A trigonometric polynomial ( INMO 2020 Problem 2)

The problem

Suppose $P(x)$ is a polynomial with real coefficients satisfying the condition
$$
P(\cos \theta+\sin \theta)=P(\cos \theta-\sin \theta),
$$
for every real $\theta$. Prove that $P(x)$ can be expressed in the form
$$
P(x)=a_0+a_1\left(1-x^2\right)^2+a_2\left(1-x^2\right)^4+\cdots+a_n\left(1-x^2\right)^{2 n},
$$
for some real numbers $a_0, a_1, a_2, \ldots, a_n$ and nonnegative integer (n).

Hint 1

Using a very standard trigometric identity, we can easily convert the following ,
$$
\begin{aligned}
P(\cos \theta+\sin \theta) & =P(\cos \theta-\sin \theta
\Longrightarrow P\left(\sqrt{2} \sin \left(\frac{\pi}{4}+\theta\right)\right) & =P\left(\sqrt{2} \cos \left(\frac{\pi}{4}+\theta\right)\right) \
\Longrightarrow P(\sqrt{2} \sin x) & =P(\sqrt{2} \cos x)
\end{aligned}
$$
⟹ \(P(\sqrt{2} \sin x)=P(\sqrt{2} \cos x) \quad\)

Assuming,

$\left(\frac{\pi}{4}+\theta\right)=x$ for all reals $x$. So,

$P(-\sqrt{2} \sin (x))=P(\sqrt{2} \sin (-x))=P(\sqrt{2} \cos (-x))=P(\sqrt{2} \cos (x))=P(\sqrt{2} \sin (x))$ for all $x \in \mathbb{R}$. Since $P(x)=P(-x)$ holds for infinitely many $x$, it must hold for all $x$ (since $P(x)$ is a polynomial). so we get that, $P(x)$ is a even polynomial.

Hint 2

$P(\sqrt{2} \cos (x))=P(\sqrt{2} \sin (x))$ implies that
$$
P(t)=P\left(\sqrt{2} \sin \left(\cos ^{-1}(t / \sqrt{2})\right) \text { putting }, x=\cos ^{-1}(t / \sqrt{2})\right.
$$
for infinitely many $t \in[-\sqrt{2}, \sqrt{2}]$.
$$
\sqrt{2} \sin \left(\cos ^{-1}(t / \sqrt{2})\right)=\sqrt{2-t^2} \text { so we get, } P(x)=P\left(\sqrt{2-t^2}\right)
$$

Again as it is a polynomial function we can extend it all $\mathbb{R}$. And we get, $P(x)=P\left(\sqrt{2-x^2}\right)$ for all reals (x)

Hint 3

Since $P(x)$ is even, we can choose an even polynomial $Q(x)$ such that, $Q(x)=P(\sqrt{x+1}) \cdot P(\sqrt{1+x}$=$Q(x)=a_0+a_1 x^2+a_2 x^4+\cdots+a_n x^{2 n}$ now take, $\sqrt{1+x}=y$ and you get the polynomial of required form.

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Hidden triangular inequality (PRMO Problem 23, 2019)

Problem

Let ABCD be a convex cyclic quadrilateral . Suppose P is a point in the plane of the quadrilateral such that the sum of its distances from the vertices of ABCD is the least .If {PA,PB,PC,PD} = {3,4,6,8}.What is the maximum possible area of ABCD?

Topic

Geometry

Contest

Pre Regional Math Olympiad
2019

Reading
Mathematical circle .

Sequential Hints

The first step is very clear, you have to some how locate the point P .

claim : The intersection of the diagonal is the required point .
Proof : suppose that there is a point P' other than P (the intersection of diagonal) .
now try to apply triangular inequality in \(\triangle\)PBD & \(\triangle\)APC.
(do the rest yourself , and complete the proof)

so , the area of [ABCD]=\(\frac{1}{2}\)\(\sin\theta\)(PA*PB+PB*PC+PC*PD+PD*PA)
now, the area will maximum when \(\sin\theta\)=1
so , just put the value of PA, PB, PC,PD
You get the answer 55

Watch video

I.S.I Entrance-2013 problem 2

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]For x ≥ 0 define
f(x) =$latex \frac{1}{x+2cos x}$
.
Determine the set {y ∈ R : y = f(x), x ≥ 0}.

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I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2013. Subjective Problem no. 2.

