I have following equation $$(1-x)-(1-x)\frac{a}{(1-p)}x^{-\frac{k}{2}}=g$$ where $k>2$, $0<p<1$ $a>0$ and $g>0$. I want to know how will the roots behave with $p$. In essence will the roots become higher and higher with decreasing value of $p$? Any help will be much appreciated. Thanks in advance.
My Comment:
I think the very first root will definitely decrease with the increase in $p$. But I am not sure about the other roots.
The main idea in this answer is to have a form $$F(x)=p$$ and consider the graphs of $y=F(x)$ and $y=p$ which is a line parallel to $x$-axis.
Let us separate it into cases. In each case, drawing graphs of functions should help.
Note that $x\not=0$ and $x\not=1-g$.
Case 1 : $g=1$ $$1+(1-x)ax^{-\frac{k}{2}-1}=p$$ Let $f(x)=1+(1-x)ax^{-\frac{k}{2}-1}$. Then, we have $$\lim_{x\to 0^+}f(x)=\infty,\quad \lim_{x\to\infty}f(x)=1,\quad f'(x)=\frac{x-\left(1+\frac{2}{k}\right)}{\frac{2}{ak}x^{\frac k2+2}},\quad f\left(1+\frac 2k\right)\lt 1$$ Case 1-1 : $f(1+\frac 2k)\le 0$
There are two distinct positive roots $\alpha\lt\beta$ for any $0\lt p\lt 1$. With the increase in $p$, we see that $\alpha$ is decreasing and that $\beta$ is increasing.
Case 1-2 : $f(1+\frac 2k)\gt 0$
If $0\lt p\lt f(1+\frac 2k)$, then there are no positive roots. If $p=f(1+\frac 2k)$, then there is only one positive root $x=1+\frac 2k$. If $f(1+\frac 2k)\lt p\lt 1$, then there are two distinct positive roots $\alpha\lt\beta$. With the increase in $p$, we see that $\alpha$ is decreasing and that $\beta$ is increasing.
In the following, $g\not=1$.
We have $$\frac{(x-1)a}{(1-x-g)x^{\frac k2}}+1=p$$ Let $h(x)$ be the LHS.
Then, we have $$\lim_{x\to\infty}h(x)=1,\quad h'(x)=\frac{i(x)}{\frac 2a(1-x-g)^2x^{\frac k2+1}}$$ where$$i(x)=kx^2+(-2g-2k+kg)x+k(1-g),\quad \Delta :=(-2g-2k+kg)^2-4k^2(1-g)$$
$$(\text{axis of the parabola $y=i(x)$})=1-\frac{g(k-2)}{2k}\lt 1$$
Case 2 : $g\gt 1$ $$\lim_{x\to 0^+}h(x)=\infty,\quad i(0)=k(1-g)\lt 0$$
Case 2-1 : $0\lt h(\delta)\lt 1$ where $\delta$ is the only positive root of $i(x)$
If $0\lt p\lt h(\delta)$, then there are no positive roots. If $p=h(\delta)$, then there is only one positive root $x=\delta$. If $h(\delta)\lt p\lt 1$, then there are two distinct positive roots $\alpha\lt\beta$. With the increase in $p$, we see that $\alpha$ is decreasing and that $\beta$ is increasing.
Case 2-2 : $h(\delta)\le 0$
There are two distinct positive roots $\alpha\lt\beta$ for any $0\lt p\lt 1$. With the increase in $p$, we see that $\alpha$ is decreasing and that $\beta$ is increasing.
Case 3 : $g\lt 1$ $$\lim_{x\to 0^+}h(x)=-\infty,\quad\lim_{x\to (1-g)^{\pm}}h(x)=\mp\infty,\quad i(0)=k(1-g)\gt 0$$
Case 3-1 : $\Delta\gt 0$
Let $r_1\lt r_2$ be the roots of $i(x)$. Then, we have $$r_1\lt 1,\quad h\left(r_1\right)\gt 1,\quad h\left(r_2\right)\lt 1$$
Case 3-1-1 : $0\lt h\left(r_2\right)\lt 1$
If $0\lt p\lt h\left(r_2\right)$, then there are two positive roots $\alpha\lt\beta$ such that $0\lt\alpha\lt 1-g\lt \beta$. With the increase in $p$, we see that both $\alpha,\beta$ are increasing. If $p=h\left(r_2\right)$, then there are three positive roots $\alpha\lt\beta\lt\gamma$ such that $0\lt\alpha\lt 1-g\lt \beta\lt \gamma=r_2$. If $h\left(r_2\right)\lt p\lt 1$, then there are four positive roots $\alpha\lt\beta\lt\gamma\lt\psi$ such that $0\lt\alpha\lt1-g\lt\beta\lt r_1\lt\gamma\lt r_2\lt\psi$. With the increase in $p$, we see that $\alpha,\beta,\psi$ are increasing and that $\gamma$ is decreasing.
Case 3-1-2 : $h\left(r_2\right)\le 0$
There are four positive roots $\alpha\lt\beta\lt\gamma\lt\psi$ such that $0\lt\alpha\lt1-g\lt\beta\lt r_1\lt\gamma\lt r_2\lt\psi$ for any $0\lt p\lt 1$. With the increase in $p$, we see that $\alpha,\beta,\psi$ are increasing and that $\gamma$ is decreasing.
Case 3-2 : $\Delta\le 0$
Since $i(x)\ge 0$, we see that $y=h(x)$ is increasing. So, there are two distinct positive roots $\alpha\lt\beta$ such that $0\lt\alpha\lt 1-g\lt \beta$ for any $0\lt p\lt 1$. With the increase in $p$, we see that both $\alpha$ and $\beta$ are increasing.