Let f: R arrow R be a twice differentiable function satisfying the...

Expanding the given determinant, we get:
$f(x+y)(\sin 2x - \sin 2y) - f(x-y)(\sin 2x + \sin 2y) = 0$

Using the sum-to-product trigonometric identities:
$\sin 2x - \sin 2y = 2 \cos(x+y) \sin(x-y)$
$\sin 2x + \sin 2y = 2 \sin(x+y) \cos(x-y)$

Substituting these into the equation:
$f(x+y)(2 \cos(x+y) \sin(x-y)) = f(x-y)(2 \sin(x+y) \cos(x-y))$

Dividing both sides by $2 \sin(x+y) \cos(x+y) \sin(x-y) \cos(x-y)$, we obtain:
$\dfrac{f(x+y)}{\sin(x+y) \cos(x+y)} = \dfrac{f(x-y)}{\sin(x-y) \cos(x-y)}$

Multiplying the numerator and denominator by 2:
$\dfrac{f(x+y)}{\sin 2(x+y)} = \dfrac{f(x-y)}{\sin 2(x-y)}$

Since this holds for all $x, y \in \mathbf{R}$, the ratio must be a constant $K$.
Thus, $f(x) = K \sin 2x$.

Differentiating with respect to $x$:
$f'(x) = 2K \cos 2x$

Given $f'(0) = 4$, we have:
$2K \cos(0) = 4 \implies 2K = 4 \implies K = 2$

So, the function is $f(x) = 2 \sin 2x$.

Differentiating again to find the second derivative:
$f''(x) = -4 \sin 2x \cdot 2 = -8 \sin 2x$

Evaluating at $x = \dfrac{\pi}{12}$:
$f''\left(\dfrac{\pi}{12}\right) = -8 \sin\left(2 \cdot \dfrac{\pi}{12}\right) = -8 \sin\left(\dfrac{\pi}{6}\right)$
$f''\left(\dfrac{\pi}{12}\right) = -8 \left(\dfrac{1}{2}\right) = -4$

Therefore, \left|f''\left(\dfrac{\pi}{12}\right)\right| = |-4| = 4$.<br><br>Answer:$4