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Medium 

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Problems In CALCULUS OF ONE VARIABLE

by I.A. Maron

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Start with hints

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Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]

This problem simply ask for the range of the function defined by f(x)=$latex \frac {1}{x+2cosx}$ compute the derivative of the function = $latex \frac {2sinx-1}{(x+2cosx)^2}$      

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]

 First extrema occurs at $latex x$= $latex \frac{\pi}{6}$ The first derivative is negetive in the interval [ 0, $latex \frac{\pi}{6}$] hence the function is decreasing in this interval f(0)=$latex \frac{1}{2}$ ; f($latex \frac{\pi}{6}$)=$latex \frac{1}{\frac{\pi}{6}+ {\sqrt{3}}}$    

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.4"]

 For x>$latex \frac{\pi}{6}$ the derivative becomes positive , and remain so upto x=$latex \frac{5\pi}{6}$ after which it becomes negative  thus we have minima at x= $latex \frac{\pi}{6}$ and maxima at x= $latex \frac{5\pi}{6}$ f($latex \frac{5\pi}{6}$)= $latex \frac{1}{\frac{5\pi}{6}+\sqrt{3}}$ note that as $latex x\rightarrow \infty$the denominator of the function increases 

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]

Hence we can conclude that    $latex f(x)\rightarrow0$ clearly x=$latex \frac{5\pi}{6}$ gives the global maxima  so , the range is (0,$latex \frac{1}{\frac{5\pi}{6}+\sqrt{3}}$]       

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Connected Program at Cheenta

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Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

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Similar Problem

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Extremal Principle : I.S.I Entrance 2013 problem 4

[et_pb_section fb_built="1" _builder_version="3.22.4"][et_pb_row _builder_version="3.22.4"][et_pb_column type="4_4" _builder_version="3.22.4"][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_padding="20px|20px|20px|20px"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]In a badminton singles tournament, each player played against all the others
exactly once and each game had a winner. After all the games, each player
listed the names of all the players she defeated as well as the names of all the
players defeated by the players defeated by her. For instance, if A defeats B
and B defeats C, then in the list of A both B and C are included. Prove that
at least one player listed the names of all other players.

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.22.4"][et_pb_column type="4_4" _builder_version="3.22.4"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.22.4" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]

I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2013. Subjective Problem no. 4.

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7.5 out of 10

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Problem Solving Strategies by Engel

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Start with hints

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]

 Do you know what is Well-ordering principle ? it says that Every nonempty set A of nonnegative integers has a minimal element and a maximal element , which need not to be unique . now some time this simple property help us to get very nice solution to a problem , can you some how apply this property .        

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]

Use the method of contradiction , first of all assume that there is no player which have the given property .  Now try to use the property of hint 1 .  

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.4"]

 If there is no such list , A's list has the maximum no. of players  Now , if  A does not have the certain property then there exist another another player B , who has won against A . Now B's list contain the name of A [ by the 1st condition ] and all the names of the players defeated by A [ by the 2nd condition]  

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]

Now , can you find out some contradiction ,  yes exactly ..... B's list contain more number of element than A So, A's list must have the certain property .            

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Connected Program at Cheenta

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Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

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Similar Problem

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Lattice point inside a triangle

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]Given a triangle ABC with three lattice vertices . it is known that no more lattice point lies on the edges . only one lattice point D is inside the triangle . prove that D is centroid of that triangle .    

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.22.4"][et_pb_column type="4_4" _builder_version="3.22.4"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.22.4" title_text_shadow_horizontal_length="0em" title_text_shadow_vertical_length="0em" title_text_shadow_blur_strength="0em" closed_title_text_shadow_horizontal_length="0em" closed_title_text_shadow_vertical_length="0em" closed_title_text_shadow_blur_strength="0em"]
Iran Maths olympiad 
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Start with hints

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.23.3"]

   Do you know Pick’s theorem ?  NO!  then read this post first and return to the problem again  https://cheenta.com/a-proof-from-my-book/  

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.23.3"]

 Join AD , BD , & CD  NOW , can you do it ?

