le k-m \le 3$ which gives us 240 cases.
When $n = 1$,
$38.5 \le k-m \le 45.5$ which again gives us 6 cases.
Total such cases $= 252$
But since there's no distinction between $\alpha$ , $\beta$ we are overcounting by a factor of 2.
Total favorable cases $= 126$

Total ways of selecting two distinct roots = $\dbinom{42}{2} = 861$
Probability = $\dfrac{126}{861} = \dfrac{6}{41}$
$a + b = 47$

- 4 years, 5 months ago

What I mean by the distinction statement is if we choose the two roots to be say,
$e^{i\frac{\pi}{21}} , e^{i\frac{2\pi}{21}}$.
I can call the first one $\alpha$ or the second one , doesn't make a difference.

- 4 years, 5 months ago

Yeah. Exactly!

You could've reduced the counting part by putting $r=k-m$ and then saying that $0 and $r$ is a integer. As $k$ and $m$ are random, so is $r$, and hence all the values of $r$ are equally likely.

You can post the next problem.

- 4 years, 5 months ago

Problem 60:

$\int_{x=0}^\pi \sin(3x)\cos(5x) d x$

This problem has been solved by Vignesh S

- 4 years, 4 months ago

Answer is 0. The above integral can be written as $0.5\int_{0}^{\pi}\sin{8x}-\sin{2x}dx=0.5\int_{0}^{\pi}\sin{(8\pi-8x)}-\sin{(2\pi-2x)}dx=I$ $\implies2I=0.5\int_{0}^{\pi}\sin{8x}+\sin{(8\pi-8x)}dx-0.5\int_{0}^{\pi}\sin{2x}+\sin{(2\pi-2x)}dx$ $2I=0 ;I=0$

- 4 years, 4 months ago

Very good! The answer can also be reached by observing that $\sin(3(\pi-x))=\sin(\pi-3x)=\sin(3x)$ and $\cos(5(\pi-x))=\cos(\pi-5x)=-\cos(5x)$, so that the integral from $0$ to $\frac{\pi}{2}$ is the negative of the integral from $\frac{\pi}{2}$ to $\pi$.

Your turn to post a problem!

- 4 years, 4 months ago

Problem 61

If $f(x)=x^{3}+ax^{2}+bx+c=0$ has 3 integral roots and $(x^{2}+2x+2)^{3}+a(x^{2}+2x+2)^{2}+b(x^{2}+2x+2)+c=0$ has no real roots, then find $a,b,c$.

Note:$|c|<3$

Mark C has provided a solution to this problem.

- 4 years, 4 months ago

$f(x) = x^3+ax^2+bx+c = (x-n_1)(x-n_2)(x-n_3)=0$

$a=-(n_1+n_2+n_3), b=n_1n_2+n_1n_3+n_2n_3, c=-n_1n_2n_3$

$(x^2 +2x+2)^3 +(x^2+2x+2)^2*a +(x^2+2x+2)*b +c = 0$

has no real roots, so it is always positive (it can't be always negative because the positive addend $(x^2+2x+2)^3$ will obviously exceed the sum of the others for large positive and negative values of $x$.)

$x^2 +2x +2$ has a minimum of 1 for $x=-1$. So:

$a+b+1+c > 0$ $a+b+c>-1$

Since $|n_1n_2n_3| < 3$, $\{n_1,n_2,n_3\} \subset \{-2,-1,0,1,2\}$. Of the possibilities, only one works: $\{n_1,n_2,n_3\}=\{-2,-1,0\}$ yields $(a,b,c)=\boxed{(3,2,0)}$, which sum to 5.

- 4 years, 4 months ago

I did the the same solution.

- 4 years, 4 months ago

Problem 62:

$V_1$ is the unit sphere with equation $x^2+y^2+z^2=1$, and $V_2$ is the volume formed by subtracting, from the volume enclosed by the cylinder $x^2+y^2=1$, $-1\leq z\leq 1$, the volume enclosed by the cones $x^2 +y^2 =z^2$. At what heights $z$ is the maximum absolute difference between the horizontal cross-sections of $V_1$ and $V_2$, and what is that difference?

This problem has been solved by Sarvesh Nalawade.

- 4 years, 4 months ago

Zero, the cross sections of V1 and V2 are always same .

Consider a plane $z=k$ where $|k| \leq 1$ .

