Question

Suppose the polynomial \(p(x)=a+bx+cx^{2}+dx^{3}\) interpolates the data \((-1,1),\ (0,3),\ (1,2),\ (2,4)\). Then which one of the following statements is correct?

• A. \(a=-2c,\ d=-2b\)
• B. \(a=2c,\ d=2b\)
• C. \(b=3c,\ a=2d\)
• D. \(b=2c,\ a=3d\)

Hint:

When a cubic interpolates four points, plug them directly to form four linear equations in $a,b,c,d$. Solve systematically—often one value (like $a$ from $x=0$) drops out immediately and simplifies the system.

Answer 1

The Correct Option is A

Step 1: Set up equations from the four points.
\[ \begin{aligned} x=0:& a=3.
x=1:& a+b+c+d=2 \ \Rightarrow\ b+c+d=-1 (1)
x=-1:& a-b+c-d=1 \ \Rightarrow\ -b+c-d=-2 \ \Rightarrow\ b-c+d=2 (2)
x=2:& a+2b+4c+8d=4 \ \Rightarrow\ 2b+4c+8d=1 (3) \end{aligned} \]

Step 2: Solve for $b,c,d$.
From (1): \(b= -1-c-d\). Substitute in (2): \[ (-1-c-d)-c+d=2 \ \Rightarrow\ -1-2c=2 \ \Rightarrow\ c=-\frac{3}{2}. \] Then \(b= -1 - (-\tfrac{3}{2}) - d = \tfrac{1}{2} - d\). Use (3): \[ 2b+4c+8d=(1-2d)+(-6)+8d = -5+6d = 1 \ \Rightarrow\ d=1. \] Hence \(b=\tfrac{1}{2}-1=-\tfrac{1}{2}\) and \(a=3\).

Step 3: Check the listed relations.
\[ a=-2c \ \ (\text{since } -2\cdot(-\tfrac{3}{2})=3=a), \qquad d=-2b \ \ (\text{since } -2\cdot(-\tfrac{1}{2})=1=d). \] Thus (A) holds; the others do not. Final Answer:\ \(\boxed{\text{Option (A)}}\)