Let's find the extreme values of the following function, where "a" is a positive constant.

$$
f(x)=\sin 2 x+\sqrt{\sin ^{2} 2 x+a \cos ^{2} x}
$$
I. solution. The function is defined for all $x$ values because $a>0$ ensures that the expression under the square root cannot be negative. Therefore,

$$
\sqrt{\sin ^{2} 2 x+a \cos ^{2} x} \geqq \sqrt{\sin ^{2} 2 x}=|\sin 2 x|
$$

which means the function's value is never negative, and it takes the value 0 where $\cos x=0$, that is,

$$
x=x_{m}=\frac{\pi}{2}+k \pi \quad(k \text { integer })
$$

Thus, the minimum value of the function is 0.

Let's denote the function value by $y$. From the equation defining it, we form an equation that contains only one trigonometric function. Squaring and rearranging $y-\sin 2 x$,

$$
y^{2}-2 y \sin 2 x=a \cos ^{2} x
$$

From here, squaring $2 y \sin 2 x=4 y \sin x \cos x$, we get an equation that contains only $\cos x$:

$$
16 y^{2}\left(1-\cos ^{2} x\right) \cos ^{2} x=y^{4}-2 a y^{2} \cos ^{2} x+a^{2} \cos ^{4} x
$$

or, rearranging in powers of $\cos x$,

$$
\left(a^{2}+16 y^{2}\right) \cos ^{4} x-2 y^{2}(a+8) \cos ^{2} x+y^{4}=0
$$

This is a quadratic equation in $\cos ^{2} x$. $y$ can only take values for which this equation has a solution, so the discriminant, or a quarter of it, must be non-negative:

$$
y^{4}(a+8)^{2}-y^{4}\left(a^{2}+16 y^{2}\right)=16 y^{4}\left(a+4-y^{2}\right) \geqq 0
$$

This holds if $y=0$ or if the second factor is non-negative, thus

$$
y \leqq \sqrt{a+4}
$$

The value $\sqrt{a+4}$ can be taken by $y$ for the $x$ value where (from the above equation)

$$
\cos ^{2} x=\frac{a+4}{a+8}
$$

This is a positive value less than 1, since $a>0$, so there is an $x$ for which

$$
\cos x=\sqrt{\frac{a+4}{a+8}}, \text { or for which } \cos x=-\sqrt{\frac{a+4}{a+8}}
$$

Substitution shows that the function indeed takes the value $\sqrt{a+4}$ at the first mentioned points (not at the latter). Thus, the maximum value of the function is $\sqrt{a+4}$.

II. solution. To determine the maximum from the zeros of the derivative, we introduce the variable $t / 2$ instead of $x$, so we are looking for the maximum of the function $f(x)=f(2 t)=g(t)$. Thus,

$$
\begin{gathered}
f(x)=g(t)=\sin t+\sqrt{\sin ^{2} t+\frac{a}{2}(1+\cos t)} \\
g^{\prime}(t)=\cos t+\frac{\sin t \cos t-\frac{a}{4} \sin t}{\sqrt{\sin ^{2} t+\frac{a}{2}(1+\cos t)}}
\end{gathered}
$$

The denominator cannot be 0 at the maximum, so we exclude the points $t=\pi+2 k \pi$, and since $g(t)$ is periodic with $2 \pi$, we only need to consider $0 \leqq t0$ because the first term and the denominator of the second term are positive, while the numerator is

$$
-\frac{a(a+4)}{4(a+8)} \sin t
$$

which is positive in the $(3 \pi / 2,2 \pi)$ interval, so $g^{\prime}(t)=0$ has no roots there. For the $t_{M}$ value in the $(0, \pi / 2)$ interval, where

$$
\cos t_{M}=\frac{a}{a+8}, \quad \sin t_{M}=\frac{4 \sqrt{a+4}}{a+8}
$$

$g^{\prime}\left(t_{M}\right)=0$, since the value of the denominator is $\frac{\sqrt{(a+4)^{3}}}{a+8}$.

Since $g^{\prime}(0)=1>0$ and $g^{\prime}(\pi / 2)<0$ at the points $0$ and $\pi / 2$ that bracket $t_{M}$, there is a maximum there, and its value is

$$
\frac{4 \sqrt{a+4}}{a+8}+\frac{(a+4) \sqrt{a+4}}{a+8}=\sqrt{a+4}
$$

Zoltán Papp (Debrecen, Fazekas M. Gymnasium)