Problem: Let the sequence \( { a_n} _{n \ge 1 } \) be defined by \(a_n = \tan n \theta \) where \( \tan \theta = 2 \). Show that for all n \( a_n \) is a rational number which can be written with an odd denominator. Discussion: This is simple induction. The claim […]
Here, you will find all the questions of ISI Entrance Paper 2017 from Indian Statistical Institute's B.Stat Entrance. You will also get the solutions soon of all the previous year problems. Problem 1 : Let the sequence \( \{ a_n\} _{n \ge 1 } \) be defined by $$ a_n = \tan n \theta $$ […]
Here are the problems and their corresponding solutions from BStat Hons Objective Admission Test 2005. Try it yourself and then read the solutions.
Here, you will find all the questions of ISI Entrance Paper 2016 from Indian Statistical Institute's B.Stat Entrance. You will also get the solutions soon of all the previous year problems. Problem 1: In a sports tournament of $n$ players, each pair of players plays exactly one match against each other. There are no draws. […]
Try this problem from TOMATO Objective 288, useful for ISI BStat, BMath Entrance Exam based on finding big remainder in a small way. Problem: Tomato objective 288 The remainder R(x) obtained by dividing the polynomial [latex]x^{100}[/latex] by the polynomial [latex]x^2-3x+2[/latex] is (A) [latex]2^{100}-1[/latex] (B) [latex](2^{100}-1)x-(2^{99}-1)[/latex] (C) [latex]2^{100}x-3(2^{100})[/latex] (D) [latex](2^{100}-1)x+(2^{99}-1)[/latex] SOLUTION: (B) The the divisor is […]
Problem: If the roots of the equation ${(x-a)(x-b)}$+${(x-b)(x-c)}$+${(x-c)(x-a)}$=$0$, (where a,b,c are real numbers) are equal , then (A) $b^2-4ac=0$ (B) $a=b=c$ (C) a+b+c=0 (D) none of foregoing statements is correct Answer: $(B)$ ${(x-a)(x-b)}$+${(x-b)(x-c)}$+${(x-c)(x-a)}$=$0$ => $x^2-{(a+b)}x$+$ab+x^2-{(b+c)}x$+$bc+x^2-{(c+a)}x+ca$=$0$ => $3x^2-2{(a+b+c)}x$+$(ab+bc+ca)$=$0$ discriminant, of the equation is => $4{(a+b+c)^2}$-$4.3{(ab+bc+ca)}$=$0$ => $a^2+b^2+c^2+2(ab+bc+ca)$-$3(ab+bc+ca)$=$0$ => $a^2+b^2+c^2$-$(ab+bc+ca)$=$0$ => $a=b=c$ So, option (B) is correct.
Try this beautiful problem from TOMATO Objective no. 258 based on Real Roots of a Cubic Polynomial. Problem: Real Roots of a Cubic Polynomial Let a,b,c be distinct real numbers. Then the number of real solution of [latex](x-a)^3+(x-b)^3+(x-c)^3=0[/latex] is (A) 1 (B) 2 (C) 3 (D) depends on a,b,c Solution: Ans: (A) Let [latex]f(x)=(x-a)^3+(x-b)^3+(x-c)^3[/latex] [latex]=> f'(x)=3(x-a)^2+3(x-b)^2+3(x-c)^2=0[/latex] [latex]=> […]
Try this beautiful problem from TOMATO Objective no. 257 based on Roots of a Quintic Polynomial. Problem: Roots of a Quintic Polynomial The number of real roots of [latex] x^5+2x^3+x^2+2=0[/latex] is (A) 0 (B) 3 (C) 5 (D) 1 Solution: Answer: (D) [latex] x^5+2x^3+x^2+2=0[/latex] [latex] \implies x^3(x^2+2)+(x^2+2)=0[/latex] [latex] \implies (x^3+1)(x^2+2)=0[/latex] [latex] \implies (x+1)\bold{\underline{(x^2-x+1)(x^2+2)}}=0[/latex] The expression in underline doesn't have any […]
Problem: The number of terms in the expression of $latex [(a+3b)^2 (a-3b)^2]^2 $ A) 4; B) 5; C) 6; D) 7; Solution: $latex [(a+3b)^2 (a-3b)^2]^2 $ $latex = [\{(a+3b)(a-3b)\}^2]^2 $ $latex = \{ (a^2 -9b^2)^2\}^2 = (a^2 - 9b^2)^4 $ By Binomial Theorem, the given expression contains 5 terms (since $latex (x +y)^n $ has […]
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on combinatorics in Tournament.
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Algebraic value.
