Prove that function has only one maximum

Setting $x=\cos 2t $ we get $1-x =2\sin^2 t ,$ $1+x =2\cos^2 t ,$ $1 -x -\sqrt{1-x^2 } =2\sin^2 t -\sin 2t =2\sin t (\sin t -\cos t ),$ $1 +x -\sqrt{1-x^2 } =2\cos^2 t -\sin 2t =2\cos t (\cos t -\sin t )$ So we get a equation $$\sin^{2n} t + 2n (\cos^2 t -\sin^2 t ) -\cos^{2n} t =0$$ Hence $$(\sin t -\cos t )\cdot [(\cos t + \sin t) \sum_{k=0}^{n-1} \sin^{2k} t\cos^{2n -2 -2k} t + 2n (\cos t + \sin t )]=0$$ But since $t\in [0, 0.25 \pi ]$ the abowe expression i square bracket is positive hence $\sin t = \cos t $ thus $x=\cos 2t =\cos (0.5 \pi )=0$ is the only zero of derivative.

MotylaNogaTomkaMazura has found a very good parametrization $x=\cos(2t)$. (but it seems that his/her answer has an error) This answer uses this parametrization.

If you parametrize $x=\cos(2t)$ with $t \in [0,\pi/4]$, the original function can be written as $$ g_k(t)=\frac{\cos ^k(t)-\sin ^k(t)}{\cos (t)-\sin (t)},$$ with $k=2n+1$ and omitting a factor $2^{1 + n}$.

True. Let $c:=\cos(t),s:=\sin(t)$. For $k=2n+1\ge 7$,

$$\begin{align}\frac{d}{dt}\left(\frac{c^k-s^k}{c-s}\right)&=\frac{(kc^{k-1}(-s)-ks^{k-1}c)(c-s)-(c^k-s^k)(-s-c)}{(c-s)^2}\\&=\frac{-k(c^{k-1}s+s^{k-1}c)(c-s)+(c^k-s^k)(s+c)}{(c-s)^2}\end{align}$$

Here, let $$F(t):=-k(c^{k-1}s+s^{k-1}c)(c-s)+(c^k-s^k)(s+c)$$

Then,

$$F'(t)=-k((k-1)c^{k-2}(-s)s+c^{k-1}c+(k-1)s^{k-2}c^2+s^{k-1}(-s))(c-s)$$$$-k(c^{k-1}s+s^{k-1}c)(-s-c)+(kc^{k-1}(-s)-ks^{k-1}c)(s+c)+(c^k-s^k)(c-s)$$ $$\begin{align}&=-k(-(k-1)c^{k-2}s^2+c^{k}+(k-1)s^{k-2}c^2-s^{k})(c-s)+(c^k-s^k)(c-s)\\&=(c-s)(k(k-1)c^{k-2}s^2-k(k-1)s^{k-2}c^2+c^k-kc^{k}-s^k+ks^{k})\\&=(k-1)(c-s)(kc^{k-2}s^2-ks^{k-2}c^2+s^k-c^k)\\&=c^k(k-1)(c-s)(k\tan^2(t)-k\tan^{k-2}(t)+\tan^k(t)-1)\end{align}$$

Here, let $$G(u):=ku^2-ku^{k-2}+u^k-1$$ Then, $$G'(u)=2ku-k(k-2)u^{k-3}+ku^{k-1}$$ $$G''(u)=2k-k(k-2)(k-3)u^{k-4}+k(k-1)u^{k-2}$$ $$\begin{align}G'''(u)&=-k(k-2)(k-3)(k-4)u^{k-5}+k(k-1)(k-2)u^{k-3}\\&=k(k-2)u^{k-5}((k-1)u^2-(k-3)(k-4))\end{align}$$

This is negative, so $G''(u)$ is decreasing with $G''(0)=2k\gt 0$ and $G''(1)=-k(k-1)(k-5)\lt 0$.

So, there exists only one real $0\lt\alpha\lt 1$ such that $G''(\alpha)=0$, and $G'(u)$ is increasing for $0\lt u\lt\alpha$ and is decreasing for $\alpha\lt u\lt 1$ with $G'(0)=0$ and $G'(1)=-k(k-5)\lt 0$.

So, there exists only one real $0\lt \beta\lt 1$ such that $G'(\beta)=0$, and $G(u)$ is increasing for $0\lt u\lt\beta$ and is decreasing for $\beta\lt u\lt 1$ with $G(0)=-1\lt 0$ and $G(1)=0$.

So, there exists only one real $0\lt \gamma\lt 1$ such that $G(\gamma)=0$.

It follows from this that there exists only one real $0\lt\arctan\gamma\lt\pi/4$ such that $F'(\arctan\gamma)=0$. So, $F(t)$ is decreasing for $0\lt t\lt\arctan\gamma$ and is increasing for $\arctan\gamma\lt t\lt \pi/4$ with $F(0)=1\gt 0$ and $F(\pi/4)=0$.

It follows from this that $f_n(x)$ where $x\in ]0,1[$ has only one critical point for every value of $n \in \{3,4,5,\dots\}$. $\blacksquare$