Alan

You've lost a sign in there Anonymous!

The minimum can only be -∞ and the maximum +∞

(I assumed k was positive in my original answer, but positive or negative the max/min are still +∞/-∞)

Percent change is given by 100*(new number - original number)/original number

$${\frac{{\mathtt{100}}{\mathtt{\,\times\,}}\left({\mathtt{\,-\,}}{\mathtt{1\,739\,000}}{\mathtt{\,-\,}}{\mathtt{2\,018\,000}}\right)}{{\mathtt{2\,018\,000}}}} = -{\mathtt{186.174\: \!430\: \!128\: \!840\: \!436\: \!1}}$$

or ≈ -186%   (the negative sign indicates a reduction)

When x = 0 the function goes to +∞ (the maximum).  When x tends to -1 from x>-1 the function goes to -∞ (the minimum). (When x tends to -1 from x<-1 the function goes to +∞).

Here's a way of generating more sets of values:

Choose a Pythagorean triple, say 3, 4, 5.  Let a = 3m, b = 4m, c = 3n, d = 4n, and e = 5√(m² + n²) 

a² + b² + c² + d² = (3² + 4²)m² + (3² + 4²)n² = (3² + 4²)(m² + n²) = 5²(m² + n²) = e²

Now choose m and n to be part of another Pythagorean triple, say m = 5 and n = 12, and we have e as an integer (=5*13).

So a = 3*5 = 15;  b = 4*5 = 20;  c = 3*12 = 36;  d = 4*12 = 48; e = 5*13 = 65;

$${{\mathtt{15}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{20}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{36}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{48}}}^{{\mathtt{2}}} = {\mathtt{4\,225}}$$

$${{\mathtt{65}}}^{{\mathtt{2}}} = {\mathtt{4\,225}}$$

(I don't remember seeing Chris's last line when I gave my previous answer.  Hmm!)

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Write this as 

1/4 - a - 1/4 + a² = -1/2   remembering that two minuses make a plus, which is why the a² term is + not -.

Simplify:

a² - a = -1/2

Add 1/2 to both sides

a² - a + 1/2 = 0

Solve using the quadratic formula:

$${{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{a}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{a}} = {\mathtt{\,-\,}}{\frac{\left({i}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
{\mathtt{a}} = {\frac{\left({i}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{a}} = {\mathtt{\,-\,}}\left({\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{i}\right)\\
{\mathtt{a}} = {\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{i}\\
\end{array} \right\}$$

You can see that the solutions are complex (that is, they involve i, the square root of -1); there are no real- number solutions.

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No!  It's such a well known question I must have seen it a hundred times.

A googol is 10^100 so a googol times a googol is (10^100)*(10^100) = 10^200.

What I did was to divide both sides by ncr(20,5) and then take the fifth root of both sides.  This leaves a quartic in p.  There are four solutions to this quartic; two real and two complex.  The 0.0414 was one of the real solutions (the other was around 0.6, so I doubt this was the desired result!).

(I used Mathcad to do the calculations).

Q3.3

Acceleration is a vector quantity, so it is non-zero if either the magnitude or direction (or both) of the velocity is changing.  Point C is the only marked point where neither the magnitude nor direction of velocity is changing, so the answer is B.

Q3.7

I'm not sure what the symbols above v and a mean, but, ignoring these:

There is a component of acceleration in the i-direction which is making the object deviate from its path in the j-direction, so the object is on a curved path.  The component of acceleration in the j-direction is in the opposite direction to the direction of the velocity, so tending to slow the object, so the answer is C.

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