JEE-Advanced Maths Contest '16 (Continued)

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.

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

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.

1

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 $x>0$, $\ln(x) \leq x - 1$:) $\leq \exp (-\sum_{i=1}^\infty(\frac{1}{2i+1})) \leq \exp (-\sum_{i=1}^\infty(\frac{1}{i})) = e^{-\infty} = \boxed{0}.$

Solution to problem 52:

@Rishi Sharma No coordinate bash!

@Rishi Sharma

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

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:

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

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:

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

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

$\alpha , \beta$ 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 = $\dfrac{126}{861} = \dfrac{6}{41}$
$a + b = 47$

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

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

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

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

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.

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

Are repetitions allowed?

1872 , is it correct ?

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!

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!

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

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 $\dfrac{AB}{AF} + \dfrac{AF}{AB}=\boxed{4}$

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

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,

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

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 $\ln(L)=\lim_{x\to \infty} \frac{x^2}{2x+1} \cdot \frac{4x-4}{(x-2)^2}=2\\\implies \boxed{ L=e^2 }$

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

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

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

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

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

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 .