It will intersect the sphere , i.e $V_{1}$ in a circle with radius $\sqrt{1-k^{2}}$

Thus its area will be given by :

$A_{1}= \pi[1-k^{2}]$

Also , it will intersect the cone in another circle with radius $k$

Therefore , Area of horizontal cross-section of $V_{2}$ will be given by :

$A_{2} = \pi [ 1^{2} - k^{2} ]$

Therefore $A_{1}=A_{2} \quad \forall k \in (-1,1)$.

Hence their difference will always be zero

- 4 years, 4 months ago

Correct!

- 4 years, 4 months ago

Problem 63:

The number of 7-digit numbers formed from the digits 1 to 7 taken all at a time such that sum of the digits at the first three places is smaller than the sum of the digits at the last three places is ?

Vighnesh Shenoy has provided a complete solution for this problem

- 4 years, 4 months ago

Are repetitions allowed?

- 4 years, 4 months ago

No, each number has to be taken exactly once.

- 4 years, 4 months ago

M be number of arrangements where sum of first 3 is greater than last 3, and K be number of arrangements where sum of last 3 is equal to sum of first 3.
$\therefore 2M + K = 7!$
For evaluating K,
If the middle digit is P,
Sum of first 3 digits $(S)$ = Sum of last 3 digits = $\dfrac{28-P}{2}$
$\therefore$ P is even.
Case 1:

$P = 2$
$S = 13$
Only possible grouping for this,
$(7,1,5) ,(6,4,3)$
Number of such arrangements $= (3!)^{2} \times 2! = 72$

Case 2:
$P = 4$
$S = 12$
Only possible grouping = $(7,2,3) , (6,5,1)$
Number of such arrangements $=(3!)^{2} \times 2! = 72$

Case 3:

$P = 6$.
$S = 11$
Only possible grouping $(7,3,1) , (5,4,2)$
Number of such arrangements $= (3!)^{2} \times 2! = 72$

$\therefore K = 3 \cdot 72 = 216$
$2M = 5040 - K = 5040 - 216 = 4824$
$M = 2412$

- 4 years, 4 months ago

For the cases, how'd you end up with the grouping? Trial and error?

- 4 years, 4 months ago

Yes.
For case 1 : I wanted sum of one group to be 13.
$13 = 7 + 6 = 7 + 1 + 5 = 7 + 2 + 4 = 7 + 3 + 3$
Out of this only one is possible due to given constraints.
For case 2 :
$12 = 7 + 5 = 7 + 1 + 4 = 7 + 2 + 3$
For case 3 :
$11 = 7 + 4 = 7 + 1 + 3 = 7 + 2 + 2$
I used the largest number to make the grouping easier.

- 4 years, 4 months ago

Ok. I was put off by counting the cases. Didn't realize it'd be so simple :P

- 4 years, 4 months ago

Correct!!

- 4 years, 4 months ago

1872 , is it correct ?

- 4 years, 4 months ago

No, that's not correct

- 4 years, 4 months ago

Problem 66:

What is the expected number of die rolls to get three sixes in a row?

- 4 years, 4 months ago

Solution to Problem 66:

We are looking for $E$. If your first non-six comes up right away ($\frac{5}{6}$ chance), then you are starting fresh on the second roll, so the expectation is now $E+1$. If your first non-six comes at the second roll ($\frac{1}{6}\cdot\frac{5}{6}$ chance), then you are starting fresh on the third roll, so the expectation is now $E+2$. If your first non-six comes on the third roll ($(\frac{1}{6})^2\cdot\frac{5}{6}$ chance) then the expectation is $E+3$. Otherwise ($(\frac{1}{6})^3$ chance) you get three sixes in three rolls. So:

$E = \frac{5}{6}*(E+1)+\frac{1}{6}\cdot\frac{5}{6}(E+2) + (\frac{1}{6})^2\cdot\frac{5}{6}(E+3) + (\frac{1}{6})^3\cdot3 =$ $= \frac{258}{216} + \frac{215}{216}E$ $\frac{1}{216}E = \frac{258}{216}$ $E = \boxed{258}$

Good luck to all at the JEE!

- 4 years, 4 months ago

There seems to be no activity on this thread anymore, so I will not be visiting it again. Anyone who wants to should feel free to post problems. Thanks for the fun problems, everyone!

- 4 years, 4 months ago

The result for inactivity being many of us have our exams close. Can you post the solution to this problem though, I am intrigued.

- 4 years, 4 months ago

@Mark C By close,he means only one week. It is the same exam for which this note was made.

- 4 years, 4 months ago

Problem 56 : For whole numbers $m$ and $n$, determine $S = \displaystyle \sum_{i=0}^{n-1} { m+i \choose i }$.

