1. Write the determinant in a simpler form

Given

$$f(x)=\begin{vmatrix} 1+\sin^2x & \cos^2x & \sin 2x\\ \sin^2x & 1+\cos^2x & \sin 2x\\ \sin^2x & \cos^2x & 1+\sin 2x \end{vmatrix},\qquad x\in\left[\frac\pi6,\frac\pi3\right].$$

Let $s=\sin^2x,\qquad c=\cos^2x,\qquad t=\sin 2x.$ Then $s+c=1.$ So the matrix becomes

$$\begin{vmatrix} 1+s & c & t\\ s & 1+c & t\\ s & c & 1+t \end{vmatrix}.$$

2. Use row operations to evaluate the determinant

Apply:

• $R_1\to R_1-R_2$
• $R_2\to R_2-R_3$

Then the determinant becomes

$$\begin{vmatrix} 1 & -1 & 0\\ 0 & 1 & -t\\ s & c & 1+t \end{vmatrix}.$$

Now expand along the first row:

$$f(x)=1\cdot\begin{vmatrix}1&-t\\ c&1+t\end{vmatrix}-(-1)\cdot\begin{vmatrix}0&-t\\ s&1+t\end{vmatrix}.$$

Hence

$$f(x)=\big(1+t+ct\big)-\big(ts\big).$$

So

$$f(x)=1+t+t(c-s).$$

Since

$$$$

we get

$$f(x)=1+\sin 2x+\sin 2x\cos 2x.$$

Also, $\sin 2x\cos 2x=\frac12\sin 4x,$ so

$$f(x)=1+\sin 2x+\frac12\sin 4x.$$

3. Convert to a single-variable expression

Let $y=2x.$ Then $y\in\left[\frac\pi3,\frac{2\pi}3\right],$ and $f=1+\sin y+\frac12\sin 2y=1+\sin y(1+\cos y).$

Now set $u=\sin y.$ For $y\in\left[\frac\pi3,\frac{2\pi}3\right]$, we have $u\in\left[\frac{\sqrt3}{2},1\right].$ Also on this interval, $\cos y=\pm\sqrt{1-u^2},$ so direct substitution is less convenient. Better differentiate with respect to $y$.

4. Find critical points

From $f(y)=1+\sin y+\frac12\sin 2y,$ we get $f'(y)=\cos y+\cos 2y.$ Using $\cos 2y=2\cos^2y-1,$ $f'(y)=2\cos^2y+\cos y-1=(2\cos y-1)(\cos y+1).$

Thus critical points satisfy $\cos y=\frac12\quad\text{or}\quad \cos y=-1.$ On $y\in\left[\frac\pi3,\frac{2\pi}3\right],$ we have:

• $\cos y=\frac12$ at $y=\frac\pi3$ (left endpoint),
• $\cos y=-1$ is not in the interval.

So there is no interior critical point except endpoint behavior. Therefore extrema occur at endpoints, or by checking monotonicity.

5. Check monotonicity

On $\left(\frac\pi3,\frac\pi2\right)$, we have $\cos y>0$, so $2\cos y-1<0$ after crossing $\pi/3$, and $\cos y+1>0$. Hence $f'(y)<0$.

On $\left(\frac\pi2,\frac{2\pi}3\right)$, still $\cos y+1>0$ and $2\cos y-1<0$, so again $f'(y)<0$.

Thus $f$ is decreasing on the whole interval $\left[\frac\pi3,\frac{2\pi}3\right].$ So:

• maximum at $y=\frac\pi3$ i.e. $x=\frac\pi6$,
• minimum at $y=\frac{2\pi}3$ i.e. $x=\frac\pi3$.

6. Compute maximum and minimum values

At $x=\frac\pi6$: $\sin 2x=\sin\frac\pi3=\frac{\sqrt3}{2},\qquad \sin 4x=\sin\frac{2\pi}3=\frac{\sqrt3}{2}.$ Therefore

$$\alpha=f\left(\frac\pi6\right)=1+\frac{\sqrt3}{2}+\frac12\cdot\frac{\sqrt3}{2} =1+\frac{3\sqrt3}{4}.$$

At $x=\frac\pi3$: $\sin 2x=\sin\frac{2\pi}3=\frac{\sqrt3}{2},\qquad \sin 4x=\sin\frac{4\pi}3=-\frac{\sqrt3}{2}.$ Therefore

$$\beta=f\left(\frac\pi3\right)=1+\frac{\sqrt3}{2}-\frac12\cdot\frac{\sqrt3}{2} =1+\frac{\sqrt3}{4}.$$

So $\alpha=1+\frac{3\sqrt3}{4},\qquad \beta=1+\frac{\sqrt3}{4}.$

7. Test the options

First compute squares:

$$\alpha^2=\left(1+\frac{3\sqrt3}{4}\right)^2 =1+\frac{9\sqrt3}{8}+\frac{27}{16} =\frac{43}{16}+\frac{9\sqrt3}{8},$$ $$\beta^2=\left(1+\frac{\sqrt3}{4}\right)^2 =1+\frac{\sqrt3}{2}+\frac{3}{16} =\frac{19}{16}+\frac{\sqrt3}{2}.$$

Also,

$$\sqrt{\alpha}=\sqrt{1+\frac{3\sqrt3}{4}}.$$

Notice $\left(\frac12+\frac{\sqrt3}{2}\right)^2=\frac14+\frac34+\frac{\sqrt3}{2}=1+\frac{\sqrt3}{2},$ not equal to $\alpha$, so let us instead evaluate the option expression directly by a smarter identity.