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.23.3"]

To prove D is centroid it is suficient to prove that [ ABD] = [ BDC] = [ ADC ]   where [.] denotes the bounded area . but why ? { think yourslfe , use the fact that medians  devide the triagle in six equal part }

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]

Now applying pick's theorem we get , [ABD]= [ BDC] = [ ADC ] = 0.5

Because in each of these three  triangle , three are 3 lattice vertex and lattice point is inside the triangle . so  area = 0 + 3/2 - 1 = 0.5 

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Connected Program at Cheenta

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Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

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Combinatorics - I.S.I 2016 SUBJECTIVE PROBLEM - 1

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Introduction

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2" locked="off"]Suppose that in a sports tournament featuring n players, each pair
plays one game and there is always a winner and a loser (no draws).
Show that the players can be arranged in an order P1, P2, . . . , Pn such
that player Pi has beaten Pi+1 for all i = 1, 2, . . . , n.  

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2" locked="off"]Suppose that in a sports tournament featuring n players, each pair
plays one game and there is always a winner and a loser (no draws).
Show that the players can be arranged in an order P1, P2, . . . , Pn such
that player Pi has beaten Pi+1 for all i = 1, 2, . . . , n.  

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.22.4"]

I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2016. Subjective Problem no.1.

[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.22.4" open="off"]Combinatorics

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.23.3" open="off"]

7 out of 10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.23.3" open="off"][/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Start with hints

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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]

we define a sequence \(p_1\)>\(p_2\)>\(p_3\)...........>\(p_n\) . such that \(p_i>p_{i+1}\) implies \(p_i\) has beaten \(p_{i+1}\)  

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]

let assume that the statement is true for n=k  i.e \(p_1> p_2>p_3.........p_k\) now look into the case of \(p_{k+1}\)    

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.23.3"]

if \(p_{k+1}\) > \(p_1\) or \(p_k\)>\(p_{k+1}\) we are done (why?)[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"] if \(p_k\)<\(p_{k+1}\) or \(p_{k+1}\)<\(p_1\) then  we can find out at least one pair \(p_i\) >\(p_{i+1}\) such that \(p_i\)> \(p_{k+1}\) >\(p_{i+1}\) , we can just find it by shifting \(p_{k+1}\) term by term [ you can make some small experiment ]. other wise it contradict our initial assumption that \(p_k\)<\(p_{k+1}\) or \(p_{k+1}\)<\(p_1\) [ why?]   hence we are done 

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Connected Program at Cheenta

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Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

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Polynomials (Algebra) - I.S.I. 2019 : Problem #7

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"] Let $f$ be a polynomial with integer coefficients. Define$$a_1 = f(0)~,~a_2 = f(a_1) = f(f(0))~,$$ and $~a_n = f(a_{n-1})$ for $n \geqslant 3$.

If there exists a natural number $k \geqslant 3$ such that $a_k = 0$, then prove that either $a_1=0$ or $a_2=0$.  

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="off" _builder_version="3.22.4"]
I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2019. Subjective Problem no. 7.
[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.22.7" open="off"]Polynominals (Algebra)

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.7" open="off"]

8 out of 10

[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.3.4" hover_enabled="0" open="on"]

Polynomials - Edward Barbeau [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Start with hints

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Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]

Do you know this lemma , Lemma: If $p, q \in \mathbb{Z}$ and $p \neq q$, then $p - q \mid f(p) - f(q)$ . 

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]

To prove this, let $f(x) = a_nx^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \cdots + a_0$. Then$$f(p) - f(q) = a_n(p^n - q^n) + a_{n-1}(p^{n-1} - q^{n-1}) + a_{n-2}(p^{n-2} - q^{n-2}) + \cdots + (p - q).$$Each bracket is divisible by $p - q$, proving the statement.  

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.7"]

We use the fact that the sequence $a_1, a_2, a_3, \cdots$ consists of only integers.
We'll first prove that we cannot have three distinct integers $p$, $q$, and $r$ such that $f(p) = q$, $f(q) = r$, and $f(r) = p$ (In other words, the variables cannot come in a cycle of 3). Assume that there does exist such numbers. Then we should have $p - q \mid f(p) - f(q) = q - r$, which means $\mid p - q \mid \le \mid q - r \mid$ . Similarly we can get $\mid p - q \mid \le \mid q - r \mid \le \mid r - p\mid \le \mid p - q \mid$ , which implies equality. Ultimately, it leads to two equal variables, contradiction. In a similar manner we can prove that these variables cannot come in cycles of more than 3.

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.7"]

Therefore, we conclude that the variables of $f$ can only come in cycles of most two. We realize that since $a_{k+1} = f(0) = a_1$, we have a cycle $a_1, a_2, a_3, \cdots, a_k$. Since the minimal cycle has length at most 2, one of $a_1$ or $a_2$ must be equal to 0, and we are done.

