The curve is defined by $y=\sqrt{6x+5}$. The shaded area is enclosed by the curve, the $x$-axis and the lines $x=a$ and $x=2a$, where $a$ is a positive constant. When this shaded area is rotated through $360^\circ$ about the $x$-axis, a solid is produced. The solid has volume, $V$, with $V>46\pi$.
(a)[4]
Show that the inequality $9a^2+5a-46>0$ holds.
(b)[3]
Determine the range of possible values of $a$.
Worked solution & mark scheme
This 7-mark question has a full step-by-step worked solution and mark scheme. One marking point: “$V=\pi\int_a^{2a}(6x+5)\,dx$” …