Question

If \[ f(x) = \begin{vmatrix} 2 \cos^4 x & 2 \sin^4 x & 3 + \sin^2 2x \\ 3 + 2 \cos^4 x & 2 \sin^4 x & \sin^2 2x \\ 2 \cos^4 x & 3 + 2 \sin^4 x & \sin^2 2x \end{vmatrix} \] then \( \frac{1}{5} f'(0) \) is equal to

• A. 0
• B. 1
• C. 2
• D. 6

Answer 1

The Correct Option is A

To determine \( \frac{1}{5} f'(0) \), we first evaluate the derivative of \( f'(x) \) at \( x = 0 \).

The given function is:

\( f(x) = \begin{vmatrix} 2 \cos^4 x & 2 \sin^4 x & 3 + \sin^2 2x \\ 3 + 2 \cos^4 x & 2 \sin^4 x & \sin^2 2x \\ 2 \cos^4 x & 3 + 2 \sin^4 x & \sin^2 2x \end{vmatrix} \)

We must differentiate this determinant with respect to \( x \) and then evaluate it at \( x = 0 \).

The terms \( \sin^2 2x \) and \( \cos^4 x \) can be expanded as:

• \(\sin^2 2x = 4 \sin^2 x \cos^2 x\)
• \(\cos^4 x = (\cos^2 x)^2\)
• \(\sin^4 x = (\sin^2 x)^2\)

Evaluate \( f(x) \) at \( x = 0 \):

• Since \( \cos 0 = 1 \) and \( \sin 0 = 0 \).

The matrix at \( x = 0 \) is:

\( f(0) = \begin{vmatrix} 2 & 0 & 3 \\ 5 & 0 & 0 \\ 2 & 3 & 0 \end{vmatrix} \)

The determinant at \( x = 0 \) is calculated as:

• Determinant, \( \begin{vmatrix} 2 & 0 & 3 \\ 5 & 0 & 0 \\ 2 & 3 & 0 \end{vmatrix} = 2(0 \cdot 0 - 0 \cdot 3) - 0(5 \cdot 0 - 0 \cdot 2) + 3(5 \cdot 3 - 0 \cdot 2) = 0 - 0 + 3(15) = 45 \)

To find \( f'(0) \), we differentiate each element of the determinant with respect to \( x \) and then substitute \( x = 0 \).

Applying the Leibniz rule for the derivative of a determinant, which is complex, we observe a simplification: If the determinant is a non-zero constant or a linear function of \( x \) at \( x = 0 \), its derivative at \( x = 0 \) will be 0.

At \( x = 0 \), the determinant remains stable under differentiation, as the terms do not introduce a linear dependence on \( x \) at that point. Therefore, \( f'(0) = 0 \).

Finally, we compute \( \frac{1}{5} f'(0) \). Given that \( f'(0) = 0 \):

\(\frac{1}{5} f'(0) = \frac{1}{5} \times 0 = 0\).

The result is: 0