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) :

[Post your solution here]

PROBLEM xxx (number of problem) :

[Post your problem here]

Please try to post problems from all the spheres of JEE syllabus and share this note so that maximum users come to know about this contest and participate in it. (>‿◠)✌

Note by Sandeep Bhardwaj
3 years, 3 months ago

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Problem 56 : For whole numbers m m and n n , determine S=i=0n1(m+ii) S = \displaystyle \sum_{i=0}^{n-1} { m+i \choose i } .

Karthik Venkata - 3 years, 2 months ago

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in1(m+ii)=in1(m+im)Wehavetofindthecoefficientofxmintheexpansionof(1+x)m+(1+x)m+1+(1+x)m+2...........................(1+x)m+n1ThatsGPsum!=(1+x)((1+x)n1)xcoefficientofxmintheexpressionis(m+nm+1).Whichistheanswer.\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.

Mayank Chaturvedi - 3 years, 2 months ago

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Problem 66:

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

Mark C - 3 years, 3 months ago

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Solution to Problem 66:

We are looking for EE. If your first non-six comes up right away (56\frac{5}{6} chance), then you are starting fresh on the second roll, so the expectation is now E+1E+1. If your first non-six comes at the second roll (1656\frac{1}{6}\cdot\frac{5}{6} chance), then you are starting fresh on the third roll, so the expectation is now E+2E+2. If your first non-six comes on the third roll ((16)256(\frac{1}{6})^2\cdot\frac{5}{6} chance) then the expectation is E+3E+3. Otherwise ((16)3(\frac{1}{6})^3 chance) you get three sixes in three rolls. So:

E=56(E+1)+1656(E+2)+(16)256(E+3)+(16)33=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 = =258216+215216E= \frac{258}{216} + \frac{215}{216}E 1216E=258216\frac{1}{216}E = \frac{258}{216} E=258E = \boxed{258}

Good luck to all at the JEE!

Mark C - 3 years, 3 months ago

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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!

Mark C - 3 years, 3 months ago

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

A Brilliant Member - 3 years, 3 months ago

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

Saarthak Marathe - 3 years, 3 months ago

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Posting the proboem by Deepraj since he can't post due to bug.
Problem 65:
Let f:[a,b]R f:[a,b] \to \mathbb{R} be a differentiable function satisfying following conditions :
f(a)=0f(x)Af(x)x[a,b] \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 abf(x)dx \displaystyle \int_{a}^{b} f(x)dx

A Brilliant Member - 3 years, 3 months ago

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

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

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

A(xa)lnf(x)f(a)A(xa)\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)=0f(a)=0

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

Thus, the answer is zero .

Sarvesh Nalawade - 3 years, 3 months ago

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1/f(a)1/f(a) isn't defined. The solution is not correct also because 0/00/0 is indeterminate form.

Deeparaj Bhat - 3 years, 3 months ago

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@Deeparaj Bhat Fix Sarvesh's solution as follows. We assume for reductio that f(x)0f(x) \neq 0, and we replace the constant aa in Sarvesh's integrals with a variable ww, and ask for the the value of the integrals as nonzero f(w)0f(w) \rightarrow 0 (which we know happens, if not in the neighborhood of aa then somewhere else). Since f(w)f(w) and (by assumption) f(x)f(x) are nonzero, everything is defined and the assumptions of the problem entail that the inequalities hold. But for sufficiently small f(w)f(w), the inequality A(xw)lnf(x)f(w)A(xw)-A(x-w) \leq \ln\frac{f(x)}{f(w)} \leq A(x-w) must be false. Contradiction; therefore f(x)=0f(x)=0.

Mark C - 3 years, 3 months ago

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@Mark C You can post the next question

Sarvesh Nalawade - 3 years, 3 months ago

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@Mark C That is correct! You two decide who'll post the next problem.

Deeparaj Bhat - 3 years, 3 months ago

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Yeah. Post the solution.

