Problem source

\begin{questionparts}
\item Show that the equation 
\[
(x-1)^{4}+(x+1)^{4}=c
\]
has exactly two real roots if $c>2,$ one root if $c=2$ and no roots
if $c<2$. 
\item How many real roots does the equation $\left(x-3\right)^{4}+\left(x-1\right)^{4}=c$
have?
\item How many real roots does the equation $\left|x-3\right|+\left|x-1\right|=c$
have?
\item How many real roots does the equation $\left(x-3\right)^{3}+\left(x-1\right)^{3}=c$
have?
\end{questionparts}
{[}The answers to parts \textbf{(ii)}, \textbf{(iii)} and \textbf{(iv)}
may depend on the value of $c$. You should give reasons for your
answers.{]}

Solution source

\begin{questionparts}
\item $\,$
\begin{align*}
&& c &= (x-1)^4+(x+1)^4 \\
&&&= 2x^4+12x^2+2 \\
\Rightarrow && 0 &= (x^2+6)^2-\frac{c}{2} - 35 \\
\Rightarrow && \underbrace{x^2+6}_{\geq 6} &= \pm \sqrt{35 + \frac{c}{2}}\\
\end{align*}

Therefore there are two solutions if $c > 2$, one solution if $c = 2$ and no solutions otherwise.

\item $\,$

This equation is the same equation if $y = x-2$, ie there are two solutions if $c > 2$, one solution if $c = 2$ and no solutions otherwise. 

\item Rewriting as $|x-1|+|x+1| = c$ we have

For $x < -1$: $1-x-1-x = -2x$
For $-1 \leq x \leq 1$: $1-x+x+1 = 2$
For $x > 1$: $x-1+x+1 = 2x$

Therefore there are infinitely many solutions if $c = 2$ (the interval $[-3,-1]$), two solutions if $c > 2$ and none otherwise.

\item Rewriting as $(x-1)^3+(x+1)^3$ we have $x^3+6x = c$. Notice that $3x^2+6 > 0$ so the function is increasing, ie there is one solution for all $c$
\end{questionparts}