ques for all

- anonymous

ques for all

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- jamiebookeater

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- anonymous

Madhava Mathematics Competition
(A Mathematics Competition for Undergraduate Students)
Organized by Department of Mathematics, S.P. College, Pune and Homi Bhabha Centre for Science Education, T.I.F.R, Mumbai
Date:13/12/2015
Time:12-3pm
Part I 20 marks, Part II 30 marks, Part III 50 marks
Part I each question carries 2 marks in Part I
Q1) Let A(t) denote the area bounded by the curve
\[y=e^{-|x|}\], the X-axis and the straight lines x=-t and x=t, then
\[\lim_{t \rightarrow \infty} A(t)\]
is
a) 2
b) 1
c) 0.5
d) e
Q2) How many triples of real numbers(x,y,z) are common solutions of the equations \[x+y=2\]\[xy-z^2=1\]
a)0
b)1
c)2
d)infinitely many
Q3)For non negative integers x,y the function f(x,y) satisfies the relations
\[f(x,0)=x\]
and
\[f(x,y+1)=f(f(x,y),y)\].
Then which if the following is the largest?
a)f(10,15)
b)f(12,13)
c)f(13,12)
d)f(14,11)
q4) Suppose p,q,r,s are 1,2,3,4 in some order
Let
\[x=\frac{1}{p+\frac{1}{q+\frac{1}{r+\frac{1}{s}}}}\]
We choose p,q,r,s so that x is as large as possible, then s is
a)1
b)2
c)3
d)4
q5)Let
|dw:1450002044699:dw|
then
f''(0) is
a)0
b)2
c)3
d)none of these
More in a minute.

- bubblegum.

for 2nd ques
and y both should be positive
and thus both must be less than 2
if they are in fraction like x=1.5 and y=0.5
this won't work because multiplying them will always give a number less than 1
and thus 2nd eq won't satisfy
only clear sol we can see is (1,1,0)
so B

- ParthKohli

Q4 seems nice :)
It looks like we're playing with dominoes actually.
Minimise \(p\), so \(p =1\). Also maximise\[q+\frac{1}{r + \frac{1}{s}}\]For that \(q = 4\) as we want to maximise this and we want to minimise \(r + 1/s\). For that, \(r = 2\) and \(s = 3\).
Maybe it can get a lot more complicated because it is an undergraduate competition and I'm treating this way too one-dimensionally but...

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## More answers

- anonymous

U know it 2:36 in the morning

- anonymous

I'm still on airplanes for free

- anonymous

for 2nd question
take the value of y from 1st eqn as 2-x
put it in 2nd eqn
and use the condition D>0 as x is real
this will give as -z^2 > 0 so z=0
and AM and GM of first eqn is same so x=1 y=1 z=0

- ParthKohli

@Astrophysics discriminants

- anonymous

Yes 2:37am I'm from California going to North Carolina

- anonymous

Q6) There are 8 teams in pro-kabaddi league. Each team plays against every other exactly once. Suppose every game results in a win, that is, there is no draw. Let w1,w2,....,w8 be number of wins and l1,l2,....,l8 be number of loses by teams T1,T2,....,T8, then
a)\[w_{1}^2+...+w_{8}^2=49+(l_{1}^2+...+l_{8}^2)\]
b)
\[w_{1}^2+...+w_{8}^2=l_{1}^2+...+l_{8}^2\]
c)\[w_{1}^2+...+w_{8}^2=49-(l_{1}^2+...+l_{8}^2)\]
d) none of these
q7)The remainder when m+n is divided by 12 is 8, and the remainder when m-n is divided by 12 is 6. If m>n, then the remainder when mn is divided by 6 is
a)1
b)2
c)3
d)4
q8)
let
|dw:1450002970074:dw|
Select any entry and call it x1. Delete row and column containing x1 to get an
(n-1) *(n-1) matrix. Then select any entry from the remaining entries and call it x2. Delete row and colum containing x2 to get (n-2)*(n-2) matrix. Perform n such steps. Then x1+x2+...+xn is
a)n
b)(n(n+1))/2
c)(n(n^2+1))/2
d)none of these
q9)The maximum of the areas of the rectangles inscribed in the regions bounded by the curve y=3-x^2 and X-axis is
a)4
b)1
c)3
d)2
q10)how many factors of
\[2^53^65^2\]
are perfect squares?
a)24
b)20
c)30
d)36

