William Lowell Putnam Mathematical Competition

Given (f(x) = x^3 - 3x^2 + 2x + 1), we have (f(1) = 1^3 - 3(1)^2 + 2(1) + 1 = -1). Therefore, (f(f(1)) = f(-1) = (-1)^3 - 3(-1)^2 + 2(-1) + 1 = 1).

Using the Law of Sines, we have (\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}). Therefore, (\frac{a}{\sin 60^\circ} = \frac{b}{\sin 75^\circ}). Solving for (\frac{a}{b}), we get (\frac{a}{b} = \frac{\sin 60^\circ}{\sin 75^\circ} = \frac{\sqrt{3}}{2}).

To find the area of the region, we need to find the points of intersection of the two curves. Solving (x^2 - 4x + 3 = x - 1), we get (x^2 - 5x + 4 = 0). Factoring, we get ((x - 1)(x - 4) = 0). Therefore, the points of intersection are ((1, 0)) and ((4, 3)). The area of the region is given by the integral (\int_1^4 (x^2 - 4x + 3) - (x - 1) \ dx). Evaluating the integral, we get (\frac{8}{3}).

The elements of (S) are the multiples of 3 and 5 less than 100. The multiples of 3 are (3, 6, 9, ..., 99), and the multiples of 5 are (5, 10, 15, ..., 95). However, some numbers are counted twice, such as 15, which is a multiple of both 3 and 5. To find the sum of all the elements of (S), we need to subtract the sum of the numbers that are counted twice. The sum of the multiples of 3 is (3 + 6 + 9 + ... + 99 = \frac{33(100)}{2} = 1650), and the sum of the multiples of 5 is (5 + 10 + 15 + ... + 95 = \frac{20(100)}{2} = 1000). The sum of the numbers that are counted twice is (15 + 30 + 45 + ... + 90 = \frac{15(100)}{2} = 750). Therefore, the sum of all the elements of (S) is (1650 + 1000 - 750 = 1350).

We can factor the numerator of (f(x)) as (x^2 - 1 = (x + 1)(x - 1)). Therefore, (f(x) = \frac{(x + 1)(x - 1)}{x - 1} = x + 1). Substituting (x = 1) into (f(x)), we get (f(1) = 1 + 1 = 2). Therefore, (\lim_{x \to 1} f(x) = 2).

The determinant of a matrix is the product of its eigenvalues. Therefore, the determinant of (A) is (1 \cdot 2 \cdot 3 = 6). The determinant of (A^2) is the square of the determinant of (A), so the determinant of (A^2) is (6^2 = 18).

The numbers that are divisible by 7 but not by 11 are the multiples of 7 that are not multiples of 11. The multiples of 7 less than 1000 are (7, 14, 21, ..., 994), and the multiples of 11 less than 1000 are (11, 22, 33, ..., 990). The numbers that are multiples of both 7 and 11 are (77, 154, 231, ..., 931). Therefore, the number of positive integers less than 1000 that are divisible by 7 but not by 11 is (\frac{994}{7} - \frac{990}{11} + \frac{931}{77} = 132).

We can use the quotient rule to find the derivative of (f(x)). The quotient rule states that if (f(x) = \frac{g(x)}{h(x)}), then (f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{h(x)^2}). In this case, (g(x) = x^3 - 1) and (h(x) = x - 1). Therefore, (g'(x) = 3x^2) and (h'(x) = 1). Substituting these values into the quotient rule, we get (f'(x) = \frac{(x - 1)(3x^2) - (x^3 - 1)(1)}{(x - 1)^2} = \frac{3x^3 - 3x^2 - x^3 + 1}{(x - 1)^2} = \frac{2x^3 - 3x^2 + 1}{(x - 1)^2}). Substituting (x = 2) into (f'(x)), we get (f'(2) = \frac{2(2)^3 - 3(2)^2 + 1}{(2 - 1)^2} = \frac{16 - 12 + 1}{1} = 2).

The trace of a matrix is the sum of its eigenvalues. Therefore, the trace of (A) is (1 + 2 + 3 = 6). The trace of (A^3) is the trace of (A \cdot A \cdot A), which is equal to the sum of the eigenvalues of (A \cdot A \cdot A). The eigenvalues of (A \cdot A \cdot A) are the cubes of the eigenvalues of (A), which are (1^3, 2^3, 3^3). Therefore, the trace of (A^3) is (1^3 + 2^3 + 3^3 = 1 + 8 + 27 = 36).

