IITJEE Foundation Math IITJEE Main and Advanced Math and RMO/INMO of (TIFR and Homibhabha)

Category Archives: IITJEE Foundation Math IITJEE Main and Advanced Math and RMO/INMO of (TIFR and Homibhabha)

getting there…about education, parenting, kids

February 10, 2020 – 8:22 am

via getting there

By Nalin Pithwa | Comments (0)

Mathematician Dr Neena Gupta shines as the youngest Shanti Swarup Bhatnagar awardee

December 11, 2019 – 3:56 am

https://wordpress.com/post/madhavamathcompetition.com/4969

By Nalin Pithwa | Comments (0)

Set theory, functions, relations: part VI

October 8, 2019 – 6:18 am

What follows are some more practice questions on functions. The questions are not challenging but we can say that they do lead to conceptual clarity and present some standard set of questions on this topic (it behooves every beginner in calculus or IITJEE mains or RMO or pre RMO to try these set of questions):

Find the domain and range of the function: $f(x)=\frac{x-2}{3-x}$
If $f(x)=3x^{3}-5x^{2}+9$ , find $f(x-1)$ .
If $f(x)=x^{3}-\frac{1}{x^{3}}$ , show that $f(x)+f(\frac{1}{x})=0$
If $f(x)=\frac{x+1}{x-1}$ show that $f(f(x))=x$ .
Find the domain and range of the real valued function $f(x)=\frac{x^{2}+2x+1}{x^{2}-8x+12}$
Find the domain of the real valued function of a real variable: $f(x)=\frac{x-2}{2-x}$
Find the domain and range of the real valued function $f(x)=\frac{1}{1-x^{2}}$ .
A function $f: \Re \longrightarrow \Re$ is defined by $f(x)=\frac{3x}{5}+2$ where $x \in \Re$ . Does the inverse of f exist? If so, find it. Also, find the domain and range of the inverse.
A function is defined piece-wise as follows: $f(x)=3x+5$ for $- 4 \leq x \leq 0$ and $f(x)=5-3x$ for $0 < x \leq 4$ , find $f(f(\frac{5}{2}))$ ; the domain and range of f; and the value of x for which $f(x)=-4$
If $f: \Re \longrightarrow \Re$ and $g: \Re \longrightarrow \Re$ given by $f(x)=x-5$ and $g(x)=x^{2}-1$ , find (a) $f \circ g$ (b) $g \circ f$ (c) $f \circ f$ and (d) $g \circ g$
Find $f(g(x))$ and $g(f(x))$ if (a) $f(x)=3x-1$ and $g(x)=x^{2}+1$ (b) $f(x)=2x$ and $g(x)=4x+1$
If $f(x)=\frac{3x+4}{5x-7}$ and $g(x)=\frac{7x+4}{5x-3}$ prove that $f(g(x))=g(f(x))=x$
Find the domain and range of the following functions: (a) $f(x)=x^{2}$ (b) $f(x)=\sqrt{(x-1)(3-x)}$ (c) $f(x)=\frac{1}{\sqrt{x^{2}-1}}$ (d) $f(x)=\frac{x+3}{x-3}$ (e) $f(x)=\sqrt{9-x^{2}}$ (f) $f(x)=\sqrt{\frac{x-2}{3-x}}$
Find the range of each of the following functions: (a) $f(x)=3x-4$ , when $-1 \leq x <3$ (b) $f(x)=9-2x^{2}$ for $-5 \leq x \leq 3$ (c) $f(x)=x^{2}-6x+11$ for all $x \in \Re$ .
Solve the following: (a) if $f(x)=\frac{x^{3}+1}{x^{2}+1}$ , find $f(-3)$ , and $f(-1)$ . (b) If $f(x)=(x-1)(2x+1)$ , find $f(1)$ , $f(2)$ , $f(-3)$ . (c) If $f(x)=2x^{2}-3x-1$ , find $f(x+2)$ .
Which of the following relations are functions? Justify your answer. If it is a function, determine its range and domain. (a) $\{ (2,1),(4,2),(6,3),(8,4), (10,5), (12,6),(14,7)\}$ (b) $\{ (2,1),(3,1),(5,2)\}$ (c) $\{ (2,3),(3,2),(2,5),(5,2)\}$ (d) $\{ (0,0),(1,1),(1,-1),(4,2),(4,-2),(9,3),(9,-3),(16,4),(16,-4)\}$
Find a, if $f(x)=ax+5$ , and $f(1)=8$
If $f(x)=f(3x-1)$ for $f(x)=x^{2}-4x+11$ , find x.
If $f(x)=x^{2}-3x+4$ , then find the value of x satisfying $f(x)=f(2x+1)$ .
Let $A = \{ 1,2,3,4 \}$ and $Z$ be the set of integers. Define $f:A \longrightarrow Z$ by $f(x)=3x+7$ . Show that f is a function from A to Z. Also, find the range of f.
Find if the following functions are one-one or onto or bijective: (a) $f: \Re \longrightarrow \Re$ (b) $f: Z \longrightarrow Z$ given by $f(x)=x^{2}+4$ for all $x \in Z$ .
Find which of the following functions are surjective, injective or bijective or none of these : (a) $f: \Re \longrightarrow \Re$ as $f(x)=3x+7$ for all $x \in \Re$ (b) $f: \Re \longrightarrow \Re$ given as $f(x)=x^{2}$ for all $x \in \Re$ (c) $f = \{ (1,3),(2,6),(3,9),(4,12)\}$ defined from A to B where $A = \{ 1,2,3,4\}$ and $B = \{ 5,6,9,12,15\}$
Let f and g be two real valued functions defined by $f(x)=x+1$ and $g(x)=2x-9$ . Find $f+g$ and $f-g$ and $\frac{f}{g}$ .
Find $g \circ f$ and $f \circ g$ where (a) $f(x)=x-2$ and $g(x)=x^{2}+3x+1$ (b) $f(x)=\frac{1}{x}$ and $g(x)=\frac{x-2}{x+2}$ .
If $f(x)= \frac{2x+3}{3x-2}$ prove that $f \circ f$ is an identity function.
If $f(x)=\frac{3x+2}{4x-1}$ and $g(x)=\frac{x+2}{4x-3}$ , prove that $(g \circ f)(x)=(f \circ g)(x)=x$ .
If $f = \{ (2,4),(3,6),(4,8),(5,10),(6,12)\}$ and $g = \{ (4,13),(6,19),(8,25),(10,31),(12,37)\}$ find $g \circ f$ .
Show that $f:\Re \longrightarrow \Re$ given by $f(x)=3x-4$ is one-one and onto also. Find its inverse function also. Also, find the domain and range of the inverse function. Also find $f^{-1}(9)$ and $f^{-1}(-2)$
Let $f: \Re-\{ 2\} \longrightarrow \Re$ be defined by $f(x)=\frac{x^{2}-4}{x-2}$ and $g: \Re \longrightarrow \Re$ be defined by $g(x)=x+2$ . Find whether the two functions f and g are same, or not same. Justify your answers.

