mscroggs.co.uk
mscroggs.co.uk

subscribe

Puzzles

14 December

The function \(f(x)=ax+b\) (where \(a\) and \(b\) are real constants) satisfies
$$-x^3+2x^2+6x-9\leqslant f(x)\leqslant x^2-2x+3$$
whenever \(0\leqslant x\leqslant3\). What is \(f(200)\)?

Show answer

Find them all

Find all continuous positive functions, \(f\) on \([0,1]\) such that:
$$\int_0^1 f(x) dx=1\\ \mathrm{and }\int_0^1 xf(x) dx=\alpha\\ \mathrm{and }\int_0^1 x^2f(x) dx=\alpha^2$$

Show answer & extension

Odd and even outputs

Let \(g:\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}\) be a function.
This means that \(g\) takes two natural number inputs and gives one natural number output. For example if \(g\) is defined by \(g(n,m)=n+m\) then \(g(3,4)=7\) and \(g(10,2)=12\).
The function \(g(n,m)=n+m\) will give an even output if \(n\) and \(m\) are both odd or both even and an odd output if one is odd and the other is even. This could be summarised in the following table:
\(n\)
oddeven
\(m\)oddevenodd
eoddeven
Using only \(+\) and \(\times\), can you construct functions \(g(n,m)\) which give the following output tables:
\(n\)
oddeven
\(m\)oddoddodd
eoddodd
\(n\)
oddeven
\(m\)oddoddodd
eoddeven
\(n\)
oddeven
\(m\)oddoddodd
eevenodd
\(n\)
oddeven
\(m\)oddoddodd
eeveneven
\(n\)
oddeven
\(m\)oddoddeven
eoddodd
\(n\)
oddeven
\(m\)oddoddeven
eoddeven
\(n\)
oddeven
\(m\)oddoddeven
eevenodd
\(n\)
oddeven
\(m\)oddoddeven
eeveneven
\(n\)
oddeven
\(m\)oddevenodd
eoddodd
\(n\)
oddeven
\(m\)oddevenodd
eoddeven
\(n\)
oddeven
\(m\)oddevenodd
eevenodd
\(n\)
oddeven
\(m\)oddevenodd
eeveneven
\(n\)
oddeven
\(m\)oddeveneven
eoddodd
\(n\)
oddeven
\(m\)oddeveneven
eoddeven
\(n\)
oddeven
\(m\)oddeveneven
eevenodd
\(n\)
oddeven
\(m\)oddeveneven
eeveneven

Show answer & extension

Tags: functions

Bézier curve

A Bézier curve is created as follows:
1) A set of points \(P_0\), ..., \(P_n\) are chosen (in the example \(n=4\)).
2) A set of points \(Q_0\), ..., \(Q_{n-1}\) are defined by \(Q_i=t P_{i+1}+(1-t) P_i\) (shown in green).
3) A set of points \(R_0\), ..., \(R_{n-2}\) are defined by \(R_i=t Q_{i+1}+(1-t) Q_i\) (shown in blue).
.
.
.
\(n\)) After repeating the process \(n\) times, there will be one point. The Bézier curve is the path traced by this point at \(t\) varies between 0 and 1.

What is the Cartesian equation of the curve formed when:
$$P_0=\left(0,1\right)$$ $$P_1=\left(0,0\right)$$ $$P_2=\left(1,0\right)$$

Show answer & extension

Archive

Show me a random puzzle
 Most recent collections 

Advent calendar 2023

Advent calendar 2022

Advent calendar 2021

Advent calendar 2020


List of all puzzles

Tags

3d shapes sum to infinity menace percentages division tiling fractions functions games square roots axes digital products products symmetry geometric mean quadrilaterals people maths even numbers dates bases complex numbers circles wordplay digits time cube numbers scales perfect numbers spheres coordinates grids pentagons sums folding tube maps trigonometry rugby means indices floors ave square numbers volume triangles 2d shapes taxicab geometry gerrymandering factors probability angles proportion dodecagons range christmas tournaments sets cards pascal's triangle numbers number perimeter logic unit fractions remainders clocks crosswords averages shapes matrices cubics chocolate squares factorials chalkdust crossnumber parabolas multiplication integration money tangents chess palindromes addition coins graphs hexagons irreducible numbers elections speed integers polygons multiples sequences routes median area quadratics differentiation polynomials digital clocks crossnumbers decahedra planes consecutive integers lines odd numbers triangle numbers books the only crossnumber arrows binary star numbers cryptic clues partitions doubling probabilty consecutive numbers calculus expansions geometry determinants dice prime numbers algebra dominos balancing rectangles surds colouring albgebra sport crossnumber ellipses regular shapes cryptic crossnumbers combinatorics advent geometric means mean shape

Archive

Show me a random puzzle
▼ show ▼
© Matthew Scroggs 2012–2024