Deeparaj Bhat - 3 years, 3 months ago

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Problem 64:
In a problem of differentiation of f(x)g(x) \dfrac{f(x)}{g(x)} one student writes the derivative as f(x)g(x) \dfrac{f'(x)}{g'(x)} and he finds the correct result. If g(x)=x2,limxf(x)=4 g(x) = x^{2} , \displaystyle \lim_{x \to \infty} f(x) = 4 .
Find the value of limx(f(x)4)x22x+1 \displaystyle \lim_{x \to \infty} \left(\dfrac{f(x)}{4}\right)^{\dfrac{x^{2}}{2x+1}}

A Brilliant Member - 3 years, 3 months ago

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Solution to problem 64:

The condition given is equivalent to the following differential equation:

(fgfgg2)(x)=(fg)(x)\left(\frac{f'g-fg'}{g^2} \right) (x) = \left(\frac{f'}{g'} \right)(x)

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

f(x)4=(xx2)2\frac{f(x)}{4}=\left( \frac{x}{x-2} \right)^2

Call the required limit L. Then, we have ln(L)=limxx22x+14x4(x2)2=2    L=e2\ln(L)=\lim_{x\to \infty} \frac{x^2}{2x+1} \cdot \frac{4x-4}{(x-2)^2}=2\\\implies \boxed{ L=e^2 }

Deeparaj Bhat - 3 years, 3 months ago

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Correct.

A Brilliant Member - 3 years, 3 months ago

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Is the answer e1e^{-1}?

Deeparaj Bhat - 3 years, 3 months ago

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No.

A Brilliant Member - 3 years, 3 months ago

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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

Sarvesh Nalawade - 3 years, 3 months ago

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1872 , is it correct ?

Akash Shukla - 3 years, 3 months ago

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No, that's not correct

Sarvesh Nalawade - 3 years, 3 months ago

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Are repetitions allowed?

Deeparaj Bhat - 3 years, 3 months ago

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No, each number has to be taken exactly once.

Sarvesh Nalawade - 3 years, 3 months ago

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@Sarvesh Nalawade 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.
2M+K=7! \therefore 2M + K = 7!
For evaluating K,
If the middle digit is P,
Sum of first 3 digits (S) (S) = Sum of last 3 digits = 28P2 \dfrac{28-P}{2}
\therefore P is even.
Case 1:

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

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

Case 3:

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

K=372=216 \therefore K = 3 \cdot 72 = 216
2M=5040K=5040216=4824 2M = 5040 - K = 5040 - 216 = 4824
M=2412 M = 2412

A Brilliant Member - 3 years, 3 months ago

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@A Brilliant Member Correct!!

Sarvesh Nalawade - 3 years, 3 months ago

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

Deeparaj Bhat - 3 years, 3 months ago

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@Deeparaj Bhat 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 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 12 = 7 + 5 = 7 + 1 + 4 = 7 + 2 + 3
For case 3 :
11=7+4=7+1+3=7+2+2 11 = 7 + 4 = 7 + 1 + 3 = 7 + 2 + 2
I used the largest number to make the grouping easier.

A Brilliant Member - 3 years, 3 months ago

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

Deeparaj Bhat - 3 years, 3 months ago

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Problem 62:

V1V_1 is the unit sphere with equation x2+y2+z2=1x^2+y^2+z^2=1, and V2V_2 is the volume formed by subtracting, from the volume enclosed by the cylinder x2+y2=1x^2+y^2=1, 1z1-1\leq z\leq 1, the volume enclosed by the cones x2+y2=z2x^2 +y^2 =z^2. At what heights zz is the maximum absolute difference between the horizontal cross-sections of V1V_1 and V2V_2, and what is that difference?

This problem has been solved by Sarvesh Nalawade.

Mark C - 3 years, 3 months ago

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

Consider a plane z=k z=k where k1 |k| \leq 1 .

It will intersect the sphere , i.e V1 V_{1} in a circle with radius 1k2 \sqrt{1-k^{2}}

Thus its area will be given by :

A1=π[1k2] A_{1}= \pi[1-k^{2}]

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

Therefore , Area of horizontal cross-section of V2 V_{2} will be given by :

A2=π[12k2] A_{2} = \pi [ 1^{2} - k^{2} ]

Therefore A1=A2k(1,1) A_{1}=A_{2} \quad \forall k \in (-1,1) .

Hence their difference will always be zero

Sarvesh Nalawade - 3 years, 3 months ago

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Correct!