Regards,

Nalin Pithwa

By Nalin Pithwa | Comments (0)

Various proofs of important algebraic identity: a^{3}+b^{3}+c^{3}-3abc

September 19, 2019 – 10:51 am

We know the following factorization: $a^{3}+b^{3}+c^{3}-3abc= (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

Proof 1:

Let a, b, c be roots of a polynomial P(X) Then by fundamental theorem of algebra

$P(X)=(X-a)(X-b)(X-c) = X^{3}-(a+b+c)X^{2}+(ab+bc+ca)X-abc$ .

Now, once again basic algebra says that as a, b, c are roots/solutions of the above:

$P(a)=a^{3}-(a+b+c)a^{2}+(ab+bc+ca)a-abc=0$

$P(b)=b^{3}-(a+b+c)b^{2}+(ab+bc+ca)b-abc=0$

$P(c)=c^{3}-(a+b+c)c^{2}+(ab+bc+ca)c-abc$ =0$

Adding all the above:

$0= a^{3}+b^{3}+c^{3}-(a+b+c)(a^{2}+b^{2}+c^{2})+(ab+bc+ca)(a+b+c)-3abc$

So, we get $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

Also, the above formula can be written as $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(\frac{1}{2})((a-b)^{2}+(b-c)^{2}+(c-a)^{2})$

Proof 2:

Consider the following determinant D: $\left| \begin{array}{ccc} a & b & c\\c & a & b\\ b & c & a \end{array} \right|$

On adding all three columns to the first column: we know that the value of the determinant is unchanged: So we get the following:

$D = \left| \begin{array}{ccc} a+b+c & b & c \\a+b+c & a & b\\a+b+c & c & a \end{array} \right|$ . Note that columns 2 and 3 of the three by three determinant do not change.

On expanding the original determinant D, we get

$D = a(a^{2}-bc)-b(ac-b^{2})+c(c^{2}-ab)$

$D= a^{3}-abc-bac+b^{3}+c^{3}-cab$

$D= a^{3}+b^{3}+c^{3}-3abc$

Whereas we get from the other transformed but equal D:

$D =(a+b+c) \left| \begin{array}{ccc} 1 & b & c \\ 1 & a & b \\ 1 & c & a \end{array}\right|$

$D=(a+b+c)((a^{2}-bc)-b(a-b)+c(c-a))$

So, that we again get $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

Proof 3:

Now, let us consider $E=a^{2}+b^{2}+c^{2}-ab-bc-ca$ as a quadratic in a with b and c as parameters.

