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Papers (30)
2025
pgmath 18
2024
pgmath 4 ugmath 20
2023
pgmath 12 ugmath 16
2022
pgmath 18 ugmath_22may 16 ugmath_23may 16
2021
pgmath 11 ugmath 15
2020
pgmath 18 ugmath 16
2019
pgmath 20 ugmath 15
2018
pgmath 4 ugmath 6
2017
ugmath 15
2016
pgmath 10 ugmath 16
2015
pgmath 5 ugmath 13
2014
ugmath 9
2013
pgmath 11 ugmath 12
2012
pgmath 24 ugmath 14
2011
pgmath 15 ugmath 12
2010
pgmath 4 ugmath 19
2012 ugmath

14 maths questions

QA1 6 marks Standard trigonometric equations Count zeros or intersection points involving trigonometric curves View
Find the number of real solutions to the equation $x = 99 \sin ( \pi x )$.
QA2 6 marks Proof Existence Proof View
A differentiable function $f : \mathbb { R } \rightarrow \mathbb { R }$ satisfies $f ( 1 ) = 2 , f ( 2 ) = 3$ and $f ( 3 ) = 1$. Show that $f ^ { \prime } ( x ) = 0$ for some $x$.
QA3 6 marks Proof Proof Involving Combinatorial or Number-Theoretic Structure View
Show that $\frac { \ln ( 12 ) } { \ln ( 18 ) }$ is irrational.
QA4 6 marks Sequences and series, recurrence and convergence Convergence proof and limit determination View
Show that $$\lim _ { x \rightarrow \infty } \frac { x ^ { 100 } \ln ( x ) } { e ^ { x } \tan ^ { - 1 } \left( \frac { \pi } { 3 } + \sin x \right) } = 0$$
QA5 6 marks Discrete Probability Distributions Probability Computation for Compound or Multi-Stage Random Experiments View
(a) $n$ identical chocolates are to be distributed among the $k$ students in Tinku's class. Find the probability that Tinku gets at least one chocolate, assuming that the $n$ chocolates are handed out one by one in $n$ independent steps. At each step, one chocolate is given to a randomly chosen student, with each student having equal chance to receive it.
(b) Solve the same problem assuming instead that all distributions are equally likely. You are given that the number of such distributions is $\binom { n + k - 1 } { k - 1 }$. (Here all chocolates are considered interchangeable but students are considered different.)
QB1 10 marks Roots of polynomials Divisibility and minimal polynomial arguments View
a) Find a polynomial $p ( x )$ with real coefficients such that $p ( \sqrt { 2 } + i ) = 0$. b) Find a polynomial $q ( x )$ with rational coefficients and having the smallest possible degree such that $q ( \sqrt { 2 } + i ) = 0$. Show that any other polynomial with rational coefficients and having $\sqrt { 2 } + i$ as a root has $q ( x )$ as a factor.
QB2 10 marks Straight Lines & Coordinate Geometry Area Computation in Coordinate Geometry View
a) Let $\mathrm { E } , \mathrm { F } , \mathrm { G }$ and H respectively be the midpoints of the sides $\mathrm { AB } , \mathrm { BC } , \mathrm { CD }$ and DA of a convex quadrilateral ABCD. Show that EFGH is a parallelogram whose area is half that of ABCD. b) Let $\mathrm { E } = ( 0,0 ) , \mathrm { F } = ( 0 , - 1 ) , \mathrm { G } = ( 1 , - 1 ) , \mathrm { H } = ( 1,0 )$. Find all points $\mathrm { A } = ( p , q )$ in the first quadrant such that $\mathrm { E } , \mathrm { F } , \mathrm { G }$ and H respectively are the midpoints of the sides $\mathrm { AB } , \mathrm { BC } , \mathrm { CD }$ and DA of a convex quadrilateral ABCD.
QB3 10 marks Proof Proof Involving Combinatorial or Number-Theoretic Structure View
a) We want to choose subsets $A _ { 1 } , A _ { 2 } , \ldots , A _ { k }$ of $\{ 1,2 , \ldots , n \}$ such that any two of the chosen subsets have nonempty intersection. Show that the size $k$ of any such collection of subsets is at most $2 ^ { n - 1 }$. b) For $n > 2$ show that we can always find a collection of $2 ^ { n - 1 }$ subsets $A _ { 1 } , A _ { 2 } , \ldots$ of $\{ 1,2 , \ldots , n \}$ such that any two of the $A _ { i }$ intersect, but the intersection of all $A _ { i }$ is empty.