- ParthKohli

Q10 is not too bad...
For 2, select from 2^0, 2^2, 2^4
For 3, select from 3^0, 3^2, 3^4, 3^6
For 5, select from 5^0, 5^2

- anonymous

Part 2 each question is of 6 marks
q1) How many 15-digit plaindromes are these in each of which the product of the non-zero digits is 36 and the sum of the digits is equal to 15?(A string of digits is called a palindrome if it reads the same forwards and backwards. for example 04340, 6411146)
q2)Let H be a finite set of distinct positive integers none of which has a prime factor greater than 3. Show that the sum of the reciprocals of the elements of H is smaller than 3. Find two different such sets with sum of the reciprocals equal to 2.5
q3) Let A be an n*n matrix with real entries such that each row sum is equal to one.
Find the sum of all entries of \[A^{2015}\]
q4)Let \[f: \mathbb{R} \rightarrow \mathbb{R}\]
be a differentiable function such that \[f(0)=0\]\[f'(x)>f(x) \forall x \in \mathbb{R}\]
Prove that \[f(x)>0 \forall x>0\]
q5) Give an example of a function which is continuous on [0,1], differentiable on (0,1) and not differentiable at end points. Justify.

- ParthKohli

wow this is getting real...
for Q5, |x| + |x-1|

- ganeshie8

Has anyone found answer for this
Q3)For non negative integers x,y the function f(x,y) satisfies the relations
\[f(x,0)=x\]
and
\[f(x,y+1)=f(f(x,y),y)\].
Then which if the following is the largest?
a)f(10,15)
b)f(12,13)
c)f(13,12)
d)f(14,11)

- anonymous

q4
f'(x) - f(x) >0
put x=0
means f'(0) >0
i.e. x->0+
f(0+)> f(0) >0
the same can be extended

- ganeshie8

Day and Date of Examination: Sunday, December 13, 2015.

- ganeshie8

Hey @Nishant_Garg
hows the exam ?

- anonymous

Part 3
q1) There are some marbles in a bowl. A,B and C take turns removing one or two marbles from the bowl, with A going first, then B,then C,then A again and so on. The player who takes the last marble from the bowl is the loser and the other two players are the winners. If the game starts with N marbles in the bowl, for what values of N can B and C work together and force A to lose?(12 marks)
q2)Let \[f: \mathbb{R} \rightarrow \mathbb{R}\]
be a function such that f'(0) exists. Suppose \[\alpha_{n} \neq \beta_{n}, \forall n \in \mathbb{N}\]
and both sequences \[\left\{ \alpha_{n} \right\} \space \space \space , \space \space \space \left\{ \beta_{n} \right\}\] converge to 0
Define
\[D_{n}=\frac{f(\beta_{n})-f(\alpha_{n})}{\beta_{n}-\alpha_{n}}\]
Prove that
\[\lim_{n \rightarrow \infty }D_{n}=f'(0)\]
under EACH of the following conditions:
(a) \[\alpha_{n}<0<\beta_{n} \forall n \in \mathbb{N}\]
(b)
\[0<\alpha_{n}<\beta_{n}\]
and \[\frac{\beta_{n}}{\beta_{n}-\alpha_{n}}\leq M, \space \forall n \in \mathbb{N} , \space M>0\]
for some M
(c)
f'(x) exists and is continuous for all x in (-1,1)
(13 marks)
q3)
Let \[f(x)=x^5\]
for \[x_{1}>0\]
let
\[P_{1}=(x_{1},f(x_{1}))\]
Draw a tangent at the point P1 and let it meet the graph again at point P2. then draw a tangent at P2 and so on
Show that the ratio
\[\frac{A(\triangle P_{n}P_{n+1}P_{n+2})}{A(\triangle P_{n+1}P_{n+2}P_{n+3})}\]
is constant.
(12 marks)
q4) Let p(x) be a polynomial with positive integers conefficients. You can ask the question: What is p(n) for any positive integer n? What is the minimum number of questions to be asked to determine p(x) completely?Justify. (13 marks)

- ParthKohli

I've done Q4 before! Wow.

- anonymous

Haha, I did pretty bad, I had completely forgotten about this exam :P, didn't even knew it would be so different. I'm just sharing because I know you guys love this stuff.

- ganeshie8

Happens haha.. have they announced the results yet ?

- ganeshie8

Q3)For non negative integers x,y the function f(x,y) satisfies the relations
\[f(x,0)=x\]
and
\[f(x,y+1)=f(f(x,y),y)\].
Then which if the following is the largest?
a)f(10,15)
b)f(12,13)
c)f(13,12)
d)f(14,11)
\(f(x,0)=x\)
\(f(x,1)=f(f(x,0),0) = f(x,0)=x\)
\(f(x,2)=f(f(x,1),1) = f(x,1)=x\)
\(\vdots\)
\(f(x,y) = x\)
therefore \(f(14,11)\) is greatest

- anonymous

Nah, the exam was just today, at 12

- anonymous

Wow, nice one!

- bubblegum.

part 2 ques 2
all the numbers will be of the form-> \(N=2^a3^b\)
we can write the summation of all the terms as-> \(\sum_{a=0}^{\infty}\frac{1}{2^a} \sum_{b=0}^{\infty}\frac{1}{3^b}\)
thats just and infinite GP summation
so that will be-> \(1 \times 1/2 =1/2\)
now we take 2 different such sets
what ever the sets may be after all they hav to be added
even if u double the summation of all the terms of set H then also you get 1
:/

- anonymous

I'm really bad at pure mathematics stuff, how hard are these questions, anyone knows?

- ParthKohli

q1) How many 15-digit palindromes are these in each of which the product of the non-zero digits is 36 and the sum of the digits is equal to 15?(A string of digits is called a palindrome if it reads the same forwards and backwards. for example 04340, 6411146)
are we allowed to begin a number with zero?

- anonymous

I don't know, It's not given, I've written the question exactly as it was.

- anonymous

i dont think so @ParthKohli

- ParthKohli

according to their examples, I'm a little more inclined to yes.

- ParthKohli

if the middle digit is 9 then we have to adjust 1, 2 in the first seven digits (the rest being zeroes)
if the middle digit is 1 then we have to adjust 6, 1 in the first seven digits
if the middle digit is 3 then we have to adjust ???
if the middle digit is 6 then we have to adjust ???

- bubblegum.

Q9)
the x intercepts of the curve y=3-x^2
are \(+ \sqrt{3} ~and -\sqrt{3}\)
so |dw:1450005172063:dw|
so area = \( 2x \times (3-x^2)\)
1st derivative must be 0 to maximize area
\(A=6x-2x^3\)
\(A'=0=6-6x^2\)
\(x= \pm1\)
so the one dimension that is 2x is 2
and other dimension is 3-x^2 =2
so area =2x2 =4

- anonymous

@bubblegum. That's exactly how I've done that question

- anonymous

second derivative test

- bubblegum.

:)

- ParthKohli

if the middle digit is 0, we should adjust ???

- bubblegum.

in the 1st ques i donno what area they are asking for :/
are they asking for this-http://prntscr.com/9dluih ?

- ParthKohli

no it's just an improper integral, I think

- ParthKohli

\[\int \limits_{-\infty}^{\infty} e^{-|x|}dx \]

- anonymous

q7 part 1
write m+n = 12a+8
m-n = 12b + 6
square and subtract them
you will be able to see the answer

- bubblegum.

lol u gotta work too much for that why don't directly integrate from -t to t and then subtract the unwanted area

- ParthKohli

it's not that bad though, I'm getting 2.

- anonymous

|dw:1450005902701:dw|

- anonymous

its the 1st ques

- ParthKohli

\[I=2\int \limits_{0}^{\infty}e^{-x}dx =2\times 1 \]

- anonymous

\[\int\limits_{-\infty}^{\infty }e^{-|x|}dx=\int\limits_{-\infty}^{0}e^xdx+\int\limits_{0}^{\infty}e^{-x}dx\]
Both integrals evaluate to 1 each, so the sum is 2.

- bubblegum.

