Solution to “Possible HKAL Pure Problems” « Mathematics in HKCEE, HKAL, and Beyond

Solution to “Possible HKAL Pure Problems”

四月 1, 2009

1(a) Expanding along the last row, we have:

$f(x) = 1 \det{\begin{pmatrix}x & x^3\\y & y^3 \end{pmatrix}} - z \det{\begin{pmatrix} 1 & x^3 \\1 & y^3 \end{pmatrix}} + z^3 \det{\begin{pmatrix} 1 & x \\1 & y \end{pmatrix}}.$

Therefore, $f(z)$ is a cubic in $z$ with leading coefficient $\det \begin{pmatrix} 1 & x \\1 & y \end{pmatrix} = y - x.$

1(b) Note that $f(x)$ and $f(y)$ have two identical rows, and therefore by properties of determinants, we have $f(x) = f(y) = 0.$ By Factor theorem, we have $z - x, z - y$ are factors of $f(z).$ Since $f(z)$ is a cubic, we must have $f(z) = A(z - x)(z - y)(z - r)$ , where $r$ is some root to be determined, and $A$ being the leading coefficient. By part (a), we have $A = y - x.$ Hence $f(z) = (y - x)(z - x)(z - y)(z - r)$ as claimed.

1(c) Since the coefficient of $z^2$ of $f(z)$ is zero, we have the sum of roots is zero. This implies $x + y + r = 0$ , and therefore we have $r = -x - y$ as claimed.

2) Suppose $g(x) = \log(x)$ is a polynomial. Then $deg(g(x^2)) = 2deg(g(x))$ , but $\log(x^2) = 2\log(x)$ , which implies $\deg(\log(x^2)) = deg(\log(x)).$ Hence we have $2deg(g(x)) = deg(g(x)),$ which implies $deg(g(x)) = 0$ . i.e. $\log(x)$ is constant, which is absurd. Hence $\log(x)$ is not a polynomial.