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Question: If f(θ) = \(\left| \begin{matrix} 1 & 1 & - 1 \\ 1 & e^{i\theta} & 1 \\ 1 & - 1 & - e^{\_ i\theta} \...

If f(θ) = ∣11−11eiθ11−1−e_iθ∣\left| \begin{matrix} 1 & 1 & - 1 \\ 1 & e^{i\theta} & 1 \\ 1 & - 1 & - e^{\_ i\theta} \end{matrix} \right|then

A

∫−π/2π/2f(θ)dθ=2∫0π/2f(θ)dθ\int_{- \pi/2}^{\pi/2}{f(\theta)d\theta = 2\int_{0}^{\pi/2}{f(\theta)d\theta}}

B

f(θ) is purely real

C

f(π/2) = 2

D

None of these

Answer

f(π/2) = 2

Explanation

Solution

On operating [R1 → R1 − R2 and R3 → R3 − R2],

f(θ) = ∣01−eiθ−21eiθ10−1−eiθ−1−eiθ∣\left| \begin{matrix} 0 & 1 - e^{i\theta} & - 2 \\ 1 & e^{i\theta} & 1 \\ 0 & - 1 - e^{i\theta} & - 1 - e^{i\theta} \end{matrix} \right|= (-1) [(1-eiθ)

(-1 - e-iθ) – 2(−1 − eiθ] = 2 when θ = π2\frac{\pi}{2}