Question

A cubic polynomial $P(x)$ is such that $P(1)=1,P(2)=2,P(3)=3$ and $P(4)=5$, then $P(6)$ is equal to

• A. 7
• B. 10
• C. 13
• D. 16

Answer 1

Answer: (D)

Let $P(x)=A(x-1)(x-2)(x-3)+x$
Now, $P(4)=5$
$\Rightarrow 5=A(3)(2)(1)+4$
$\Rightarrow A=\frac{1}{6}$
So, $P(x)=\frac{1}{6}(x-1)(x-2)(x-3)+x$
$\therefore P(6)=16$