Question

A cubic polynomial P(x) is such that $P\left(1\right)=1$, $P\left(2\right)=2$, $P\left(3\right)=3$ and $P\left(4\right)=5$. The value of $P\left(6\right)$ is:

• A. 7
• B. 10
• C. 13
• D. 16

Answer 1

Answer: (D)

16

Given $P\left(1\right)=1,P\left(2\right)=2,P\left(3\right)=3,P\left(4\right)=5$

Let $f\left(x\right)=\left(P\right(x)-x)$

$f\left(1\right)=P\left(1\right)-1=1-1=0$

$f\left(2\right)=P\left(2\right)-2=2-2=0$

$f\left(3\right)=P\left(3\right)-3=3-3=0$

$\therefore f\left(x\right)=0,x=1,2,3$

$\Rightarrow f\left(x\right)=a(x-1)(x-2)(x-3)$

$P\left(x\right)=a(x-1)(x-2)(x-3)+x$

Put $x=4$

$5=a\left(3\right)\left(2\right)\left(1\right)+4$

$\Rightarrow a=\frac{1}{6}$
$\therefore P\left(x\right)=\frac{1}{6}(x-1)(x-2)(x-3)+x$
$\therefore P\left(6\right)=\frac{1}{6}\left(5\right)\left(4\right)\left(3\right)+6=16$