Observe

$$\alpha=1+\frac{3\sqrt3}{4}=\frac{4+3\sqrt3}{4}.$$

Then

$$4\alpha=4+3\sqrt3=(\tfrac{3}{2}+\tfrac{\sqrt3}{2})^2?$$

Check:

$$\left(\frac32+\frac{\sqrt3}{2}\right)^2=\frac94+\frac34+\frac{3\sqrt3}{2}=3+\frac{3\sqrt3}{2},$$

not equal. So compute numerically if needed.

Instead test the simpler algebraic options first.

Option D

$$\alpha^2+\beta^2= \left(\frac{43}{16}+\frac{9\sqrt3}{8}\right)+\left(\frac{19}{16}+\frac{\sqrt3}{2}\right) =\frac{62}{16}+\frac{13\sqrt3}{8} =\frac{31}{8}+\frac{13\sqrt3}{8}\neq \frac92.$$

So D is false.

Option A

$$\alpha^2-\beta^2= \left(\frac{43}{16}-\frac{19}{16}\right)+\left(\frac{9\sqrt3}{8}-\frac{\sqrt3}{2}\right) =\frac{24}{16}+\frac{5\sqrt3}{8} =\frac32+\frac{5\sqrt3}{8}\neq 4\sqrt3.$$

So A is false.

Now check B and C.

We compute

$$\beta^2=\frac{19}{16}+\frac{\sqrt3}{2}.$$

Also

$$\alpha=1+\frac{3\sqrt3}{4}=\frac{4+3\sqrt3}{4}.$$

Numerically,

$$\alpha\approx 2.2990,\qquad \sqrt\alpha\approx 1.5163,$$

so $2\sqrt\alpha\approx 3.0326.$ Then

$$\beta^2-2\sqrt\alpha\approx 2.0535-3.0326\approx -0.9791\neq \frac{19}{4},$$

so B is false.

And

$$\beta^2+2\sqrt\alpha\approx 2.0535+3.0326\approx 5.0861\neq \frac{19}{4}=4.75,$$

so C is also false.

8. Recheck determinant simplification carefully

Let us expand the determinant directly:

$$f=(1+s)\big((1+c)(1+t)-ct\big)-c\big(s(1+t)-st\big)+t\big(sc-(1+c)s\big).$$

Now simplify each bracket:

$$(1+c)(1+t)-ct=1+c+t,$$ $$s(1+t)-st=s,$$ $$sc-(1+c)s=-s.$$

Thus

$$f=(1+s)(1+c+t)-cs-st.$$

Since $s+c=1$,

$$(1+s)(1+c+t)=(1+s)(1+c)+(1+s)t.$$

Also

$$(1+s)(1+c)=1+s+c+sc=2+sc.$$

Hence

$$f=2+sc+(1+s)t-cs-st=2+t.$$

Because $sc-cs=0$ and $(1+s)t-st=t$. So in fact

$$f(x)=2+\sin 2x.$$

This is much simpler and correct.

9. Now find extrema correctly

Since $x\in\left[\frac\pi6,\frac\pi3\right],$ we have $2x\in\left[\frac\pi3,\frac{2\pi}3\right].$ On this interval, $\sin 2x$ attains

• maximum $1$ at $2x=\frac\pi2\Rightarrow x=\frac\pi4$,
• minimum $\frac{\sqrt3}{2}$ at endpoints.

Therefore

$$\alpha=2+1=3,$$ $$\beta=2+\frac{\sqrt3}{2}.$$

10. Check the options with correct values

Compute

$$\beta^2=\left(2+\frac{\sqrt3}{2}\right)^2=4+2\sqrt3+\frac34=\frac{19}{4}+2\sqrt3.$$

Also

$$$$

Thus

$$\beta^2-2\sqrt\alpha=\left(\frac{19}{4}+2\sqrt3\right)-2\sqrt3=\frac{19}{4}.$$

So B is true.

Check others quickly:

$$\beta^2+2\sqrt\alpha=\frac{19}{4}+4\sqrt3\neq \frac{19}{4},$$

so C false.

$$\alpha^2-\beta^2=9-\left(\frac{19}{4}+2\sqrt3\right)=\frac{17}{4}-2\sqrt3\neq 4\sqrt3,$$

so A false.

$$\alpha^2+\beta^2=9+\frac{19}{4}+2\sqrt3\neq \frac92,$$

so D false.

11. Conclusion

The correct option is $\boxed{\text{B}}.$