Mark C - 3 years, 3 months ago

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Problem 61

If f(x)=x3+ax2+bx+c=0f(x)=x^{3}+ax^{2}+bx+c=0 has 3 integral roots and (x2+2x+2)3+a(x2+2x+2)2+b(x2+2x+2)+c=0(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,ca,b,c.

Note:c<3 |c|<3

Mark C has provided a solution to this problem.

Vignesh S - 3 years, 3 months ago

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f(x)=x3+ax2+bx+c=(xn1)(xn2)(xn3)=0f(x) = x^3+ax^2+bx+c = (x-n_1)(x-n_2)(x-n_3)=0

a=(n1+n2+n3),b=n1n2+n1n3+n2n3,c=n1n2n3a=-(n_1+n_2+n_3), b=n_1n_2+n_1n_3+n_2n_3, c=-n_1n_2n_3

(x2+2x+2)3+(x2+2x+2)2a+(x2+2x+2)b+c=0(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 (x2+2x+2)3(x^2+2x+2)^3 will obviously exceed the sum of the others for large positive and negative values of xx.)

x2+2x+2x^2 +2x +2 has a minimum of 1 for x=1x=-1. So:

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

Since n1n2n3<3|n_1n_2n_3| < 3, {n1,n2,n3}{2,1,0,1,2}\{n_1,n_2,n_3\} \subset \{-2,-1,0,1,2\}. Of the possibilities, only one works: {n1,n2,n3}={2,1,0}\{n_1,n_2,n_3\}=\{-2,-1,0\} yields (a,b,c)=(3,2,0)(a,b,c)=\boxed{(3,2,0)}, which sum to 5.

Mark C - 3 years, 3 months ago

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I did the the same solution.

Vignesh S - 3 years, 3 months ago

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Problem 60:

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

This problem has been solved by Vignesh S

Mark C - 3 years, 3 months ago

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Answer is 0. The above integral can be written as 0.50πsin8xsin2xdx=0.50πsin(8π8x)sin(2π2x)dx=I0.5\int_{0}^{\pi}\sin{8x}-\sin{2x}dx=0.5\int_{0}^{\pi}\sin{(8\pi-8x)}-\sin{(2\pi-2x)}dx=I     2I=0.50πsin8x+sin(8π8x)dx0.50πsin2x+sin(2π2x)dx\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=02I=0 ;I=0

Vignesh S - 3 years, 3 months ago

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Very good! The answer can also be reached by observing that sin(3(πx))=sin(π3x)=sin(3x)\sin(3(\pi-x))=\sin(\pi-3x)=\sin(3x) and cos(5(πx))=cos(π5x)=cos(5x)\cos(5(\pi-x))=\cos(\pi-5x)=-\cos(5x), so that the integral from 00 to π2\frac{\pi}{2} is the negative of the integral from π2\frac{\pi}{2} to π\pi.

Your turn to post a problem!

Mark C - 3 years, 3 months ago

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Problem 59:

We have bundle of nn (for nn 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 nn, of the probability that we have created just one loop composed of all of the ropes?

Mark C - 3 years, 3 months ago

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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 nn 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 n2n-2 of the n1n-1 other ropes that will avoid failure (failure is creating a mini-loop). So the probability of avoiding failure with my first knot is n2n1\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 n2n-2 ropes: there are n2n-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 n4n3\frac{n-4}{n-3}.

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

Pn=n2n1n4n34523P_n = \frac{n-2}{n-1}\cdot\frac{n-4}{n-3}\cdots\frac{4}{5}\cdot\frac{2}{3}

limn(Pn)=234567=i=1(112i+1)\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(i=1(112i+1)))=exp(i=1ln(112i+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)\exp(x) means exe^x. Since for x>0x>0, ln(x)x1\ln(x) \leq x - 1:) exp(i=1(12i+1))exp(i=1(1i))=e=0.\leq \exp (-\sum_{i=1}^\infty(\frac{1}{2i+1})) \leq \exp (-\sum_{i=1}^\infty(\frac{1}{i})) = e^{-\infty} = \boxed{0}.

Mark C - 3 years, 3 months ago

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Problem 57 :

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

f(x+y)f(xy)yy2x,yR \left| f\left( x+y \right) -f\left( x-y \right) -y \right| \le { y }^{ 2 } \quad \forall x,y \in R .

