The least value of '$a$' for which the equation $\frac{4}{\sin x}+\frac{1}{1-\sin x}=a$ has atleast one solution on the interval $\left(0, \frac{\mathrm{\pi }}{2}\right)$ is -

$\mathrm{I}$. A differentiable function '$f$' with maximum at $x=c \Rightarrow f^{\text{'}\text{'}}\left(c\right)<0$.

$\mathrm{II}$. Antiderivative of a periodic function is also a periodic function.

$\mathrm{III}$. If $f$ has a period $T$ then for any $a\in R,{\int }_0^{T}f\left(x\right)dx={\int }_0^{T}f\left(x+a\right)dx$

$\mathrm{IV}$. If $f\left(x\right)$ has a maxima at $x=c$, then '$f$' is increasing in $\left(c-h, c\right)$ and decreasing in $\left(c, c+h\right)$ as $h\to 0$ for $h>0$.

Now, indicate the correct alternative.