I need to find the number of roots on $\left[k,1\right]$ of the following function:
$$f\left(x\right)=\mu\frac{1-2x}{x(1-x)} + \frac{2(2k+x)\sqrt{\tau\left(x-k\right)}}{3\sqrt{\pi}\rho}$$
for $\mu>0$, $\tau>0$, $\rho>0$ and $k\in\left(\frac{1}{2},1\right]$. Thanks for helping!
Hint: The function is concave, goes to $-\infty$ as $x \to 1-$ and is negative at $x=k$. So the number of roots is either $0$, $1$ or $2$, depending on the sign of the maximum value of the function.
EDIT: Dividing by $\mu$ and consolidating the constants into one constant, we're looking at roots of
$$g(x) = \frac{1-2x}{x(1-x)} + c (2k+x)\sqrt{x-k} $$
where $c > 0$. It may be easier to deal with rational functions, so let $x = k + t^2$, $0 \le t \le \sqrt{1-k}$, so
$$ g(x) = h(t) = \frac{1-2k-2t^2}{(t^2+k)(1-k-t^2)} + c t (t^2 + 3 k) $$ Where there is one solution, $h(t)$ is tangent to the $t$ axis, so both $h(t)=0$ and $h'(t) = 0$. The resultant of the numerator of $h(t)$ and the numerator of $h'(t)$ is
$$ R = {c}^{5} \left( 442368\,{c}^{8}{k}^{17}-442368\,{c}^{8}{k}^{16}-663552 \,{c}^{8}{k}^{15}+442368\,{c}^{8}{k}^{14}+470016\,{c}^{8}{k}^{13}- 82944\,{c}^{8}{k}^{12}-884736\,{c}^{6}{k}^{14}-138240\,{c}^{8}{k}^{11} +884736\,{c}^{6}{k}^{13}-27648\,{c}^{8}{k}^{10}+3833856\,{c}^{6}{k}^{ 12}-1806336\,{c}^{6}{k}^{11}-1539072\,{c}^{6}{k}^{10}+599040\,{c}^{6}{ k}^{9}+442368\,{c}^{4}{k}^{11}+864000\,{c}^{6}{k}^{8}-442368\,{c}^{4}{ k}^{10}-43776\,{c}^{6}{k}^{7}-5677056\,{c}^{4}{k}^{9}-188928\,{c}^{6}{ k}^{6}+2340864\,{c}^{4}{k}^{8}-39168\,{c}^{6}{k}^{5}+4999424\,{c}^{4}{ k}^{7}-2474368\,{c}^{4}{k}^{6}-1095408\,{c}^{4}{k}^{5}+497344\,{c}^{4} {k}^{4}+2506752\,{c}^{2}{k}^{6}+494836\,{c}^{4}{k}^{3}-976896\,{c}^{2} {k}^{5}-18300\,{c}^{4}{k}^{2}-7068160\,{c}^{2}{k}^{4}-72500\,{c}^{4}k+ 3516800\,{c}^{2}{k}^{3}-12500\,{c}^{4}+588240\,{c}^{2}{k}^{2}-418144\, {c}^{2}k+27407\,{c}^{2}+3200000\,k-1600000 \right) $$
Ignoring the $c^5$ factor, we have a rather ugly polynomial in $k$ and $c$ which is $0$ when there is one solution. As this is a quartic in $c^2$, in principle there is a "closed-form" solution for $c$ in terms of radicals, but it's going to be very ugly. Here's a plot obtained using Maple:
Below this curve, there are no solutions, above it there are two.