If \(\ell < m < n\), then <br> <img src="https://latex.codecogs.com/svg.image?\begin{vmatrix} 1 & \ell & \ell^4 \\ 1 & m & m^4 \\ 1 & n & n^4 \end{vmatrix}" alt="\begin{vmatrix} 1 & \ell & \ell^4 \\ 1 & m & m^4 \\ 1 & n & n^4 \end{vmatrix}"> will always be greater than -
Step-by-Step Solution
Key Concept: The determinant is a Vandermonde-like determinant. Specifically, it is (m-l)(n-l)(n-m)(l+m+n)(l^2+m^2+n^2). Since l < m < n, the factors (m-l), (n-l), and (n-m) are positive. The expression is positive if (l+m+n)(l^2+m^2+n^2) > 0.
The determinant is given by $D = \begin{vmatrix} 1 & \ell & \ell^4 \\ 1 & m & m^4 \\ 1 & n & n^4 \end{vmatrix}$. This is a known determinant form. Expanding it, we get $D = (m-\ell)(n-\ell)(n-m)(\ell+m+n)(\ell^2+m^2+n^2)$. Since $\ell < m < n$, the terms $(m-\ell) > 0$, $(n-\ell) > 0$, and $(n-m) > 0$. Thus, the sign of the determinant depends on $(\ell+m+n)(\ell^2+m^2+n^2)$. Since the question asks what it is always greater than, and the determinant can be positive or negative depending on the values of $\ell, m, n$, the options C and D are correct as it is always greater than -1 and 0.
Correct Answer: C,D