Find the gradient function of the curve $y = 2x^2 - \ln x + c$. The curve touches the positive $x$-axis at point $T$. Find the coordinates of $T$. Show that $c = -\ln 2\sqrt{e}$. Find the value of $\frac{\text{d}^2y}{\text{d}x^2}$ at $T$ and hence determine the nature of point $T$.

1. Differentiate the given equation with respect to $x$

$\frac{\text{d}y}{\text{d}x} = \frac{\text{d}}{\text{d}x}(2x^2) - \frac{\text{d}}{\text{d}x}(\ln x) + \frac{\text{d}}{\text{d}x}(c)$
$\frac{\text{d}y}{\text{d}x} = 4x - \frac{1}{x}$

1. Set the gradient function to zero to find the $x$-coordinate of $T$

Since the curve touches the $x$-axis at $T$, the gradient of the curve at $T$ is $0$.
From (i), $\frac{\text{d}y}{\text{d}x} = 4x - \frac{1}{x}$.
$4x - \frac{1}{x} = 0$
$4x^2 - 1 = 0$
$x^2 = \frac{1}{4}$
Since $T$ is on the positive $x$-axis, $x > 0$, so $x = \frac{1}{2}$.

2. State the coordinates of $T$

Since $T$ lies on the $x$-axis, its $y$-coordinate is $0$.
Therefore, the coordinates of $T$ are $\left(\frac{1}{2}, 0\right)$.

1. Substitute the coordinates of $T$ from part (ii) into the equation of the curve

From part (ii), the curve touches the $x$-axis at $T$, where $\frac{\text{d}y}{\text{d}x} = 4x - \frac{1}{x} = 0 \implies x = \frac{1}{2}$. Thus, $T$ is $(\frac{1}{2}, 0)$.
Substitute $x = \frac{1}{2}$ and $y = 0$ into $y = 2x^2 - \ln x + c$:
$0 = 2\left(\frac{1}{2}\right)^2 - \ln\left(\frac{1}{2}\right) + c$

2. Simplify the equation to express $c$ in terms of $\ln 2$

$0 = 2\left(\frac{1}{4}\right) - (\ln 1 - \ln 2) + c$
$0 = \frac{1}{2} + \ln 2 + c$
$c = -\frac{1}{2} - \ln 2$

3. Use logarithm laws to manipulate the expression into the required form

$c = -\left(\ln 2 + \frac{1}{2}\right)$
Since $\frac{1}{2} = \ln e^{\frac{1}{2}} = \ln \sqrt{e}$:
$c = -(\ln 2 + \ln \sqrt{e})$
$c = -\ln(2\sqrt{e})$ (Shown)

1. Find the second derivative of $y$ with respect to $x$.

From part (i), $\frac{\text{d}y}{\text{d}x} = 4x - \frac{1}{x}$.

$\frac{\text{d}^2y}{\text{d}x^2} = \frac{\text{d}}{\text{d}x}\left(4x - x^{-1}\right) = 4 + x^{-2} = 4 + \frac{1}{x^2}$

2. Evaluate the second derivative at point $T$.

From part (ii), the $x$-coordinate of $T$ is $\frac{1}{2}$.

At $x = \frac{1}{2}$,
$\frac{\text{d}^2y}{\text{d}x^2} = 4 + \frac{1}{(\frac{1}{2})^2} = 4 + 4 = 8$

3. Determine the nature of point $T$.

Since $\frac{\text{d}^2y}{\text{d}x^2} = 8 > 0$, $T$ is a minimum point.