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Click here to win prizes by solving the mscroggs.co.uk puzzle Advent calendar.

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Puzzles

10 December

For all values of \(x\), the function \(f(x)=ax+b\) satisfies
$$8x-8-x^2\leqslant f(x)\leqslant x^2.$$
What is \(f(65)\)?
Edit: The left-hand quadratic originally said \(8-8x-x^2\). This was a typo and has now been corrected.

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6 December

\(p(x)\) is a quadratic with real coefficients. For all real numbers \(x\),
$$x^2+4x+14\leq p(x)\leq 2x^2+8x+18$$
\(p(2)=34\). What is \(p(6)\)?

Two tangents

Source: Reddit
Find a line which is tangent to the curve \(y=x^4-4x^3\) at 2 points.

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Between quadratics

Source: Luciano Rila (@DrTrapezio)
\(p(x)\) is a quadratic polynomial with real coefficients. For all real numbers \(x\),
$$x^2-2x+2\leq p(x)\leq 2x^2-4x+3$$
\(p(11)=181\). Find \(p(16)\).

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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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Parabola

On a graph of \(y=x^2\), two lines are drawn at \(x=a\) and \(x=-b\) (for \(a,b>0\). The points where these lines intersect the parabola are connected.
What is the y-coordinate of the point where this line intersects the y-axis?

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Two lines

Let A and B be two straight lines such that the gradient of A is the y-intercept of B and the y-intercept of A is the gradient of B (the gradient and y-intercept of A are not the same). What are the co-ordinates of the point where the lines meet?

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Sunday Afternoon Maths LXVII

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