Question

A cubic polynomial $\mathit{p}\left(x\right)$ such that $p\left(1\right)=1,p\left(2\right)=2,p\left(3\right)=3andp\left(4\right)=5$, then the value of $\mathit{p}\left(6\right)$ is:

• A. $16$
• B. $13$
• C. $10$
• D. $7$

Answer 1

The correct option is A

$16$

Solution:

Step1: Find the roots:

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

$$\begin{gathered} Letg\left(x\right)=p\left(x\right)-x \\ \Rightarrow g\left(1\right)=p\left(1\right)-1 \\ \Rightarrow g\left(1\right)=0 \\ \Rightarrow g\left(2\right)=p\left(2\right)-2 \\ \Rightarrow g\left(2\right)=2-2[Substitutegivenvaluesofp(2)=2] \\ \Rightarrow g\left(2\right)=0 \\ \Rightarrow g\left(3\right)=p\left(3\right)-3 \\ \Rightarrow g\left(3\right)=0[Substitutegivenvaluesofp(3)=3] \end{gathered}$$

Thus, $g\left(x\right)$ has three roots $1,2,and3$. Also $p\left(x\right)$ is cubic which means $g\left(x\right)=p\left(x\right)-x$ is also cubic.

Step 2: Find the cubic equation and value of k:

Hence, $g\left(x\right)=k(x-1)(x-2)(x-3)...\left(1\right)$ for some non-zero real number $k$.

Now,

$$\begin{gathered} g\left(4\right)=p\left(4\right)-4 \\ \Rightarrow g\left(4\right)=5-1[Substitutegivenvaluesofp(4)=5] \\ \Rightarrow g\left(4\right)=1 \\ \Rightarrow 1=k\left(3\right)\left(2\right)\left(1\right)[Substitutex=4ineqn(1\left)\right] \\ \Rightarrow k=\frac{1}{6} \end{gathered}$$

Step 3: Find the value of $\mathbf{p}\mathbf{\left(}\mathbf{6}\mathbf{\right)}$:

$$\begin{gathered} \therefore g\left(x\right)=\frac{1}{6}(x-1)(x-2)(x-3) \\ \Rightarrow p\left(x\right)=g\left(x\right)+x \\ \Rightarrow p\left(x\right)=\frac{1}{6}(x-1)(x-2)(x-3)+x \\ \Rightarrow p\left(6\right)=\frac{1}{6}(6-1)(6-2)(6-3)+6 \\ \Rightarrow p\left(6\right)=(\frac{1}{6}\times 5\times 4\times 3)+6 \\ \Rightarrow p\left(6\right)=10+6 \\ \Rightarrow p\left(6\right)=16 \end{gathered}$$

Final answer: Hence, correct option is A.