- 4 years, 4 months ago

$\sum _{ i }^{ n-1 }{ \left( \begin{matrix} m+i \\ i \end{matrix} \right) } =\sum _{ i }^{ n-1 }{ \left( \begin{matrix} m+i \\ m \end{matrix} \right) } \\ We\quad have\quad to\quad find\quad the\quad coefficient\quad of\quad { x }^{ m }\quad in\quad the\quad expansion\quad of\\ { (1+x) }^{ m }+{ (1+x) }^{ m+1 }+{ (1+x) }^{ m+2 }...........................{ (1+x) }^{ m+n-1 }\\ That's\quad GP\quad sum\quad !\quad =\frac { (1+x)({ (1+x) }^{ n }-1) }{ x } \\ coefficient\quad of\quad { x }^{ m }\quad in\quad the\quad expression\quad is\quad \left( \begin{matrix} m+n \\ m+1 \end{matrix} \right) .\quad Which\quad is\quad the\quad answer.$

- 4 years, 4 months ago

is this contest still on??

- 11 months, 2 weeks ago

Problem 55 :
ABCDEFGHIJKL is a regular dodecagon with vertices in order. Find the value of $\dfrac{AB}{AF} + \dfrac{AF}{AB}$.

Saarthak Marathe has provided the correct solution. Feel free to add new approaches. Thanks!

- 4 years, 5 months ago

Answer is $4$.

Draw a dodecagon. Let the center be $O$.

From $O$ 12 triangles can be drawn joining each vertices making ${30}^{\circ}$ with $O$.

Let the radius be $x$. Then $AB=2xcos({75}^{\circ})=\dfrac{(\sqrt{3}-1)}{\sqrt{2}}.x$.

We get $\angle{AOF}=150^{\circ},AO=OF=x$ and by cosine rule , $AF=\dfrac{(\sqrt{3}+1)}{\sqrt{2}}.x$

Solving we get $\dfrac{AB}{AF} + \dfrac{AF}{AB}=\boxed{4}$

- 4 years, 5 months ago

Problem 56:

Find the locus of point $P$ such that tangents drawn from it to the given ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ meet the coordinate axes at concyclic points.

Sarvesh Nalawade has provided a complete solution to this problem. Thanks!

- 4 years, 5 months ago

Is this the answer? $a^2-b^2=x^2-y^2$

- 4 years, 5 months ago

I too am getting the same result

Consider a point $\left( h,k \right)$outside the given ellipse .

The combined equation of pair of tangents to the ellipse : $S =\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } -1 = 0$ is given by :

$S{ S }_{ 1 }={ T }^{ 2 }$

i.e $\left[ \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } -1 \right] \left[ \frac { h^{ 2 } }{ { a }^{ 2 } } +\frac { { k }^{ 2 } }{ { b }^{ 2 } } -1 \right] ={ \left[ \frac { xh }{ { a }^{ 2 } } +\frac { yk }{ { b }^{ 2 } } -1 \right] }^{ 2 }$

On simplifying ,we get

${ x }^{ 2 }\left( { b }^{ 2 }-{ k }^{ 2 } \right) +2xyhk+{ y }^{ 2 }\left( { a }^{ 2 }-{ h }^{ 2 } \right) -2yka^{ 2 }-2xh{ b }^{ 2 }+{ h }^{ 2 }{ b }^{ 2 }+{ k }^{ 2 }a^{ 2 }=0$

Now, let $\left( { x }_{ 1 },0 \right) \left( { x }_{ 2 },0 \right) \left( 0,{ y }_{ 1 } \right) \left( 0,{ y }_{ 2 } \right)$ be the points of intersection of tangents with co-ordinate axes.

For these four points to be concyclic , ${ x }_{ 1 }{ x }_{ 2 }={ y }_{ 1 }{ y }_{ 2 } \rightarrow \left( 1 \right)$

To find ${ x }_{ 1 }{ x }_{ 2 }$ we put $y=0$in equation of pair of tangents.

We get ${ x }_{ 1 }{ x }_{ 2 } = \frac { { h }^{ 2 }{ b }^{ 2 }+{ k }^{ 2 }a^{ 2 } }{ \left( { b }^{ 2 }-{ k }^{ 2 } \right) } \rightarrow \left( 2 \right)$

Also similarly, ${ y }_{ 1 }{ y }_{ 2 }= \frac { { h }^{ 2 }{ b }^{ 2 }+{ k }^{ 2 }a^{ 2 } }{ \left( { a }^{ 2 }-{ h }^{ 2 } \right) } \rightarrow \left( 3 \right)$

On solving (1), (2) and (3) we get locus as : ${ x }^{ 2 }-{ y }^{ 2 }=a^{ 2 }-{ b }^{ 2 }$

Note : The locus will exclude part of hyperbola inside the ellipse .