The elements of (S) are the multiples of 2 and 3 less than 100. The multiples of 2 are (2, 4, 6, ..., 98), and the multiples of 3 are (3, 6, 9, ..., 99). However, some numbers are counted twice, such as 6, which is a multiple of both 2 and 3. To find the sum of all the elements of (S), we need to subtract the sum of the numbers that are counted twice. The sum of the multiples of 2 is (2 + 4 + 6 + ... + 98 = \frac{2(100)}{2} = 1000), and the sum of the multiples of 3 is (3 + 6 + 9 + ... + 99 = \frac{3(100)}{2} = 1500). The sum of the numbers that are counted twice is (6 + 12 + 18 + ... + 96 = \frac{6(100)}{2} = 300). Therefore, the sum of all the elements of (S) is (1000 + 1500 - 300 = 1800).

We can use the quotient rule to find the derivative of (f(x)). The quotient rule states that if (f(x) = \frac{g(x)}{h(x)}), then (f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{h(x)^2}). In this case, (g(x) = x^2 - 4x + 3) and (h(x) = x - 1). Therefore, (g'(x) = 2x - 4) and (h'(x) = 1). Substituting these values into the quotient rule, we get (f'(x) = \frac{(x - 1)(2x - 4) - (x^2 - 4x + 3)(1)}{(x - 1)^2} = \frac{2x^2 - 6x + 4 - x^2 + 4x - 3}{(x - 1)^2} = \frac{x^2 - 2x + 1}{(x - 1)^2}). Substituting (x = 1) into (f'(x)), we get (f'(1) = \frac{1^2 - 2(1) + 1}{(1 - 1)^2} = \frac{0}{(0)^2} = \infty).

The determinant of a matrix is the product of its eigenvalues. Therefore, the determinant of (A) is (1 \cdot 2 \cdot 3 = 6). The determinant of (A^4) is the square of the determinant of (A^2), which is equal to the square of the determinant of (A \cdot A). The eigenvalues of (A \cdot A) are the squares of the eigenvalues of (A), which are (1^2, 2^2, 3^2). Therefore, the determinant of (A^2) is (1^2 \cdot 2^2 \cdot 3^2 = 36). The determinant of (A^4) is (36^2 = 729).

The elements of (S) are the multiples of 3 and 4 less than 100. The multiples of 3 are (3, 6, 9, ..., 99), and the multiples of 4 are (4, 8, 12, ..., 96). However, some numbers are counted twice, such as 12, which is a multiple of both 3 and 4. To find the sum of all the elements of (S), we need to subtract the sum of the numbers that are counted twice. The sum of the multiples of 3 is (3 + 6 + 9 + ... + 99 = \frac{3(100)}{2} = 1500), and the sum of the multiples of 4 is (4 + 8 + 12 + ... + 96 = \frac{4(100)}{2} = 2000). The sum of the numbers that are counted twice is (12 + 24 + 36 + ... + 96 = \frac{12(100)}{2} = 600). Therefore, the sum of all the elements of (S) is (1500 + 2000 - 600 = 1500).

We can use the quotient rule to find the derivative of (f(x)). The quotient rule states that if (f(x) = \frac{g(x)}{h(x)}), then (f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{h(x)^2}). In this case, (g(x) = x^3 - 2x^2 + x - 2) and (h(x) = x - 2). Therefore, (g'(x) = 3x^2 - 4x + 1) and (h'(x) = 1). Substituting these values into the quotient rule, we get (f'(x) = \frac{(x - 2)(3x^2 - 4x + 1) - (x^3 - 2x^2 + x - 2)(1)}{(x - 2)^2} = \frac{3x^3 - 6x^2 + x - 3x^2 + 8x - 2 - x^3 + 2x^2 - x + 2}{(x - 2)^2} = \frac{2x^3 - 7x^2 + 9x}{(x - 2)^2}). Substituting (x = 2) into (f'(x)), we get (f'(2) = \frac{2(2)^3 - 7(2)^2 + 9(2)}{(2 - 2)^2} = \frac{16 - 28 + 18}{0} = \infty).