Try this beautiful problem from Probability based on dice from AMC-10A, 2011. You may use sequential hints to solve the problem
Try this beautiful problem from Geometry: Area of Region in a Circle from AMC-10A, 2011, Problem -18. You may use sequential hints to solve the problem.
Try this beautiful problem from Algebra based smallest positive value from PRMO 2019. You may use sequential hints to solve the problem.
Try this beautiful problem from combinatorics based on Regular Polygon from PRMO 2019. You may use sequential hints to solve the problem.
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Positive solution.
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Right Rectangular Prism.
Try this beautiful problem from the Pre-RMO, 2019 based on Greatest Integer. You may use sequential hints to solve the problem.
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1996 based on Parallelogram Problem.
Author: Kazi Abu Rousan Where are the zeros of zeta of s? G.F.B. Riemann has made a good guess; They're all on the critical line, saith he, And their density's one over 2 p log t. Source https://www.physicsforums.com/threads/a-poem-on-the-zeta-function.16280/ If you are a person who loves to read maths related stuff then sure you have came […]
Problem If $\sum_{n=1}^{\infty} \frac{1}{n^2} =\frac{{\pi}^2}{6}$ then $\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2}$ is equal to (A) $\frac{{\pi}^2}{24}$ (B) $\frac{{\pi}^2}{8}$ (C) $\frac{{\pi}^2}{6}$ (D) $\frac{{\pi}^2}{3}$ Hint Try to write the summation as sum of square of reciprocal of odd numbers and even numbers and take the advantage of the infinite sum Solution $\sum_{n=1}^{\infty} \frac{1}{n^2} =\frac{{\pi}^2}{6}$ $\Rightarrow \sum_{n=1}^{\infty} \frac{1}{(2n)^2} + \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2}= […]
Probability theory is nothing but common sense reduced to calculation. Pierre-Simon Laplace Today we will be discussing a problem from the second chapter of A First Course in Probability(Eighth Edition) by Sheldon Ross. Let's see what the problem says: Describing the Problem The problem(prob-48) says: Given 20 people, what is the probability that among the […]
PROBLEM Let $p$ be an odd prime.Then the number of positive integers less than $2p$ and relatively prime to $2p$ is: (A)$p-2$ (B) $\frac{p+1}{2} $(C) $p-1$(D)$p+1$ SOLUTION This is a number theoretic problem .We can solve this problem in 2 different methods. Let us see them both one by one Method -1 Let us look […]
There should be no such thing as boring mathematics. Edsger W. Dijkstra In one of our previous post, we have discussed on Mandelbrot Set. That set is one of the most beautiful piece of art and mystery. At the end of that post, I have said that we can calculate the value of $\pi $ […]
Author: Kazi Abu Rousan The pure mathematician, like the musician, is a free creator of his world of ordered beauty. Bertrand Russell Today we will be discussing one of the most fascinating idea of number theory, which is very simple to understand but very complex to get into. Today we will see how to find […]
Problem Let $f$:$\mathbb{R} \rightarrow \mathbb{R}$ be a continous function such that for all$x \in \mathbb{R}$ and all $t\geq 0$ f(x)=f(ktx) where $k>1$ is a fixed constant Hint Case-1 choose any 2 arbitary nos $x,y$ using the functional relationship prove that $f(x)=f(y)$ Case-2 when $x,y$ are of opposite signs then show that $$f(x)=f(\frac{x}{2})=f(\frac{x}{4})\dots$$ use continuity to […]
PROBLEM Let $f (0,\infty)\rightarrow \mathbb{R}$ be a continous function such that for all $x \in (0,\infty)$ $f(x)=f(3x)$ Define $g(x)= \int_{x}^{3x} \frac{f(t)}{t}dt$ for $x \in (0,\infty)$ is a constant function HINT Use leibniz rule for differentiation under integral sign SOLUTION using leibniz rule for differentiation under integral sign we get $g'(x)=f(3x)-f(x)$ $\Rightarrow g'(x)=0$ [ Because f(3x)=f(x)] […]
PROBLEM Show that there are exactly $2$ numbers $a$ in the set $\{1,2,3\dots9400\}$ such that $a^2-a$ is divisible by $10000$ HINT Use Modular arithmetic and concepts of coprime numbers SOLUTION we know $10000=2^4*5^4$ In order for $10000$ to divide $a^2-a$ both $2^4$ and $5^4$ must divide $ a^2-a $ Write $a^2-a=a(a-1)$ Note that $a$ and […]
Author: Kazi Abu Rousan $\pi$ is not just a collection of random digits. $\pi$ is a journey; an experience; unless you try to see the natural poetry that exists in $\pi$, you will find it very difficult to learn. Today we will see a python code to find the value of $\pi $ up to […]