Question

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

• A. 1
• B. -4
• C. 0
• D. 4
• E. -1

Hint:

For odd functions: $f(-x) = -f(x)$.

Answer 1

Answer: (E)

Concept: Determinant and odd function property.

Step 1: Evaluate determinant
\[ f(x) = x(2x^2) - (1)(\sin 2\pi x) = 2x^3 - \sin 2\pi x \]

Step 2: Check odd function
\[ f(-x) = -2x^3 + \sin 2\pi x = -f(x) \Rightarrow g(x) = -f(x) \]

Step 3: Evaluate at $x=1$
\[ f(1) = 2(1)^3 - \sin 2\pi = 2 \] \[ g(1) = -2 \]

Step 4: Use condition
\[ \lambda (2)(-2) = 4 \Rightarrow -4\lambda = 4 \Rightarrow \lambda = -1 \] Final Conclusion:
Option (E)