The correct answer is (B) : 4
\(\text{Let }(\sqrt3+\sqrt2)^{x^{2}−4}=t\)
\(t+\frac{1}{t}=10\)
\(⇒t=5+2\sqrt6, 5−2\sqrt6\)
\(⇒(\sqrt3+\sqrt2)^{x^{2}−4}=5+2\sqrt6, 5−2\sqrt6\)
\(⇒x^2−4=2, −2\) or \(x^2 =6, 2\)
\(⇒x=±\sqrt2, ±\sqrt6\)
Let \( (\sqrt{3} + \sqrt{2})^{x-4} = t \), then \[ t + \frac{1}{t} = 10 \] Solving for \( t \), we get \[ t = 5 + 2\sqrt{6}, \quad t = 5 - 2\sqrt{6} \] Thus, \[ (\sqrt{3} + \sqrt{2})^{x-4} = 5 + 2\sqrt{6}, \quad 5 - 2\sqrt{6} \] Squaring both sides, \[ x^2 - 4 = 2, -2 \Rightarrow x^2 = 6, 2 \] \[ x = \pm\sqrt{2}, \pm\sqrt{6} \]
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:
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