1st we calculate the area of the region bounded by y=e^-|x| and x=t and x=-t
so that will be->\(\left( int_{-t}^{t} e^{-|x|}\right)-2\left( int_{t}^{\infty}e^{\|x| \right)\)

- anonymous

maybe the t is just there to confuse you :P

- bubblegum.

\[\left( \int\limits_{-t}^{t} e^{-|x| }\right)-2\left( \int\limits_{t}^{\infty}e^{-|x|} \right)\]

- anonymous

as t tends to infinity your 2nd term would tend to 0 I think(because same limits on the integral), so it's the same thing?

- ParthKohli

so for the palindrome question we observe that the middle digit cannot be even
that leaves 1, 3, 9 as possible candidates
now as for 3, I don't think there's any possibility

- bubblegum.

sorry i forgot to add \(dx\) in the equation

- ParthKohli

the reason why 3 cannot be the digit is because 36/3 is not a perfect square

- ParthKohli

therefore the answer does seem to be \(2\times \binom{7}2\) if zeroes are allowed at the beginning

- anonymous

\[000001616100000\]
\[\frac{15!}{10! \cdot 3! \cdot 2! } ?\]
Another one I see is
\[000112312311000\]

- anonymous

Personally I'd assume 0's allowed in the beginning because they seem to be more of talking about strings of digits rather than actual numbers

- ParthKohli

I did include the first one but the second one was a good catch!

- anonymous

Oops, must be a palindrome!

- anonymous

interchange one the 2 and 3's around

- anonymous

\[000113212311000\]

- ParthKohli

ok so we need to add\[\binom{7}4\]

- ParthKohli

\[2\binom{7}2+\binom{7}4\]

- anonymous

Alright, I'm off, keep yourself busy with these questions haha! cheers

- bubblegum.

q7)
just take proper things ;^)
m+n=20
and m-n=6
from here m=13 and n=7
mn=91
when u divide by 6 u get remainder=1

- anonymous

How about m=19 and n=1, in that case when divided by 6 we get the remainder as 1

- bubblegum.

yea we can do that too :)

- thomas5267

Can someone LaTeX-fy the function in part 1 Q5? Can't see on iPad and laptop broke down.

- bubblegum.

this is the ques-http://prntscr.com/9dnjyg

- thomas5267

Part 1 Q7. Strangely easy. Am I missing something?
\[
\begin{align*}
m+n&\equiv 8\pmod{12}\\
m-n&\equiv 6\pmod{12}\\
2m&\equiv 2\pmod{12}\\
m&\equiv 1\pmod 6\\
2n&\equiv 2 \pmod {12}\\
n&\equiv 1 \pmod 6\\
mn&\equiv 1\pmod 6
\end{align*}
\]

- thomas5267

Part 2 Q3. The question seems to indicate that the sum is the same no matter which matrix A you choose. Choose the identity matrix and \(I_n^{2015}=I_n\) so the sum is n hahahahaha.

- thomas5267

Part 2 Q3
\[
\begin{align*}
\sum_{k=1}^n \left(A^2\right)_{1k}&=\sum_{k=1}^n\sum_{j=1}^n a_{1j}a_{jk}\quad \text{Sum of first row.}\\
&=a_{11}\sum_{k=1}^n a_{1k}+ a_{12}\sum_{k=1}^n a_{2k} \dots\\
&=\sum_{j=1}^n \left(a_{1j}\sum_{k=1}^n a_{jk}\right)\\
&=\sum_{j=1}^n a_{1j} \quad \text{Row sum }=1\\
&=1 \quad \text{Row sum}=1 \text{ again.}
\end{align*}
\]
The same logic could apply to any other rows and it seems like row sum is an invariant...

- welshfella

I'm not sure why 2^0,3^0 and 5^0 are included in Parth's list ??

- welshfella

@thomas5267 No I think you are perfectly correct about Q7.

- thomas5267

Part 2 Q4
\[
\begin{align*}
f'(x)&>f(x)\\
\frac{f'(x)}{f(x)}&>1 \,\forall x\, \text{ if }f(x)>0\\
\frac{d}{dx}\ln\left(f(x)\right)&>1\\
\ln\left(f(x)\right)&>x+c\\
f(x)&>e^{x+c}
\end{align*}
\]
We have a problem since \(c\) must be \(-\infty\) in order for \(f(0)>0\) to hold. If we replaced c with a function \(g(x)\) such that \(g(0)=-\infty\) the differentiability of \(e^{x+g(x)}\) would be quite questionable.
Furthermore, the linear approximation of \(f(x)\) around \(f(0)=0\) is \(f(x)\approx f'(0)x\). Using this linear approximation, \(f(0-\epsilon)\approx -f'(0)\epsilon<0\) by \(f'(0)>f(0)=0\). If the Taylor series is convergent around \(x=0\), then the higher order terms must contribute more than the first order term. I don't know whether this violates the Taylor theorem or not.

