If u = (x, t) is the solution of the initial value problem

\(\left\{\begin{array}{ll} u_{t}=u_{x x}, & x \in \mathbb{R}, t>0 \\ u(x, 0)=\sin (4 x)+x+1, & x \in \mathbb{R} \end{array}\right.\)

satisfying |u(x. t)| < \(\rm 3e^{x^2}\) for all x ∈ ℝ and t > 0, then

1
\(\rm u\left(\frac{\pi}{8}, 1\right)+u\left(-\frac{\pi}{8},1\right)=2\)
2
\(\rm u\left(\frac{\pi}{8}, 1\right)=u\left(-\frac{\pi}{8},1\right)\)
3
\(\rm u\left(\frac{\pi}{8}, 1\right)+2u\left(-\frac{\pi}{8},1\right)=2\)
4
\(\rm u\left(\frac{\pi}{8}, 1\right)=-u\left(-\frac{\pi}{8},1\right)\)

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