Let f : R $\to$ R be a function such that f(2) = 4 and f'(2) = 1. Then, the value of $\mathop {\lim }\limits_{x \to 2} {{{x^2}f(2) - 4f(x)} \over {x - 2}}$ is equal to :

This limit can be solved using L'Hopital's Rule, which states that for the limit of the form 0/0 or ±∞/±∞, the limit can be found by taking the derivative of the numerator and the derivative of the denominator separately. <br/><br/>$\mathop {\lim }\limits_{x \to 2} {{{x^2}f(2) - 4f(x)} \over {x - 2}}$ is in the indeterminate form, and we are given that f(2) = 4 and f'(2) = 1, so we can apply L'Hopital's rule. <br/><br/>Taking the derivative of the numerator and the denominator, we get : <br/><br/>Numerator: derivative of $x^2 f(2) - 4f(x)$ is $2x f(2) - 4f'(x)$. <br/><br/>Denominator: derivative of $x - 2$ is $1$. <br/><br/>So, the limit becomes : <br/><br/>$$\mathop {\lim }\limits_{x \to 2} {{{2xf(2) - 4f'(x)}} \over 1} = 2 \times 2 \times f(2) - 4 \times f'(2) = 16 - 4 = 12.$$ <br/><br/>Therefore, Option D, 12, is the correct answer.