A curve is such that $\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} = -4\sin(2x+\pi)$ and the point $P\left(\dfrac{\pi}{2},\pi\right)$ lies on the curve. The gradient of the curve at $P$ is 3.

1. Integrate the second derivative to find the expression for the first derivative, including a constant of integration.

$\frac{\mathrm{d}y}{\mathrm{d}x} = \int -4\sin(2x+\pi) \, \mathrm{d}x$
$\frac{\mathrm{d}y}{\mathrm{d}x} = -4 \left( \frac{-\cos(2x+\pi)}{2} \right) + c$
$\frac{\mathrm{d}y}{\mathrm{d}x} = 2\cos(2x+\pi) + c$

2. Use the given gradient at point $P$ to find the constant of integration $c$.

At $P\left(\frac{\pi}{2}, \pi\right)$, the gradient $\frac{\mathrm{d}y}{\mathrm{d}x} = 3$.
$3 = 2\cos\left(2\left(\frac{\pi}{2}\right)+\pi\right) + c$
$3 = 2\cos(2\pi) + c$
$3 = 2(1) + c \implies c = 1$
So, $\frac{\mathrm{d}y}{\mathrm{d}x} = 2\cos(2x+\pi) + 1$

3. Integrate the first derivative to find the expression for $y$, including a second constant of integration.

$y = \int (2\cos(2x+\pi) + 1) \, \mathrm{d}x$
$y = 2 \left( \frac{\sin(2x+\pi)}{2} \right) + x + d$
$y = \sin(2x+\pi) + x + d$

4. Use the coordinates of point $P$ to find the constant of integration $d$.

At $P\left(\frac{\pi}{2}, \pi\right)$, $x = \frac{\pi}{2}$ and $y = \pi$.
$\pi = \sin\left(2\left(\frac{\pi}{2}\right)+\pi\right) + \frac{\pi}{2} + d$
$\pi = \sin(2\pi) + \frac{\pi}{2} + d$
$\pi = 0 + \frac{\pi}{2} + d \implies d = \frac{\pi}{2}$

5. State the final equation of the curve.

$y = \sin(2x+\pi) + x + \frac{\pi}{2}$

1. State the expression for the first derivative found in part (a)

From part (a), integrating $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = -4\sin(2x+\pi)$ and using the gradient at $P\left(\frac{\pi}{2},\pi\right)$ gives:
$\frac{\mathrm{d}y}{\mathrm{d}x} = 2\cos(2x+\pi) + 1$

2. Evaluate the first derivative at $x = 3$

Substitute $x = 3$ into the derivative:
$\left. \frac{\mathrm{d}y}{\mathrm{d}x} \right|_{x=3} = 2\cos(2(3)+\pi) + 1$
$\left. \frac{\mathrm{d}y}{\mathrm{d}x} \right|_{x=3} = 2\cos(6+\pi) + 1 \approx -0.920$

3. Conclude whether the function is increasing based on the sign of the derivative

Since $\frac{\mathrm{d}y}{\mathrm{d}x} \approx -0.920 < 0$, the gradient is negative.
Therefore, $y$ is decreasing when $x = 3$.