Puzzles

At \(x=a\), \(y=a^4-4a^2\) and \(\frac{dy}{dx}=4a^3-12a^2\). Therefore the equation of the tangent at \(x=a\) will be \(y=(4a^3-12a^2)x+8a^3-3a^4\).

Taking this away from \(y=x^4-4x^3\) gives \(y=x^4-4x^3-(4a^3-12a^2)x-8a^3+3a^4\). We can now look at where this curve is tangent to \(y=0\) and look for a value of \(a\) that makes it tangent at two points.

If this curve is tangent to the \(x\)-axis at \(x=b\), then it will have a repeated root at \(x=b\). We know it is tangent at \(x=a\), so dividing \(x^4-4x^3-(4a^3-12a^2)x-8a^3+3a^4\) by \((x-a)\) twice gives \(x^2+(2a-4)x+3a^2-8a\). We want this to have a repeated root, hence the discriminant, \((2a-4)^2-4(3a^2-8a)\), must be 0.

Solving this gives \(a=1\pm\sqrt3\).

Therefore the equation of the line is \(y=-8x-4\).

POCO is 4595 and MUCHO is 68925.

The question could be written as \(POCO\times 15=MUCHO\).

For which values of \(n\) are the letters uniquely defined by \(POCO\times n = MUCHO\)?

implies that

This means that

Given that

what is the value of

?

For which values of \(k\) is the answer an integer?

So:

Therefore \(a+b+c+d=1+2+4+66=73\).

Which numbers could 2009 be replaced with so that the problem still has a unique solution?

One of the terms in the series will be \((x-x)\), so the product will be zero.

\(y=x^{x^{x^{x^{...}}}}\) so \(y=x^y=e^{y\ln{x}}\).

By the chain and product rules, \(\frac{dy}{dx}=e^{y\ln{x}}(\frac{dy}{dx}\ln{x}+\frac{y}{x})\).

Rearranging, we get \(\frac{dy}{dx}=\frac{ye^{y\ln{x}}}{x(1-e^{y\ln{x}}\ln{x})}\).

This simplifies to \(\frac{dy}{dx}=\frac{x^{x^{x^{x^{...}}}}x^{x^{x^{x^{...}}}}}{x(1-x^{x^{x^{x^{...}}}}\ln{x})}\).

What would the graph of \(y=x^{x^{x^{x^{...}}}}\) look like?

Therefore \(x=\sqrt{2}\).

If \(x^{x^{x^{x^{...}}}}\) is equal to 4, what is the value of \(x\)?