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## Supremum of function (TIFR 2014 problem 13)

Question:

Let $S$ be the set of all tuples $(x,y)$ with $x,y$ non-negative real numbers satisfying $x+y=2n$ ,for a fixed $n\in\mathbb{N}$. Then the supremum value of $x^2y^2(x^2+y^2)$ on the set $S$ is:

A. $3n^6$

B. $2n^6$

C. $4n^6$

D. $n^6$

Discussion:

Write the expression in terms of $x$ only by substituting $y=2n-x$.

Let $f(x)=x^2(2n-x)^2(x^2+(2n-x)^2)$. Here, $x\in[0,2n]$.

Note that for $x=0$ or $x=2n$ the function $f(x)=0$. Also, $f$ is positive everywhere else on the interval.

So we want to find $sup\{f(x)|x\in (0,2n)\}$. Note that it exists because the interval is compact and $f$ is continuous.

One can straightaway take derivative and compute, or one can do the following:

Take log. Note that now we are only working on the open interval $(0,2n)$.

$log(f(x))=2logx+2log(2n-x)+log(x^2+(2n-x)^2)$

Now take derivative.

$\frac{f'(x)}{f(x)}=\frac{2}{x}+\frac{-2}{2n-x}+\frac{2x-2(2n-x)}{x^2+(2n-x)^2}$

$=\frac{4n-4x}{x(2n-x)}+\frac{4n-4x}{x^2+(2n-x)^2}$

$=4(n-x)[\frac{1}{x(2n-x)}+\frac{1}{x^2+(2n-x)^2}]$.

Now, $f'(x)=0$ if and only if $4(n-x)[\frac{1}{x(2n-x)}+\frac{1}{x^2+(2n-x)^2}]=0$.

Note that $[\frac{1}{x(2n-x)}+\frac{1}{x^2+(2n-x)^2}]>0$ for all $x\in (0,2n)$.

Therefore, $f'(x)=0$ if and only if $x=n$. If now, $x$ is slightly bigger than $n$ then $\frac{f'(x)}{f(x)}<0$ and since $f(x)>0$ we have $f'(x)<0$ in that case. And if $x$ is slightly smaller than $n$ then $f'(x)>0$.

This proves that indeed the point $x=n$ is a point of maxima.

Therefore, the supremum value is $f(n)=2n^6$. So the correct answer is option B.

Categories

## Mapping properties (TIFR 2014 problem 12)

Question:

There exists a map $f:\mathbb{Z} \to \mathbb{Q}$ such that $f$

A. is bijective and increasing

B. is onto and decreasing

C. is bijective and satisfies $f(n)\ge 0$ if $n\le 0$

D. has uncountable image.

Discussion:

Option D is out of the question right away. Because the co-domain $\mathbb{Q}$ is countable, and the image set is a subset of the co-domain set, any subset of countable set is countable, so image set has to be countable.

Let’s assume $f$ is onto and decreasing. Then $…\ge f(-1) \ge f(0) \ge f(1) \ge f(2) \ge …$.

At this point, we don’t know for sure whether $f(0)=f(1)$ or not. But, for the map to be onto, we must have strict inequality somewhere in the above chain of inequalities.

Let’s say $f(a)>f(a+1)$. Then we ask what is the pre-image of the rational number $\frac{f(a)+f(a+1)}{2}$?

Let $f(n)=\frac{f(a)+f(a+1)}{2}$

Since, $f(n)=\frac{f(a)+f(a+1)}{2}>f(a)$ and $f$ is decreasing, $a>n$.

Also, since $f(n)=\frac{f(a)+f(a+1)}{2}<f(a+1)$ and $f$ is decreasing, $n>a+1$.

The two inequalities $a>n$ and $n>a+1$ together give a contradiction.

Therefore, option B is false.

What did we use here, only onto-ness and that f is decreasing. If instead our function $f$ was bijective and increasing then we will get a similar kind of contradiction:

Suppose $f$ is bijective and increasing. Then $f(0)<f(1)$ (One-to-one ness guarantees the strict inequality)

We have $\frac{f(0)+f(1)}{2}\in\mathbb{Q}$.

Therefore there exists $m\in\mathbb{Z}$ such that $f(m)= \frac{f(0)+f(1)}{2}$.

We then have $f(0)<f(m)<f(1)$ which means $0<m<1$, a contradiction because there is no natural number in between 0 and 1.

So option A is false.

We know that $\mathbb{Q}$ does have a bijection with $\mathbb{Z}$, in other words $\mathbb{Q}$ is countable.

Now, both the set $A=\{n\in \mathbb{Z}| n\le 0 \}$ and $B=\{q\in \mathbb{Q}| q\ge 0\}$ are countably infinite, therefore has a bijection between them. To see this, let $f_1:A\to \mathbb{Z}$ and $f_2:B\to \mathbb{Z}$ be bijections. Then $f_2^{-1}o f_1:A \to B$ is a bijection.

Similarly $C=\{n\in \mathbb{Z}| n> 0 \}$ has a bijection with $D=\{q\in \mathbb{Q}| q< 0\}$.

Now, we have $h:A\to B$ and $g:C\to D$ two bijections.

Then define $f(n)=h(n)$ for $n\in A$ and $f(n)=g(n)$ for $n\in C$ to get a bijection from $\mathbb{Z}$ to $\mathbb{Q}$. This is true because $A \cup B= \mathbb{Z}$ and $C \cup D= \mathbb{Q}$.

And this $f$ satisfies the condition of option C: $f$ is bijective and satisfies $f(n)\ge 0$ if $n\le 0$

Therefore, option C is true.

Categories

## Nilpotent matrix eigenvalues (TIFR 2014 problem 11)

Question:

Let $A$ be an $nxn$ matrix with real entries such that $A^k=0$ (0-matrix) for some $k\in\mathbb{N}$.

Then

A. A has to be the 0-matrix.

B. trace(A) could be non-zero.

C. A is diagonalizable.

D. 0 is the only eigenvalue of A

Discussion:

Let $v$ be an eigenvector of $A$ with eigenvalue $\lambda$.

Then $v \neq 0$ and $Av=\lambda v$.

Again, $A^2 v=A(Av)=A(\lambda v)=\lambda Av= (\lambda)^2v$.

We continue to apply A, applying it k times gives: $A^k v=(\lambda)^k v$.

By given information, the left hand side of the above equality is 0.

So $\lambda^k v=0$ and remember $v \neq 0$.

So $\lambda =0$.

Therefore $0$ is the only eigenvalue for $A$.

So D is true.

We analyse the question a little bit further, to check it satisfies no other options above.

We know $trace(A)=$ sum of eigenvalues of A= $\sum 0 =0$

So option B is false.

Take $A=\begin{bmatrix} 0 & 1 // 0 & 0 \end{bmatrix}$.

Then $A^2 =0$. But $A$ is not the zero matrix.

Also, if $A$ were diagonalizable then the corresponding diagonal matrix would be the zero matrix. Which would then imply that $A$ is the zero matrix, which in this case it is not. (See diagonalizable-nilpotent-matrix-tifr-2013-problem-8 ) So this disproves options A and C.

Categories

## Theory of Equation (TIFR 2014 problem 10)

Question:

Let $C \subset \mathbb{ZxZ}$ be the set of integer pairs $(a,b)$ for which the three complex roots $r_1,r_2,r_3$ of the polynomial $p(x)=x^3-2x^2+ax-b$ satisfy $r_1^3+r_2^3+r_3^3=0$. Then the cardinality of $C$ is

A. $\infty$

B. 0

C. 1

D. $1<|C|<\infty$

Discussion:

We have $r_1+r_2+r_3=-(-2)=2$

$r_1r_2+r_2r_3+r_3r_1=a$ and

$r_1r_2r_3=-(-b)=b$

Also, of the top of our head, we can think of one identity involving the quantities $r_1^3+r_2^3+r_3^3=0$ and the three mentioned just above.

Let’s apply that:

$r_1^3+r_2^3+r_3^3-3r_1r_2r_3=(r_1+r_2+r_3)(r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1))$ …$…(1)$

Also, note that $r_1^2+r_2^2+r_3^2=(r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1)$

So $r_1^2+r_2^2+r_3^2=(2)^2-2(a)=4-2a$.

So by equation $(1)$,

$0-3b=2(4-2a-a)$

or, $6a-3b=8$.

Now, we notice the strangest thing about the above equation: $3|(6a-3b)$ but $3$ does not divide $8$.

So, we get a contradiction.

Why did this contradiction occur? We didn’t start off by saying “assume that … something …”. Well, even though we pretend to not assume anything, we did assume that this equation has a solution for some $a,b$. So, the

ANSWER: $|C|=0$.