Generalization of the triangular distribution: Does this distribution have a name I was trying to generalize the triangular distribution over $[0,1]$ to get a distribution that has the same unimodal structure and the same density of $0$ at the bounds, but where the spread of the distribution can be controlled using a parameter. So I came up with a distribution that is defined by
$f_{c,k}(x) = 
\begin{cases}
  \frac{(k+1)x^k}{c^k} \text{ for } x \leq c\\
  \frac{(k+1)(1-x)^k}{(1-c)^k} \text{ for } x > c
\end{cases}$
with $c\in[0,1]$ defining the mode and $k\in \mathbb{R}_{>0}$ defining the spread of the values.
If I did not mess up the integrals, this should actually be a valid distribution.
It is clear that the mode of this distribution is $c$ and I was also able to get an expression for the mean (don't have my notes with me right now). It seems variance should also be defined, but I haven't worked it out, yet.
Is this distribution known and does it have a name? I couldn't find anything about this both via google or wikipedia.
EDIT:
Here are some plots to indicate what this distribution looks like.
At $k=1$ this is just the triangular distribution:

If $k>1$ then the flanks of the distribution become steeper, therefore centering the values:

If $k<1$ then the flanks of the distribution become wider, therefore allowing for a larger variance of the values:

 A: You can create a huge number of such distributional families by following the process I described recently at https://stats.stackexchange.com/a/490160/919.
Begin with any non-negative bounded integrable function $f$ on $[0,1]$ that (i) has a unique maximum and (ii) vanishes at the endpoints.  For any $n\ge 1$ define $$f_n(t)=c_n\exp(n\log f(t)) = f(t)^{1/n}$$ (setting $f_n(t)=0$ wherever $f(t)=0$) where $c_n$ makes the integral of $f_n$ equal to $1:$ it always exists under the assumptions. You may confirm that the variance of the distribution PDF given by $f_n$ decreases down to $0$ as $n$ grows, thereby controlling the spread.
For instance, you could let $f$ by any of the functions graphed in the question, thereby extending each of them to a one-dimensional family of distributions whose variances you can control.

As an example, pick $a \gt 0$ and $b\gt 0$ and define $f(t;a,b)=t^a(1-t)^b.$ Evidently
$$f_n(t;a,b) = c_n\,t^{an}(1-t)^{bn} = c_n t^{\alpha(n)-1}(1-t)^{\beta(n)-1}$$
where $\alpha(n) = 1+an$ and $\beta(n)=1+bn.$  This is the PDF of the Beta$(\alpha(n),\beta(n))$ distribution..  Its variance is
$$\begin{aligned}
\sigma^2_n(a,b) &= \frac{\alpha(n)\beta(n)}{(\alpha(n)+\beta(n))^2(\alpha(n)+\beta(n)+1)} \\
&= \frac{(1+an)(1+bn)}{(2+(a+b)n)^2(3+(a+b)n)} \\
&= \left(\frac{9ab}{(a+b)^2}-2\right)\frac{1}{(a+b)n+3} \\&\quad-
\frac{(a-b)^2}{(a+b)^2}\frac{1}{((a+b)n+2)^2}\\&\quad+
2\frac{(a-b)^2}{(a+b)^2}\frac{1}{(a+b)n+2} 
\end{aligned}$$
The last expression makes this family of rational functions of $n$ straightforward to analyze.  Because the poles of its two terms are located at $-2/(a+b)$ and $-3/(a+b),$ it must asymptotically decrease with $n$ to zero, eventually being dominated by the last term.   We have recovered all the Beta distributions with zero densities at the endpoints.
A particularly simple example arises when we require $a=b,$ for then the last two terms vanish and
$$\sigma_n^2(a,a) = \frac{1}{4} \frac{1}{3 + 2an}$$
shows that you can find such a distribution with any variance $0 \lt \sigma^2 \lt 1/12$ by choosing positive $a$ and $n\ge 1$ so that $an = (1/(4\sigma^2) - 3)/2.$
