$X \sim Beta(m,2)$ and $P \left(X\le\dfrac{1}{2}\right)=\dfrac{1}{2}$ find Variance 
Let $X$ follow a beta distribution with parameters $m(<0)$ and $2$. If
$P \left(X\le\dfrac{1}{2}\right)=\dfrac{1}{2},$ then $Var(X)$ ?
$(A)\ \ \dfrac{1}{10}\space\ \ \ (B)\ \ \dfrac{1}{20}\ \ \ (C)\ \
 \dfrac{1}{25} \ \ \ (D)\ \ \dfrac{1}{40}$

$V(X)=\dfrac{mn}{(m+n)^2(m+n+1)}=\dfrac{2m}{(m+2)^2(m+3)} \ \ \ \ \ \ ---->(1)$
$\int_{0}^{\frac{1}{2}}x^{m-1}(1-x)dx=\left(\frac{x^{m}}{m}\right)_{0}^{\frac{1}{2}}-\left(\frac{x^{m+1}}{m+1}\right)_{0}^{\frac{1}{2}}=\left(\frac{1}{2^{m}m}\right)-\left(\frac{1}{2^{m+1}(m+1)}\right)=\frac{B(m,2)}{2}\ \ \ \ \ \ ---->(2)$
Now I plugged in values by hit and trial
For $m=2$
$\dfrac{2m}{(m+2)^2(m+3)}=\dfrac{1}{20}$
also satisfies $(2)$
$\dfrac{1}{2^2\cdot2}-\dfrac{1}{24}=\dfrac{1}{12}=\dfrac{B(2,2)}{2}$
For $m=3$
$\dfrac{2m}{(m+2)^2(m+3)}=\dfrac{1}{25}$
but
$\dfrac{1}{2^3\cdot3}-\dfrac{1}{2^4 \cdot 4}=\dfrac{1}{12}\ne\dfrac{B(3,2)}{2}$
This question came $2$ marks so there must be an easy way out which I am not able to figure out. Please suggest me alternate method.

The simple approach is to say that if a $\mathrm{Beta}(a,b)$ distribution has $a>b$ then $\mathbb P(X \le \frac12) \lt \frac12$ and its median is above $\frac12$, while if $a<b$ then then $\mathbb P(X \le \frac12) \gt \frac12$ and its median is below $\frac12$.
This can be proved by comparing the densities at $x$ and $1-x$ and then integrating over the half intervals.  When $a=b$ you get $\mathbb P(X \le \frac12) = \frac12$ and a median of $\frac12$ by symmetry.
So here you have $m=2$ and $\mathrm{Beta}(2,2)$ is symmetric with a median of $\frac12$ and a variance of $\frac1{20}$.
