To solve this problem, we need to determine the angular frequencies at which the power transferred to the circuit is half of the power at the resonant frequency.$\\$Given:$\\$• Inductance $L = 5.0 \,$H$\\$• Capacitance $C = 80 \, \mu$F$= 80 \times 10^{-6} \,$F$\\$• Resistance $R = 40 \, \Omega \\$• Voltage $V = 230 \,$V$\\$The resonant angular frequency $\omega_0$ is given by:$\\ \omega_0 = \frac{1}{\sqrt{LC}} \\$Substitute the values:$\\ \omega_0 = \frac{1}{\sqrt{5.0 \times 80 \times 10^{-6}}} \\ \omega_0 = \frac{1}{\sqrt{0.0004}} \\ \omega_0 = \frac{1}{0.02} \\ \omega_0 = 50 \,$rad/s$\\$At resonance, the power $P_0$ is given by:$\\ P_0 = \frac{V^2}{2R} \\$The power at half the resonant power is $\frac{P_0}{2}. \\$The power transferred to the circuit at any angular frequency $\omega$ is:$\\ P = \frac{V^2 R}{R^2 + (L\omega - \frac{1}{C\omega})^2} \\$We need to find $\omega$ such that $P = \frac{P_0}{2}: \\ \frac{V^2 R}{R^2 + (L\omega - \frac{1}{C\omega})^2} = \frac{V^2}{4R} \\$Simplifying, we get:$\\ 4R^2 = R^2 + (L\omega - \frac{1}{C\omega})^2 \\ 3R^2 = (L\omega - \frac{1}{C\omega})^2 \\$Taking square root on both sides:$\\ \sqrt{3}R = |L\omega - \frac{1}{C\omega}| \\$This gives two equations:$\\ L\omega - \frac{1}{C\omega} = \sqrt{3}R \quad$(1)$\\ L\omega - \frac{1}{C\omega} = -\sqrt{3}R \quad$(2)$\\$Solving equation (1):$\\ L\omega - \frac{1}{C\omega} = \sqrt{3}R \\ L\omega^2 - \sqrt{3}R\omega - \frac{1}{C} = 0 \\$This is a quadratic equation in $\omega. \\$Solving using the quadratic formula:$\\ \omega = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\$Here, $a = L, \, b = -\sqrt{3}R, \, c = -\frac{1}{C} \\ \omega = \frac{\sqrt{3}R \pm \sqrt{3R^2 + \frac{4L}{C}}}{2L} \\$Substitute the values:$\\ \omega = \frac{\sqrt{3} \times 40 \pm \sqrt{3 \times 40^2 + \frac{4 \times 5.0}{80 \times 10^{-6}}}}{2 \times 5.0} \\ \omega = \frac{40\sqrt{3} \pm \sqrt{4800 + 250000}}{10} \\ \omega = \frac{40\sqrt{3} \pm \sqrt{254800}}{10} \\$Approximating the square root:$\\ \omega = \frac{40\sqrt{3} \pm 504.78}{10} \\$Calculate the two values:$\\ \omega_1 = \frac{40\sqrt{3} + 504.78}{10} \approx 54 \,$rad/s$\\ \omega_2 = \frac{40\sqrt{3} - 504.78}{10} \approx 46 \,$rad/s$\\$Therefore, the angular frequencies are $46 \,$rad/s and $54 \,$rad/s.$\\$This corresponds to Option 3.