The correct answer is 2
\(I_{R_1}=I_1+I_2+2\sqrt{I_1I_2}\cos\phi\)
\(I_A=I+4I+2\sqrt{I⋅4I}\cos90^∘=5I\)
\(I_B=I+4I+2\sqrt{I⋅4I}\cos60^∘=7I\)
\(I_B–I_A=2I\)
\(\therefore\) value of x is 2
Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is: