1. Given expression $A=\sin^2 x+\cos^4 x$

2. Rewrite in one trigonometric function Let $t=\cos^2 x$ Since for real $x$, $0\le t\le 1$ and $\sin^2 x=1-\cos^2 x=1-t$ Therefore, $A=(1-t)+t^2=t^2-t+1$

3. Find the range of $A=t^2-t+1$ for $0\le t\le 1$

   Complete the square: $A=t^2-t+1=\left(t-\frac12\right)^2+\frac34$

   Since $\left(t-\frac12\right)^2\ge 0$, $A\ge \frac34$ Equality occurs when $t=\frac12 \quad \Rightarrow \quad \cos^2 x=\frac12$ which is possible for real $x$.

4. Maximum value On the interval $0\le t\le 1$, the quadratic opens upward, so maximum occurs at an endpoint.

   At $t=0$: $A=0^2-0+1=1$

   At $t=1$: $A=1^2-1+1=1$

   Hence, $A\le 1$

5. Range of $A$ Therefore, $\frac34\le A\le 1$

6. Check options

   • A: $\frac{13}{16}\le A\le 1$ — false, because minimum is $\frac34$.
   • B: $1\le A\le 2$ — false.
   • C: $\frac34\le A\le \frac{13}{16}$ — false, because maximum is $1$.
   • D: $\frac34\le A\le 1$ — true.

Final answer: Option D.