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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 \(f(200)\)?

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Find them all

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

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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

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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:
$$P_0=\left(0,1\right)$$ $$P_1=\left(0,0\right)$$ $$P_2=\left(1,0\right)$$

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