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 solve the given problem, we need to evaluate the expression for \( f'(x) \) at \( x = 0 \) and find \( \frac{1}{5} f'(0) \).

The function given 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 need to differentiate this determinant with respect to \( x \) and find the value at \( x = 0 \).

Let's breakdown \( \sin^2 2x \) and \( \cos^4 x \) as:

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

First, substitute \( x = 0 \) into \( f(x) \) to find \( f(0) \):

• \(\cos 0 = 1\), \(\sin 0 = 0\).

Thus, the matrix at \( x = 0 \) becomes:

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

Calculate the determinant at \( x = 0 \):

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

Now we need \( f'(0) \). For this, differentiate each function inside the determinant with respect to \( x \) and substitute \( x = 0 \):

Using Leibniz rule for the derivative of the determinant, which is quite complicated, we notice that:

• If the determinant evaluates to a non-zero constant at \( x = 0 \) or it's a linear function having \( x \) as a term, then its derivative at \( x = 0 \) will relate to \( 0 \), indicating no change at this point, producing \( f'(0) = 0 \).

Thus, at \( x = 0 \), it leads to a stable determinant even after differentiations considering levels of polynomial multiplication by terms having no linear variations at that instant \( x \). Hence, \( f'(0) = 0 \).

Finally, calculate \( \frac{1}{5} f'(0) \). Since \( f'(0) = 0 \), we have:

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

Therefore, the correct answer is: 0

Answer 2

By simplifying the determinant using row operations:

\( R_2 \rightarrow R_2 - R_1 \),    \( R_3 \rightarrow R_3 - R_1 \)

we find that \( f(x) \) is constant. Therefore, \( f'(x) = 0 \).

Thus,

\[ \frac{1}{5} f'(0) = 0 \]