If f(0)=52 f\left( 0 \right) =\frac { 5 }{ 2 } find f(3) f\left( 3 \right)

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

Sarvesh Nalawade - 3 years, 3 months ago

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Solution of Problem 57

f((x+Δx2)+Δx2)f((x+Δx2)Δx2)Δx2Δx24|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+Δx)f(x)Δx2=Δf(x)Δx2Δx24|f(x+\Delta x)-f(x) - \frac{\Delta x}{2}| = |\Delta f(x) - \frac{\Delta x}{2}| \leq \frac{\Delta x^2}{4}

As Δx\Delta x approaches zero, Δx2\Delta x^2 is negligible and so Δf(x)\Delta f(x) approaches Δx2\frac{\Delta x}{2}. Thus the slope is everywhere 12\frac{1}{2} and the function is:

f(x)=12x+52f(x) = \frac{1}{2}x + \frac{5}{2}

f(3)=82f(3) = \boxed{\frac{8}{2}}

Mark C - 3 years, 3 months ago

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That is correct!!

Sarvesh Nalawade - 3 years, 3 months ago

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Problem 56:

Find the locus of point PP such that tangents drawn from it to the given ellipse x2a2+y2b2=1\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!

Saarthak Marathe - 3 years, 3 months ago

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Equation of tangent to ellipse is y=mx+(a2m2+b2)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=±(a2m2+b2)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/(a2m2+b2)x=± m/√(a^2 m^2+b^2 ). As these are the radius,

m/(a2m2+b2) m/√(a^2 m^2+b^2 ) = (a2m2+b2)√(a^2 m^2+b^2 ).

Thus we get the value of m w.r.t 'a and b' m=(1±(14a2b2))/(2a2)m=(1±√(1-4a^2 b^2 ))/(2a^2 )

putting the value of m in (1)

we get required eqn,

[(y(1±(14a2b2))(x)/(2a2)]2[(y- (1±√(1-4a^2 b^2 )) (x)/(2a^2 )]^2 = a2m2+b2a^2 m^2+b^2

Akash Shukla - 3 years, 3 months ago

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This is incorrect @Akash Shukla.

Saarthak Marathe - 3 years, 3 months ago

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Consider any circle of the form x2+y2+2λy+c=0;λ0x^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

Deeparaj Bhat - 3 years, 3 months ago

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@Deeparaj Bhat 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).

Akash Shukla - 3 years, 3 months ago

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@Akash Shukla Right. My bad.

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

@Akash Shukla

Deeparaj Bhat - 3 years, 3 months ago

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Is this the answer? a2b2=x2y2a^2-b^2=x^2-y^2

Deeparaj Bhat - 3 years, 3 months ago

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I too am getting the same result

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

The combined equation of pair of tangents to the ellipse : S=x2a2+y2b21=0 S =\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } -1 = 0 is given by :

SS1=T2S{ S }_{ 1 }={ T }^{ 2 }

i.e [x2a2+y2b21][h2a2+k2b21]=[xha2+ykb21]2 \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

x2(b2k2)+2xyhk+y2(a2h2)2yka22xhb2+h2b2+k2a2=0{ 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 (x1,0)(x2,0)(0,y1)(0,y2)\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 , x1x2=y1y2(1){ x }_{ 1 }{ x }_{ 2 }={ y }_{ 1 }{ y }_{ 2 } \rightarrow \left( 1 \right)

To find x1x2{ x }_{ 1 }{ x }_{ 2 } we put y=0 y=0 in equation of pair of tangents.

We get x1x2=h2b2+k2a2(b2k2)(2){ 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, y1y2=h2b2+k2a2(a2h2)(3) { 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 : x2y2=a2b2 { x }^{ 2 }-{ y }^{ 2 }=a^{ 2 }-{ b }^{ 2 }

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

Sarvesh Nalawade - 3 years, 3 months ago

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

Saarthak Marathe - 3 years, 3 months ago

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

Deeparaj Bhat - 3 years, 3 months ago

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

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

A Brilliant Member - 3 years, 3 months ago

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Answer is 44.

Draw a dodecagon. Let the center be OO.

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

Let the radius be xx. Then AB=2xcos(75)=(31)2.xAB=2xcos({75}^{\circ})=\dfrac{(\sqrt{3}-1)}{\sqrt{2}}.x.