- thomas5267

Part 1 Q8
Take the sum on the diagonal.
\[
\begin{align*}
&\phantom{{}={}}1+(n+2)+(2n+3)+\cdots\\
&=\sum_{i=1}^n i +\sum_{j=0}^{n-1} jn\\
&=\frac{n(n+1)}{2}+n\times\frac{n^2}{2}\\
&=\frac{n^3+n^2+n}{2}
\end{align*}
\]
d) None of these

- thomas5267

Part 3 Q2 Part 1
Define \(D_{i,j}=\dfrac{f(\beta_i)-f(\alpha_j)}{\beta_i-\alpha_j}\). We sought to prove
\[\lim_{i\to \infty}\lim_{j\to \infty}D_{i,j}=\lim_{j\to \infty}\lim_{i\to \infty}D_{i,j}=\lim_{i\to \infty}D_{i,i}=f'(0)\]
In other words, the limiting operations commute.
Consider \(\displaystyle\lim_{j\to \infty} D_{i,j}\). We know that \(f(x)\) is continuous at \(x=0\) by the existence of \(f'(0)\). Therefore,
\[
\begin{align*}
\lim_{j\to \infty} D_{i,j}&=\frac{f(\beta_i)-f(0)}{\beta_i}\\
\lim_{i\to \infty}\lim_{j\to \infty} D_{i,j}&=\lim_{i\to \infty}\frac{f(\beta_i)-f(0)}{\beta_i}\\
\end{align*}
\]
By the convergence of \((\beta_n)>0\) to 0, we know that for all \(\epsilon>0\) there exist a \(N\) such that \(n>N\implies 0<\beta_n<\epsilon\). So we could write \(\displaystyle\lim_{\epsilon\to 0}\frac{f(\epsilon)-f(0)}{\epsilon}\), which is the definition of \(f'(0)\) and it is given that \(f'(0)\) exists.
Do the same thing to \(\displaystyle\lim_{j\to \infty}\lim_{i\to \infty}D_{i,j}\) to complete the proof.

- thomas5267

Oh crap.\[\lim_{i\to \infty}\lim_{j\to \infty}D_{i,j}=\lim_{j\to \infty}\lim_{i\to \infty}D_{i,j}\not\implies \lim_{i\to \infty}D_{i,i}\]

- thomas5267

We could try to fix this by spliting it up into cases. Combinations of \(f(x+\epsilon)\geq f(x), f(x+\epsilon)\leq f(x),f(x-\epsilon)\geq f(x), f(x-\epsilon)\leq f(x)\) and try to sqeeuze it.

- thomas5267

@ParthKohli What is your answer to the integer polynomial fitting question? For a real polynomial of degree n, n+1 points is enough to determine the whole polynomial. Since the polynomial is with positive integer coefficient, the polynomial is monotonically increasing in the postive numbers. I would argue n+2 points would be enough. n+1 points for fitting the polynomial and the last point for checking. If the polynomial is of degree n+1, then the checking will failing since the polynomial is monotonically incrsasing and adding a \(x^{n+1}\) term will only increase the value.
The algorithm would be to do it greedily. Take two random points at first. Check if it is the constant polynomial. If not, then it must be a polynomial of at least degree 1. Find a fit, check using a third point. If the third point matches, then the linear fit is correct. If not, fit a quadratic polynomial using the three points and check again. Repeat until you are done.
Is this the lower bound?

- ganeshie8

I think one question is sufficient..

- ganeshie8

or maybe atmost two questions will do

- thomas5267

I don't think one question is sufficient. You cannot distinguish any polynomial using using one point. It could be a constant function, a linear function, a quadratic function... You get my point.
Two questions? Maybe? I don't know and I have to sleep.

- anonymous

I will close this question now, feel free to look at it at closed questions section.

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