QB4 10 marks Numerical integration Riemann Sum Computation from a Given Formula View
Define $$x = \sum _ { i = 1 } ^ { 10 } \frac { 1 } { 10 \sqrt { 3 } } \frac { 1 } { 1 + \left( \frac { i } { 10 \sqrt { 3 } } \right) ^ { 2 } } \quad \text { and } \quad y = \sum _ { i = 0 } ^ { 9 } \frac { 1 } { 10 \sqrt { 3 } } \frac { 1 } { 1 + \left( \frac { i } { 10 \sqrt { 3 } } \right) ^ { 2 } } .$$ Show that a) $x < \frac { \pi } { 6 } < y$ and b) $\frac { x + y } { 2 } < \frac { \pi } { 6 }$. (Hint: Relate these sums to an integral.)
QB5 10 marks Addition & Double Angle Formulae Multi-Step Composite Problem Using Identities View
Using the steps below, find the value of $x ^ { 2012 } + x ^ { - 2012 }$, where $x + x ^ { - 1 } = \frac { \sqrt { 5 } + 1 } { 2 }$. a) For any real $r$, show that $\left| r + r ^ { - 1 } \right| \geq 2$. What does this tell you about the given $x$? b) Show that $\cos \left( \frac { \pi } { 5 } \right) = \frac { \sqrt { 5 } + 1 } { 4 }$, e.g. compare $\sin \left( \frac { 2 \pi } { 5 } \right)$ and $\sin \left( \frac { 3 \pi } { 5 } \right)$. c) Combine conclusions of parts a and b to express $x$ and therefore the desired quantity in a suitable form.
QB6 10 marks Proof Existence Proof View
For $n > 1$, a configuration consists of $2n$ distinct points in a plane, $n$ of them red, the remaining $n$ blue, with no three points collinear. A pairing consists of $n$ line segments, each with one blue and one red endpoint, such that each of the given $2n$ points is an endpoint of exactly one segment. Prove the following. a) For any configuration, there is a pairing in which no two of the $n$ segments intersect. (Hint: consider total length of segments.) b) Given $n$ red points (no three collinear), we can place $n$ blue points such that any pairing in the resulting configuration will have two segments that do not intersect. (Hint: First consider the case $n = 2$.)
QB7 10 marks Number Theory Congruence Reasoning and Parity Arguments View
A sequence of integers $c _ { n }$ starts with $c _ { 0 } = 0$ and satisfies $c _ { n + 2 } = a c _ { n + 1 } + b c _ { n }$ for $n \geq 0$, where $a$ and $b$ are integers. For any positive integer $k$ with $\operatorname { gcd } ( k , b ) = 1$, show that $c _ { n }$ is divisible by $k$ for infinitely many $n$.
QB8 10 marks Number Theory Prime Number Properties and Identification View
Let $f ( x )$ be a polynomial with integer coefficients such that for each nonnegative integer $n , f ( n ) = \mathrm { a }$ perfect power of a prime number, i.e., of the form $p ^ { k }$, where $p$ is prime and $k$ a positive integer. ($p$ and $k$ can vary with $n$.) Show that $f$ must be a constant polynomial using the following steps or otherwise. a) If such a polynomial $f ( x )$ exists, then there is a polynomial $g ( x )$ with integer coefficients such that for each nonnegative integer $n , g ( n ) =$ a perfect power of a fixed prime number. b) Show that a polynomial $g ( x )$ as in part a must be constant.
QB9 10 marks Proof Proof That a Map Has a Specific Property View
Let $N$ be the set of non-negative integers. Suppose $f : N \rightarrow N$ is a function such that $f ( f ( f ( n ) ) ) < f ( n + 1 )$ for every $n \in N$. Prove that $f ( n ) = n$ for all $n$ using the following steps or otherwise. a) If $f ( n ) = 0$, then $n = 0$. b) If $f ( x ) < n$, then $x < n$. (Start by considering $n = 1$.) c) $f ( n ) < f ( n + 1 )$ and $n < f ( n + 1 )$ for all $n$. d) $f ( n ) = n$ for all $n$.