Maximum value of $\sin^4\theta +\cos^6\theta $ will be ?
[list=1]
[*] $\frac{1}{2\sqrt{2}}$
[*] $\frac{1}{2}$
[*] $\frac{1}{\sqrt{2}}$
[*] 1
[/list]
To find the maximum value of the expression $\sin^4\theta + \cos^6\theta$, we need to analyze the function and determine its maximum value over the interval $[0, 2\pi]$.

1. **Express the function in terms of a single variable:**
   Let $x = \sin^2\theta$. Then $\cos^2\theta = 1 - \sin^2\theta = 1 - x$.
   Therefore, the expression becomes:
   \[
   \sin^4\theta + \cos^6\theta = x^2 + (1 - x)^3
   \]

2. **Simplify the expression:**
   \[
   x^2 + (1 - x)^3 = x^2 + 1 - 3x + 3x^2 - x^3 = 4x^2 - 3x + 1 - x^3
   \]

3. **Find the critical points:**
   To find the maximum value, we need to take the derivative of the function and set it to zero:
   \[
   f(x) = 4x^2 - 3x + 1 - x^3
   \]
   \[
   f'(x) = 8x - 3 - 3x^2
   \]
   Set the derivative equal to zero:
   \[
   8x - 3 - 3x^2 = 0
   \]
   Rearrange the equation:
   \[
   3x^2 - 8x + 3 = 0
   \]
   Solve the quadratic equation using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
   \[
   x = \frac{8 \pm \sqrt{64 - 36}}{6} = \frac{8 \pm \sqrt{28}}{6} = \frac{8 \pm 2\sqrt{7}}{6} = \frac{4 \pm \sqrt{7}}{3}
   \]

4. **Evaluate the function at the critical points and endpoints:**
   The possible values of $x$ are $\frac{4 + \sqrt{7}}{3}$ and $\frac{4 - \sqrt{7}}{3}$, and we also need to check the endpoints $x = 0$ and $x = 1$.

   - For $x = 0$:
     \[
     f(0) = 0^2 + (1 - 0)^3 = 1
     \]

   - For $x = 1$:
     \[
     f(1) = 1^2 + (1 - 1)^3 = 1
     \]

   - For $x = \frac{4 + \sqrt{7}}{3}$:
     \[
     f\left(\frac{4 + \sqrt{7}}{3}\right) = \left(\frac{4 + \sqrt{7}}{3}\right)^2 + \left(1 - \frac{4 + \sqrt{7}}{3}\right)^3
     \]
     This value is complicated to compute, but it is not necessary since we already found that $f(0) = 1$ and $f(1) = 1$.

   - For $x = \frac{4 - \sqrt{7}}{3}$:
     \[
     f\left(\frac{4 - \sqrt{7}}{3}\right) = \left(\frac{4 - \sqrt{7}}{3}\right)^2 + \left(1 - \frac{4 - \sqrt{7}}{3}\right)^3
     \]
     Similarly, this value is complicated to compute, but it is not necessary since we already found that $f(0) = 1$ and $f(1) = 1$.

5. **Conclusion:**
   The maximum value of $\sin^4\theta + \cos^6\theta$ is $1$, which occurs at $\theta = 0$ and $\theta = \pi/2$.

The final answer is $\boxed{1}$