Numerical Differentiation Questions and Answers

Answer: B

The formula gives \(y'(2) = 8\).

Here \(h = 1\) and \(x_{k} = 2\), so the forward differences start from \(y = 2\). From the values \(2, -3, 4, -5\):

\[ \begin{aligned} \Delta y &: -5,\ 7,\ -9 \\ \Delta^{2} y &: 12,\ -16 \\ \Delta^{3} y &: -28 \end{aligned} \]

Substituting the leading differences into the given formula:

\[ \begin{aligned} y'(2) &\approx -5 + \tfrac{1}{2}(12) - \tfrac{1}{4}(-28) \\ &= -5 + 6 + 7 = 8 \end{aligned} \]

The paper prints its own coefficients (\(+\frac{1}{2}, -\frac{1}{4}\)), and the answer follows them as given. Using the differences that start at \(x = 1\) instead, or dropping the third-difference term (which gives 1), leads away from 8; 4, 12 and 16 come from such slips.

Answer: D

The value of \(\frac{1}{\alpha\beta}\) is 12.

Differentiate the Lagrange interpolating polynomial on the nodes -1, 0, 2 and evaluate at 0. Each coefficient is the derivative at 0 of the corresponding basis polynomial:

\[ \begin{aligned} \ell_{-1}(x) &= \frac{x(x-2)}{3},\ \ \ell_{-1}'(0) = -\frac{2}{3} \\ \ell_{0}(x) &= -\frac{(x+1)(x-2)}{2},\ \ \ell_{0}'(0) = \frac{1}{2} \\ \ell_{2}(x) &= \frac{x(x+1)}{6},\ \ \ell_{2}'(0) = \frac{1}{6} \end{aligned} \]

The first coefficient matches the given -2/3, which confirms the set-up. So \(\alpha = 1/2\) and \(\beta = 1/6\), giving \(\alpha\beta = 1/12\) and \(1/(\alpha\beta) = 12\).

Check: the coefficients must sum to 0 (a constant has zero derivative): -2/3 + 1/2 + 1/6 = 0. The options 3, 6 and 9 come from reading 1/\(\alpha\) or 1/\(\beta\) alone or from a slip in one basis polynomial.