Explanation

(d)
Given equation of curve is $y = x^{2} - x + 4$
Slope of tangent at $P(1,\ 4)$ is
$\left( \frac{dy}{dx} \right) = 2x - 1\ \ \ \Rightarrow \ \ \ \ \ \ \left( \frac{dy}{dx} \right)_{(1,\ 4)} = 1$
$\therefore$ Equation of tangent is
$y - 4 = 1(x - 1) \Rightarrow \ \ \ y - x = 3$ ...(i)
And equation of normal at point $P(1,\ 4)$ is
$y - 4 = - 1(x - 1)\ \Rightarrow \ \ \ \ \ x + y = 5$ ...(ii)
Since the tangent cuts $x$-axis at $A( - 3,\ 0)$
And the normal cuts $x$-axis at $B(5,\ 0)$
$\therefore$ Area of $\mathrm{\Delta}PAB = \frac{1}{2}\ \left| \left| \begin{matrix}
1 & 4 & 1 \\
- 3 & 0 & 1 \\
5 & 0 & 1
\end{matrix} \right| \right|$
$= \frac{1}{2}\ \left| \left\lbrack - 4( - 3 - 5) \right\rbrack \right| = 16$ sq units