Let \(f(\theta) = \begin{vmatrix}1 & 1 & 1\\ 1 & 1 + \sin\theta & 1\\ 1 + \cos\theta & 1 & 1\end{vmatrix}\). Consider the following statements:
(A) \(f(\theta) = -\displaystyle\frac{1}{2}\sin 2\theta\) for every real \(\theta\)
(B) The maximum value of \(f(\theta)\) is \(\displaystyle\frac{1}{2}\)
(C) \(f\left(\displaystyle\frac{\pi}{4}\right) = \displaystyle\frac{1}{2}\)
(D) \(f\left(\displaystyle\frac{\pi}{2}\right) = 0\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (B) and (D) only
Applying \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - R_1\):
\( \displaystyle f(\theta) = \begin{vmatrix}1 & 1 & 1\\ 0 & \sin\theta & 0\\ \cos\theta & 0 & 0\end{vmatrix} \)
Expanding along the third column, only the top entry is non-zero:
\(f(\theta) = 1 \times (0 \cdot 0 - \sin\theta\cos\theta)\)
\( \displaystyle f(\theta) = -\sin\theta\cos\theta = -\frac{1}{2}\sin 2\theta \)
Since \(\sin 2\theta\) takes every value in \([-1, 1]\), \(f(\theta)\) takes every value in \(\left[-\frac{1}{2}, \frac{1}{2}\right]\).
So \(f(\theta) = -\frac{1}{2}\sin 2\theta\), with maximum \(\frac{1}{2}\), \(f(\pi/4) = -\frac{1}{2}\) and \(f(\pi/2) = 0\).
Checking the statements
(A) \(f(\theta) = -\frac{1}{2}\sin 2\theta\) — Correct. The determinant reduces to \(-\sin\theta\cos\theta\).
(B) Maximum value \(\frac{1}{2}\) — Correct. Reached when \(\sin 2\theta = -1\), for example \(\theta = \frac{3\pi}{4}\).
(C) \(f(\pi/4) = \frac{1}{2}\) — Incorrect. \(f(\pi/4) = -\frac{1}{2}\sin\frac{\pi}{2} = -\frac{1}{2}\).
(D) \(f(\pi/2) = 0\) — Correct. \(f(\pi/2) = -\frac{1}{2}\sin\pi = 0\).