x1 + 2 x2 + 3 x3 + ... + (n-1) xn-1 + n xn = n
2 x1 + 3 x2 + 4 x3 + ... + n xn-1 + xn = n-1
3 x1 + 4 x2 + 5 x3 + ... + xn-1 + 2 xn = n-2
...
(n-1) x1 + n x2 + x3 + ... + (n-3) xn-1 + (n-2) xn = 2
n x1 + x2 + 2 x3 + ... + (n-2) xn-1 + (n-1) xn = 1.
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1. Solve the equations:
x1 + 2 x2 + 3 x3 + ... + (n-1) xn-1 + n xn = n 2 x1 + 3 x2 + 4 x3 + ... + n xn-1 + xn = n-1 3 x1 + 4 x2 + 5 x3 + ... + xn-1 + 2 xn = n-2 ... (n-1) x1 + n x2 + x3 + ... + (n-3) xn-1 + (n-2) xn = 2 n x1 + x2 + 2 x3 + ... + (n-2) xn-1 + (n-1) xn = 1. |
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| 2. Find rational x in (3, 4) such that √(x-3) and √(x+1) are rational. |
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| 3. Express x13 + 1/x13 as a polynomial in y = x + 1/x. |
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| 4. f(x) is continuous on the interval [0, π] and satisfies ∫ 0π f(x) dx = 0, ∫ 0π f(x) cos x dx = 0. Show that f(x) has at least two zeros in the interval (0, π). | |
| 5. Find the smallest positive integer a such that for some integers b, c the polynomial ax2 - bx + c has two distinct zeros in the interval (0, 1). |
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| 6. Find the sharpest inequalities of the form a·AB < AG < b·AB and c·AB < BG < d·AB for all triangles ABC with centroid G such that GA > GB > GC. |
To avoid possible copyright problems, I have changed the wording, but not the substance, of the problems.
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© John Scholes
jscholes@kalva.demon.co.uk
26 September 2003
Last corrected/updated 26 Sep 03