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

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

Saarthak Marathe - 3 years, 3 months ago

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Problem 54:

Let α\alpha and β\beta be two distinct, randomly chosen roots of the following equation: z42=1z^{42}=1

If the probabilty that 2+3α+β\sqrt{2+\sqrt3}\leq|\alpha+\beta| is equal to ab\dfrac{a}{b} where aa and bb are coprimes, find a+ba+b.

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

Deeparaj Bhat - 3 years, 3 months ago

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α,β \alpha , \beta satisfy z42=1 z^{42} = 1 .
Thus, α,β \alpha, \beta are the forty second roots of unity, and lie on a unit circle centered at origin.
Let,
α=ei2πk42,β=ei2πm42 \alpha =e^{i\frac{2\pi k}{42}} , \beta = e^{i \frac{2\pi m}{42}}
α+β=(cos(kπ21)+cos(mπ21))+i(sin(kπ21)+sin(mπ21)) | \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| , 0k,m41 0 \le k,m \le 41
α+β=2+2cos((km)π21) \therefore | \alpha + \beta | = \sqrt{2 + 2\cos\left(\dfrac{(k-m)\pi}{21}\right) }
2+2cos((km)π21)2+3 \sqrt{2 + 2\cos\left(\dfrac{(k-m)\pi}{21}\right) } \ge \sqrt{2+\sqrt{3}}
cos((km)π21)32 \cos\left(\dfrac{(k-m)\pi}{21}\right) \ge \dfrac{\sqrt{3}}{2}
cos((km)π21)cos(π6) \cos\left(\dfrac{(k-m)\pi}{21}\right) \ge \cos\left(\dfrac{\pi}{6}\right)
2nππ6(km)π212nπ+π6 2n\pi - \dfrac{\pi}{6} \le \dfrac{(k-m)\pi}{21} \le 2n\pi + \dfrac{\pi}{6}
42n72km42n+72 42n - \dfrac{7}{2} \le k-m \le 42n + \dfrac{7}{2}
Also note that,
41km41 -41 \le k - m \le 41
This restricts us to n=1,0,1 n = -1,0,1
When n=1 n = -1 ,
45.5km38.5 -45.5 \le k-m \le -38.5
41km39 -41 \le k-m \le -39 which gives us 6 cases.
When n=0 n = 0 ,
3.5km3.5 -3.5 \le k-m \le 3.5
3km3 -3 \le k-m \le 3 which gives us 240 cases.
When n=1 n = 1 ,
38.5km45.5 38.5 \le k-m \le 45.5 which again gives us 6 cases.
Total such cases =252 = 252
But since there's no distinction between α \alpha , β \beta we are overcounting by a factor of 2.
Total favorable cases =126 = 126

Total ways of selecting two distinct roots = (422)=861 \dbinom{42}{2} = 861
Probability = 126861=641 \dfrac{126}{861} = \dfrac{6}{41}
a+b=47 a + b = 47

A Brilliant Member - 3 years, 3 months ago

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What I mean by the distinction statement is if we choose the two roots to be say,
eiπ21,ei2π21 e^{i\frac{\pi}{21}} , e^{i\frac{2\pi}{21}} .
I can call the first one α \alpha or the second one , doesn't make a difference.

A Brilliant Member - 3 years, 3 months ago

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@A Brilliant Member Yeah. Exactly!

You could've reduced the counting part by putting r=kmr=k-m and then saying that 0<r<42 0<r<42 and rr is a integer. As kk and mm are random, so is rr, and hence all the values of rr are equally likely.

You can post the next problem.

Deeparaj Bhat - 3 years, 3 months ago

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Problem 53:

Let Γ1\Gamma_1 be the circumcircle of ΔABC\Delta ABC and Γ2\Gamma_2 be the circumcircle of ΔDEF\Delta DEF where D,E,FD, E, F are the foot of the altitudes of ΔABC\Delta ABC.

Find the value of the following expression when the length of common chord of Γ1\Gamma_1 and Γ2\Gamma_2 is maximum for a given circumradius:

cyccos2A\sum_{cyc} \cos 2A

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

Deeparaj Bhat - 3 years, 3 months ago

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

Solution to problem 53:

First of all, note that Γ2\Gamma_2 (call it's centre O9O_9) is the nine point circle of ΔABC\Delta ABC.

