Question

Let \[ f(x) = \begin{vmatrix} \sin 3x & 1 & 2\left(\cos \frac{3x}{2} + \sin \frac{3x}{2}\right)^2 \cos 3x & -1 & 2\left(\cos \frac{3x}{2} - \sin \frac{3x}{2}\right)^2 \tan 3x & 4 & 1 + 2\tan 3x \end{vmatrix} \] Then \(f(x)\) is equal to

• A. 0
• B. 1
• C. constant
• D. none of these

Hint:

If two columns become proportional or identical, determinant simplifies heavily.

Answer 1

The Correct Option is C

Concept: Use trigonometric identities and determinant simplification.

Step 1: Simplify expressions.
\[ \left(\cos \frac{3x}{2} + \sin \frac{3x}{2}\right)^2 = 1 + \sin 3x \] \[ \left(\cos \frac{3x}{2} - \sin \frac{3x}{2}\right)^2 = 1 - \sin 3x \] So matrix becomes: \[ \begin{vmatrix} \sin 3x & 1 & 2(1+\sin 3x) \cos 3x & -1 & 2(1-\sin 3x) \tan 3x & 4 & 1 + 2\tan 3x \end{vmatrix} \]

Step 2: Column operations.
Apply: \[ C_3 \rightarrow C_3 - 2C_2 \] \[ \Rightarrow \begin{vmatrix} \sin 3x & 1 & 2\sin 3x \cos 3x & -1 & 2\cos 3x \tan 3x & 4 & 2\tan 3x \end{vmatrix} \]

Step 3: Factor column.
Take 2 common from \(C_3\): \[ = 2 \begin{vmatrix} \sin 3x & 1 & \sin 3x \cos 3x & -1 & \cos 3x \tan 3x & 4 & \tan 3x \end{vmatrix} \]

Step 4: Observe columns.
Now \(C_1\) and \(C_3\) are identical: \[ \Rightarrow \text{Determinant = constant} \]

Step 5: Conclusion.
Value is independent of \(x\), hence constant.