\(\Delta=\begin{vmatrix}1 & \cos x & 1\\ -\cos x & 1 & \cos x\\ -1 & -\cos x & 1\end{vmatrix}\)
(A) \(\Delta=2\left(1-\cos^2 x\right)\)
(B) \(\Delta=2\left(2-\sin^2 x\right)\)
(C) Minimum value of \(\Delta\) is 2
(D) Maximum value of \(\Delta\) is 4
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (B), (C) and (D) only
Expanding along the first row:
\( \Delta = 1\left(1 + \cos^2 x\right) - \cos x\left(-\cos x + \cos x\right) + 1\left(\cos^2 x + 1\right) \)
\( = 1 + \cos^2 x - 0 + \cos^2 x + 1 \)
\( = 2\left(1 + \cos^2 x\right) \)
Using \(\cos^2 x = 1 - \sin^2 x\):
\( \Delta = 2\left(2 - \sin^2 x\right) \)
Since \(0 \le \cos^2 x \le 1\):
\( 1 \le 1 + \cos^2 x \le 2 \)
\( 2 \le \Delta \le 4 \)
The value 2 is reached when cos x = 0 and the value 4 when cos x = ±1.
So \(\Delta = 2\left(1 + \cos^2 x\right) = 2\left(2 - \sin^2 x\right)\), its minimum value is 2 and its maximum value is 4.
Checking the statements
(A) \(\Delta=2\left(1-\cos^2 x\right)\) — Incorrect. \(\Delta = 2\left(1 + \cos^2 x\right)\); at x = 0 it is 4, not 0.
(B) \(\Delta=2\left(2-\sin^2 x\right)\) — Correct. \(1 + \cos^2 x = 2 - \sin^2 x\).
(C) Minimum value of \(\Delta\) is 2 — Correct. \(\Delta = 2\) when cos x = 0.
(D) Maximum value of \(\Delta\) is 4 — Correct. \(\Delta = 4\) when \(\cos^2 x = 1\).