Explanation

(a)
$\because y = \frac{a + bx^{3/2}}{x^{5/4}}$
On differentiating w.r.t. $x$, we get
$y^{'} = \frac{\frac{3}{2}bx^{7/4} - \frac{5}{4}\left( a + bx^{3/2} \right)x^{1/4}}{\left( x^{\frac{5}{4}} \right)^{2}}$
$\because y^{'} = 0\ at\ x = 5$
$\therefore\frac{3}{2}bx^{7/4} - \frac{5}{4}\left( a + bx^{\frac{3}{2}} \right)x^{1/4} = 0,\ at\ x = 5$
$\Rightarrow 6bx^{3/2} - 5\left( a + bx^{3/2} \right) = 0,\ at\ x = 5$
$\Rightarrow bx^{3/2} = 5a,\ at\ x = 5 \Rightarrow b(5)^{3/2} = 5a$
$\Rightarrow \frac{a}{b} = \frac{5^{3/2}}{5} \Rightarrow a:b = \sqrt{5}:1$