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Connected Program at Cheenta

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Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

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Similar Problem

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Number Theory - Working backward - C.M.I UG -2019

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]  If there exists a calculator with 12 buttons, 10 being the buttons for the digits and A and B being two buttons being processes where if n is displayed on the calculator if A is pressed it increases the displayed number by 1 and if B is pressed it multiplies n by 2 hence 2n. Hence find the minimum number of moves to get 260 from 1  

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Start with hints

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C.M.I (Chennai mathematical institute ) U.G- 2019 entrance  

[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.3.4" open="off"]

General concepts + Number theory

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.7" open="off"]4 out of 10

[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" open="off" _builder_version="4.3.4"]

Challenges and Thrills of Pre-College Mathematics by University Press[/et_pb_accordion_item][/et_pb_accordion][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.7" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]

DO you know how to start working backward 

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]

 working backward means that when you press A is makes -1 from the result  and pressing B you can divide 2  

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.4"]

strategies are like this , divide the no. as many step you can , when the result it not divisible by 2 just -1 

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.7"]

\(260\rightarrow130\rightarrow65\rightarrow64\rightarrow32\rightarrow16\rightarrow8\rightarrow4\rightarrow2\rightarrow1\)

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Connected Program at Cheenta

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Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

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Similar Problem

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Real numbers problem - C.M.I - U.G - 2019

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"] .Find all real numbers x for which
$\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$  

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C.M.I (Chennai mathematical institute ) U.G-2019

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Real Numbers

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.4" open="off"]8 out of 10

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Excursion in Mathematics by Bhaskaracharya Pratisthana[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]

Start with hints

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It is of the the form of  $\frac{a^3+b^3}{a^2b+b^2a}$ . Do you observe ?  where a=\(2^x\) b=\(3^x\)

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\(\frac{a^3+b^3+a^2b+b^2a}{a^2b+b^2a}\)=\(\frac{13}{6}\) \(\Rightarrow\frac{a^2+b^2}{ab}=\frac{13}{6}\) \(\frac{a}{b}+\frac{b}{a}=\frac{13}{6}\) let x=a/b then it is a quadratic equation    

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\(6x^2-13x+6=0\) on solving we get x=3/2 or 2/3 now replace x by a/b

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so again  putting value of a and b we get final ans is +1 and -1 

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Connected Program at Cheenta

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Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

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Similar Problem

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Calculus Problem - I.S.I 2019 Subjective Problem -4

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Understand the problem

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Let $f:\mathbb{R}\to\mathbb{R}$ be a twice differentiable function such that$$\frac{1}{2y}\int_{x-y}^{x+y}f(t)\, dt=f(x)\qquad\forall~x\in\mathbb{R}~\&~y>0$$Show that there exist $a,b\in\mathbb{R}$ such that $f(x)=ax+b$ for all $x\in\mathbb{R}$.
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I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2019. Subjective Problem no. 4
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Calculus

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.7" open="off"]8.5 out of 10

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Problems In CALCULUS OF ONE VARIABLE-i.a maron

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Start with hints

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Do you really need a hint? Try it first!

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Do you know Taylor series expansion good then see..... \(f(x+h) = f(x) + f'(x)h+f''(x)\frac{h^2}{2} + f'''(x)\frac{h^3}{3!}+\cdots\) \(f(x-h) = f(x) - f'(x)h+f''(x)\frac{h^2}{2} - f'''(x)\frac{h^3}{3!}+\cdots\)      

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adding previous expression we get  \(\frac{f(x+h) - 2f(x) + f(x-h)}{h^2} = f''(x) + 2\frac{f''''(x)}{4!}h^2+\cdots\)  taking the limit of the above equation as h goes to zero gives  \(\Rightarrow f''(x) = \lim_{h\to0} \frac{f(x+h) - 2f(x) + f(x-h)}{h^2} \,.\)  

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. Since $f$ is continuous, there exists $F$ such that $F' = f.$ The given identity becomes  $$F(x+y)-F(x-y) = 2yf(x).$$ Fix $x \in \mathbb{R}$ and differentiate the above identity with respect to $y$ and get $$f(x+y)+f(x-y)=2f(x).$$

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       now we already get ,  If $f$ is twice differentiable at $x$ then $$\lim_{h\to 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x).$$ by rearranging our given problem we get  that $f''(x)=0$ Therefore , $f(x)=ax+b$ for some $a,b\in\mathbb{R}.$      

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Connected Program at Cheenta

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Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

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