Options (A), (B) and (D) are all closure properties of the class of characteristic functions:

(A) If $\varphi$ is the c.f. of $X$ , then $\varphi(t/a)e^{ibt}$ is the c.f. of $\frac{X}{a}+b$ .
(B) $\varphi(-t)=\overline{\varphi(t)}$ is the c.f. of $-X$ .
(D) A countable convex combination of c.f.s is the c.f. of the corresponding mixture distribution.

(C) is not a characteristic function. It satisfies $\varphi(0)=1$ , is real, even and bounded — so the crude checks pass — but expand it as a power series near $0$ :

$\frac{1}{1+t^{4}}=1-t^{4}+t^{8}-\cdots$

There is no $t^{2}$ term, so $\varphi"(0)=0$ . For a genuine characteristic function, $\varphi"(0)$ exists and equals $-\mathbb{E}(X^{2})$ ; hence $\mathbb{E}(X^{2})=0$ , which forces $X=0$ almost surely and therefore $\varphi(t)\equiv1$ . But $1/(1+t^{4})\not\equiv1$ — a contradiction.

Equivalently: a c.f. that is twice differentiable at the origin with $\varphi"(0)=0$ must be identically $1$ . So $1/(1+t^{4})$ cannot be one. (Contrast $1/(1+t^{2})$ , which is the c.f. of the standard Laplace distribution.)