(c) $I_{\max} = I_{1} + I_{2} + 2\sqrt{I_{1}I_{2}}$ $= 4I + I + 2\sqrt{2I_{1} \times I}$ $= 9I$ $I_{\min} = I_{1} + I_{2} - 2\sqrt{2I_{1}I_{2}}$ $I_{\min} = I$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.