Discussion
Let \(f(x) = \begin{vmatrix} x & 1 \\ \sin 2\pi x & 2x^{2} \end{vmatrix}\). If f(x) is an odd function, \(f(-x) = g(x)\) and \(\lambda f(1)g(1) = 4\) Then the value of \(\lambda\) is equal to
Answer: E
\(\lambda\) is -1.
Expanding the determinant: \(f(x) = 2x^{3} - \sin 2\pi x\), which is indeed odd. Then \(g(x) = f(-x) = -f(x)\).
\[ \begin{aligned} f(1) &= 2 - \sin 2\pi = 2 \\ g(1) &= -f(1) = -2 \\ \lambda \cdot 2 \cdot (-2) &= 4 \\ \lambda &= -1 \end{aligned} \]
Forgetting that \(g(1) = -f(1)\) for an odd function gives \(4\lambda = 4\) and \(\lambda = 1\), the sign trap among the options.
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