- 4 years, 5 months ago

Even I did it in the same way, but I forgot to mention that the part of hyperbola inside the ellipse should be excluded.

- 4 years, 5 months ago

This is correct. Post the next question. @Sarvesh Nalawade

- 4 years, 4 months ago

Equation of tangent to ellipse is $y=mx+√(a^2 m^2+b^2 )$ (1) As the tangent meet X-axis and Y-axis at con-cyclic points,so putting x=0, we get, $y=±√(a^2 m^2+b^2 )$ ,two points on Y-axis . Since they lie on circle and they are at same distance from origin, so origin is the centre.

Similarly, we get $x=± m/√(a^2 m^2+b^2 )$. As these are the radius,

$m/√(a^2 m^2+b^2 )$ = $√(a^2 m^2+b^2 )$.

Thus we get the value of m w.r.t 'a and b' $m=(1±√(1-4a^2 b^2 ))/(2a^2 )$

putting the value of m in (1)

we get required eqn,

$[(y- (1±√(1-4a^2 b^2 )) (x)/(2a^2 )]^2$ = $a^2 m^2+b^2$

- 4 years, 5 months ago

Consider any circle of the form $x^2+y^2+2\lambda y+c=0; \lambda \neq 0$. This circle cuts the x-axis at two points equidistant from origin although the centre of the circle is not at origin. Hence, the proof isn't correct. @Akash Shukla

- 4 years, 5 months ago

Ya, you are right. But this circle given by you, will not cut Y axis in (0,±k). And in the soln , the circle cuts both X-axis and Y-axis in (±h,0) and (0,±k).

- 4 years, 5 months ago

Even then, you've just figured out the combined equation of the pair of tangents and not the required locus.

- 4 years, 5 months ago

This is incorrect @Akash Shukla.

- 4 years, 4 months ago

Problem 64:
In a problem of differentiation of $\dfrac{f(x)}{g(x)}$ one student writes the derivative as $\dfrac{f'(x)}{g'(x)}$ and he finds the correct result. If $g(x) = x^{2} , \displaystyle \lim_{x \to \infty} f(x) = 4$.
Find the value of $\displaystyle \lim_{x \to \infty} \left(\dfrac{f(x)}{4}\right)^{\dfrac{x^{2}}{2x+1}}$

- 4 years, 4 months ago

Solution to problem 64:

The condition given is equivalent to the following differential equation:

$\left(\frac{f'g-fg'}{g^2} \right) (x) = \left(\frac{f'}{g'} \right)(x)$

Now, substituting $g(x)$, solving the resulting (relatively simple) de, keeping in mind the boundary condition, we get,

$\frac{f(x)}{4}=\left( \frac{x}{x-2} \right)^2$

Call the required limit L. Then, we have $\ln(L)=\lim_{x\to \infty} \frac{x^2}{2x+1} \cdot \frac{4x-4}{(x-2)^2}=2\\\implies \boxed{ L=e^2 }$

- 4 years, 4 months ago

Correct.

- 4 years, 4 months ago

Is the answer $e^{-1}$?

- 4 years, 4 months ago

No.

- 4 years, 4 months ago

Posting the proboem by Deepraj since he can't post due to bug.
Problem 65:
Let $f:[a,b] \to \mathbb{R}$ be a differentiable function satisfying following conditions :
\begin{aligned} f(a)&=0 \\ |f'(x)| &\leq A|f(x)| \quad \forall x \in [a,b] \end{aligned} where A is a fixed real number.
Find $\displaystyle \int_{a}^{b} f(x)dx$

- 4 years, 4 months ago

$\left| f'(x) \right| \le A\left| f\left( x \right) \right|$

$\Longrightarrow \quad -A\le \frac { f'(x) }{ f(x) } \le A$

$\Longrightarrow \quad \int _{ a }^{ x }{ -Adx } \le \int _{ a }^{ x }{ \frac { f'(x) }{ f(x) } dx } \le \int _{ a }^{ x }{ Adx }$

$\Longrightarrow \quad -A(x-a)\le \ln { \frac { f(x) }{ f(a) } } \le A(x-a)\quad$

Now, I am not sure if I am explaining this right , but since $f(a)=0$

hence, for for the above inequality to be valid $f(x)=0\quad \forall \quad x\quad \in \quad [a,b]\quad$

Thus, the answer is zero .

- 4 years, 4 months ago