Call the centre of Γ1\Gamma_1 as OO and it's radius RR. It now follows that the radius of Γ2\Gamma_2 is R2\dfrac{R}{2}. As the radius of Γ2\Gamma_2 is less than that of Γ1\Gamma_1, the length of the common chord is maximum if and only if it (the chord) is the diameter of Γ2\Gamma_2.

So, when maximum occurs, using Pythagoras' theorem, OO9=32ROO_9=\frac{\sqrt3}{2}R

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

2OO9=OHOH2=9R2(a2+b2+c2)asinA=bsinB=csinC=2R2 \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 HH denotes the orthocentre of ΔABC\Delta ABC.

Using the above three equations, we get:

cycsin2A=32    cyccos2A=32    cyccos2A=0 \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}

Deeparaj Bhat - 3 years, 3 months ago

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PROBLEMNO52Considertwocomplexnosz1&z2satisfyingS:z5i=4suchthatarg(z1)=arg(z2)=3π10andz1<z2Ifz3isthepointofintersectionoftangentdrawnatz2onSwithrealaxis.Findarg(5iz3z2z3)BONUS:Canyoudoitwithoutcoordinatebashing.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!

Rishi Sharma - 3 years, 3 months ago

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@Rishi Sharma

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

Sandeep Bhardwaj - 3 years, 3 months ago

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Solution to problem 52:

@Rishi Sharma No coordinate bash!

Deeparaj Bhat - 3 years, 3 months ago

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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!

Sandeep Bhardwaj - 3 years, 3 months ago

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Problem 51

Let ff be a continuous real function defined on positive reals (R+\mathbb{R}^+) satisfying the following conditions: f(xy)=f(x)+f(y)x,yR+f(e)=1 \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 ff.

Note: Here ee is the base of natural logarithm.

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

Deeparaj Bhat - 3 years, 3 months ago

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1

Ben Martin - 1 year, 2 months ago

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Solution to problem 51

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

Then,

g(x+y)=g(x)+g(y)g(x+y)=g(x)+g(y) By induction, the following are true: g(i=1nxi)=i=1ng(xi)xiRg(x)=cxxQ where c is a constant  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 ana_n which satisfies the following properties: anQnNanαnNlimnan=αwhereαisanarbitraryreal 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(α)=limng(an)=(limncan)=cα g(\alpha)= \lim_{n \to \infty}g(a_n)=\left(\lim_{n \to \infty} ca_n \right)=c\alpha

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

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

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


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

Deeparaj Bhat - 3 years, 3 months ago

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

Proof: Put x=y=ex=y=e.

Then f(e2)=2f(e^{2})=2.

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

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

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

f(e2/3=2f(e1/3)f(e^{2/3}= 2f(e^{1/3}).

Also f(e4/3=2f(e2/3)=4f(e1/3)=1+f(e1/3)f(e^{4/3}=2f(e^{2/3})= 4f(e^{1/3})= 1+f(e^{1/3}).

Thus f(e1/3)=1/3f(e^{1/3})=1/3 . . . So on..

f(e1/n)=1/nf(e^{1/n})=1/n.

Hence prooved.

Aakash Khandelwal - 3 years, 3 months ago

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This proves that f(x)=ln(x)f(x)=ln(x) for all rational xx. 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.

Deeparaj Bhat - 3 years, 3 months ago

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That seems right for rational values of xx. 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.)

Mark C - 3 years, 3 months ago

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SOLUTIONOFPROBLEM51Firstwenoticethatf(1)=0sincef(e)=f(e1)=f(e)+f(1)Nowitisgiventhatf(xy)=f(x)+f(y)x,y+f(xy)y=f(x)y+f(y)yxf(xy)=f(y)insertingy=1wegetthediffrentialequation,xf(x)=f(1)dydx=f(1)xdy=f(1)dxxy=f(x)=f(1)lnx+C{Cistheconstantofintegration}asx>0wecanwritef(x)=f(1)lnx+Cnowusingf(1)=0andf(e)=1wegetf(1)=1andC=0hence,f(x)=lnxOnlyonesuchfunctionispossiblewhichsatisfiesthegivenconditionsSOLUTION\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