That is, $E= a^{2} -(b+c)a + b^{2}+c^{2}-bc$

Then, the discriminant is given by

$\triangle = (b+c)^{2}-4 \times 1 \times (b^{2}+c^{2}-bc)$ , which in turn equals:

$\triangle = (b^{2}+c^{2}+2bc)-4(b^{2}+c^{2}-bc) = -3b^{2}-3c^{2}+6bc=-3(b-c)^{2}$

$a_{1}, a_{2} = \frac{(b+c) \pm i\sqrt{3}(b-c)}{2}$

$a_{1}= b(\frac{1+i\sqrt{3}}{2})=c(\frac{1-i\sqrt{3}}{2}) = -b\omega - c\omega^{2}$

$a_{2}=b(\frac{1-i\sqrt{3}}{2})+c(\frac{1-i\sqrt{3}}{2})=-b\omega^{2}-c\omega$

Hence, the factorization of the above quadratic in a is given as:

$a^{2}+b^{2}+c^{2}-ab-bc-ca = (a+b\omega+c\omega^{2})(a+b\omega^{2}+c\omega)$

So, the other non-trivial factorization of the above famous algebraic identity is:

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a+b\omega+ c\omega^{2})(a+b\omega^{2}+c\omega)$

where $1, \omega, \omega^{2}$ are cube roots of unity.

Proof 4:

Factorize the expression $a^{3}+b^{3}+c^{3}-3abc by (a+b+c)$ .

Solution 4:

We can carry out the above polynomial division by considering the dividend to be a polynomial in a single variable, say a, (and assuming b and c are just parameters; so visualize them as arbitrary but fixed constants); further arrange the dividend in descending powers of a; so also arrange the divisor in descending powers of a (well, of course, it is just linear in a; and assume b and c are parameters also in dividend).

Proof 5:

Prove that the eliminant of

$ax+cy+bz=0$

$cx+by+az=0$

$bx+ay+cz=0$

is $a^{3}+b^{3}+c^{3}-3abc=0$

Proof 5:

By Cramer’s rule, the eliminant is given by determinant $\left| \begin{array}{ccc}a & c & b \\ c & b & a\\b & a & c \end{array}\right|=0$ .

On expansion using the first row:

$a(bc-a^{2})-c(c^{2}-ab)+b(ac-b^{2})=0$

$a^{3}+b^{3}+c^{3}-3abc=0$ upon multiplying both the sides of the above equation by (-1). Of course, we have only been able to generate the basic algebraic expression but we have done so by encountering a system of three linear equations in x, y, z. (we could append any of the above factorization methods to this further!! :-)))

Cheers,

Nalin Pithwa

By Nalin Pithwa | Comments (0)

Games, social behavior, chess, economics and maths

August 22, 2019 – 2:30 pm

The following is some trivia but in fact, not so trivia, in this age of data science, data analytics, social media platforms, on-line gaming etc…If you decide to ponder over deep…you will become a giant mathematician or applied mathematician or of course, a computer science wunderkind..

The following is “picked out as it is” from a famous biography, (which regular readers of my blog will now know, perhaps, is a favorite mathematical biography for me)…A Beautiful Mind by Sylvia Nasar, biography of mathematical genius, John Forbes Nash, Jr, Nobel Laureate (Economics) and Abel Laureate:

“It was the great Hungarian-born polymath John von Neumann who first recognized that social behaviour could be analyzed as games. Von Neumann’s 1928 article on parlor games was the first successful attempt to derive logical and mathematical rules about rivalries. Just as William Blake saw the universe in a grain of sand, great scientists have often looked for clues in vast and complex problems in the small, familiar phenomena of daily life. Isaac Newton reached insights about the heavens by juggling wooden balls. Albert Einstein contemplated a boat paddling upriver. Von Neumann pondered the game of poker.

A seemingly trivial and playful pursuit like poker, von Neumann argued, might hold the key to more serious human affairs for two reasons. Both poker and economic competition require a certain type of reasoning, namely the rational calculation of advantage and disadvantage based on some internally consistent system of values (“more is better than less’). And, in both, the outcome for any individual actor depends not only on his own actions, but on the independent actions of others.

More than a century earlier, the French economist Antoine-Augustin Cournot had pointed out that problems of economic cnoice were greatly simplified when either none or a large number of other agents were present. Alone on his island, Robinson Crusoe does not have to worry about whose actions might affect him. Neither do Adam Smith’s butchers and bakers. They live in a world with so many others that their actions, in effect, cancel each other out. But when there is more than one agent but not so many that their influence may be safely ignored, strategic behavior raises a seemingly insoluble problem:”I think that he thinks that I think that he thingks,” and so forth…

So play games but think math ! 🙂

Nalin Pithwa

By Nalin Pithwa | Comments (0)

You and your research ( You and your studies) : By Richard Hamming, AT and T, Bell Labs mathematician;

August 18, 2019 – 1:38 am

Although the title is grand (and quite aptly so)…the reality is that it can be applied to serious studies for IITJEE entrance, CMI entrance, highly competitive math olympiads, and also competitive coding contests…in fact, to various aspects of student life and various professional lifes…

Please read the whole article…apply it wholly or partially…modified or unmodified to your studies/research/profession…these are broad principles of success…

https://www.cs.virginia.edu/~robins/YouAndYourResearch.html

By Nalin Pithwa | Also posted in careers in mathematics, Cnennai Math Institute Entrance Exam, IITJEE Advanced Mathematics, IITJEE Foundation mathematics, IITJEE Mains, Information about IITJEE Examinations, INMO, ISI Kolkatta Entrance Exam, KVPY, mathematicians, memory power concentration retention, miscellaneous, motivational stuff, time management | Comments (0)

A magic trick!

June 26, 2019 – 12:02 am

You may have heard of a magic trick that goes like this. Take any number. Add 5. Double the result. Subtract 6. Divide by 2. Subtract 2. Now tell me your answer, and I will tell you what you have started with.

Pick a number and try it.

You can see what is going on if you let x be your original number, and follow the steps to make a formula $f(x)$ for the number you end up with.

Have fun !!!

Shared by Nalin Pithwa,

Ref: Calculus and Analytic Geometry by G B Thomas, 9th edition.

By Nalin Pithwa | Comments (0)

Women in Science 2019 UNESCO Awards to Ingrid Daubechies and Claire Voisin

February 22, 2019 – 1:40 pm

https://www.ams.org/news?news_id=4902

By Nalin Pithwa | Also posted in mathematicians, miscellaneous, motivational stuff | Comments (0)

Theory of Equations: III: IITJEE maths: algebra

January 9, 2019 – 4:24 am

may overlap a bit with previous lecture(s)…

By Nalin Pithwa | Also posted in algebra, IITJEE Mains, ISI Kolkatta Entrance Exam, KVPY | Tagged algebra for IITJEE, equations | Comments (0)

Applications of Derivatives: IITJEE Maths tutorial problem set: III

October 23, 2018 – 9:43 pm

Slightly difficult questions, I hope, but will certainly re-inforce core concepts:

Prove that the segment of the tangent to the curve $y=c/x$ which is contained between the co-ordinate axes, is bisected at the point of tangency.
Find all tangents to the curve $y=\cos{(x+y)}$ for $-\pi \leq x \leq \pi$ that are parallel to the line $x+2y=0$ .
Prove that the curves $y=f(x)$ , where $f(x)>0$ and $y=f(x)\sin(x)$ , where $f(x)$ is a differentiable function, have common tangents at common points.
Find the condition that the lines $x\cos{\alpha} + y \sin{\alpha}=p$ may touch the curve $(\frac{x}{a})^{m} + (\frac{y}{b})^{m}=1$ .
If $p_{1}$ and $p_{2}$ are lengths of the perpendiculars from origin on the tangent and normal to the curve $x^{2/3} + y^{2/3}=a^{2/3}$ respectively, prove that $4p_{1}^{2}+p_{2}^{2}=a^{2}$ .
Show that the curve $x=1-3t^{2}$ , $y=t-3t^{3}$ is symmetrical about x-axis and has no real points for $x>1$ . If the tangent at the point t is inclined at an angle $\psi$ to OX, prove that $3t = \tan{\psi} + \sec{\psi}$ . If the tangent at $P(-2,2)$ meets the curve again at Q, prove that the tangents at P and Q are at right angles.
A tangent at a point $P_{1}$ other than $(0,0)$ on the curve $y=x^{3}$ meets the curve again at $P_{2}$ . The tangent at $P_{2}$ meets the curve at $P_{3}$ and so on. Show that the abscissae of $P_{1}, P_{2}, \ldots, P_{n}$ form a GP. Also, find the ratio of area $\frac{\Delta P_{1}P_{2}P_{3}}{area \hspace{0.1in} P_{2}P_{3}P_{4}}$ .
Show that the square roots of two successive natural numbers greater than $N^{2}$ differ by less than $\frac{1}{2N}$ .
Show that the derivative of the function $f(x) = x \sin {(\frac{\pi}{x})}$ , when $x>0$ , and $f(x)=0$ when $x=0$ vanishes on an infinite set of points of the interval $(0,1)$ .
Prove that $\frac{x}{(1+x)} < \log {(1+x)} < x$ for $x>0$ .