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(6) is:
A. 7
B. 10
C. 13
D. 16

Hint: Here we have been given a cubic polynomial P(x) with the values of it at x=1,2,3 and 4 and we have been asked to find the value of P(6). For this, we will first assume another cubic polynomial given as $f\left( x \right)=P\left( x \right)-x$ as the value of P(x) is equal to x at x=1,2,3. So we will get those as the zeroes of f(x) and hence we can write a polynomial with zeroes $\alpha ,\beta ,\gamma $ as $f\left( x \right)=k\left( x-\alpha \right)\left( x-\beta \right)\left( x-\gamma \right)$ where k is any constant. We will keep this equal to P(x)-x and then put x=4 which will give us the value of k. Hence, we will obtain the value of P(x). Then we will put x=6 in P(x) and hence obtain the required answer.

Complete step-by-step answer:
Now, we have been given that P(x) is a cubic polynomial. Now, we know that the degree of a cubic polynomial is 3. Thus, $P\left( x \right)-x$ will also be a cubic polynomial.
Hence, let us assume a function, f(x) defined as:
$f\left( x \right)=P\left( x \right)-x$
Here, f(x) will also be a cubic polynomial.
Now, we have been given that $P\left( 1 \right)=1$
Thus, putting x=1 in f(x), we get:
$\begin{align}
  & f\left( x \right)=P\left( x \right)-x \\
 & \Rightarrow f\left( 1 \right)=P\left( 1 \right)-1 \\
 & \Rightarrow f\left( 1 \right)=1-1 \\
 & \Rightarrow f\left( 1 \right)=0 \\
\end{align}$
Thus, we can say that x=1 is a zero of the polynomial f(x).
Now, we have also been given that $P\left( 2 \right)=2$
Thus, putting x=2 in f(x) we get:
$\begin{align}
  & f\left( x \right)=P\left( x \right)-x \\
 & \Rightarrow f\left( 2 \right)=P\left( 2 \right)-2 \\
 & \Rightarrow f\left( 2 \right)=2-2 \\
 & \Rightarrow f\left( 2 \right)=0 \\
\end{align}$
Thus, we can say that x=2 is a zero of the polynomial f(x).
Now, we have also been given that $P\left( 3 \right)=3$
Thus, putting x=3 in f(x) we get:
$\begin{align}
  & f\left( x \right)=P\left( x \right)-x \\
 & \Rightarrow f\left( 3 \right)=P\left( 3 \right)-3 \\
 & \Rightarrow f\left( x \right)=3-3 \\
 & \Rightarrow f\left( x \right)=0 \\
\end{align}$
 Thus, we can say that x=3 is a zero of the polynomial f(x).
Now, we know that since the polynomial f(x) is cubic, the maximum zeroes it can have is 3.
Thus, x=1, x=2, x=3 are the three zeroes of the polynomial f(x).
Now, we know that if for a cubic polynomial, there are three zeroes $\alpha ,\beta ,\gamma $, then the polynomial f(x) is written as:
$f\left( x \right)=k\left( x-\alpha \right)\left( x-\beta \right)\left( x-\gamma \right)$
Where k is any constant.
Thus, we can write f(x) as:
$f\left( x \right)=k\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)$ .....(i)
Where k is any constant belonging to real numbers.
Now, we took f(x) as:
$f\left( x \right)=P\left( x \right)-x$ .....(ii)
Now, from equation (i) and (ii), we get that:
$P\left( x \right)-x=k\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)$ .....(iii)
Now, we have been given $P\left( 4 \right)=5$.
Now, putting x=4 in equation (iii) we will get:
$\begin{align}
  & P\left( x \right)-x=k\left( x-1 \right)\left( x-2 \right)\left( x-3 \right) \\
 & \Rightarrow P\left( 4 \right)-4=k\left( 4-1 \right)\left( 4-2 \right)\left( 4-3 \right) \\
 & \Rightarrow 5-4=k\left( 3 \right)\left( 2 \right)\left( 1 \right) \\
 & \Rightarrow 1=6k \\
 & \Rightarrow k=\dfrac{1}{6} \\
\end{align}$
Thus, the value of k is $\dfrac{1}{6}$.
Thus, our equation (iii) no becomes:
$\begin{align}
  & P\left( x \right)-x=k\left( x-1 \right)\left( x-2 \right)\left( x-3 \right) \\
 & \Rightarrow P\left( x \right)-x=\dfrac{1}{6}\left( x-1 \right)\left( x-2 \right)\left( x-3 \right) \\
\end{align}$
Now, we get P(x) as:
$\begin{align}
  & P\left( x \right)-x=\dfrac{1}{6}\left( x-1 \right)\left( x-2 \right)\left( x-3 \right) \\
 & P\left( x \right)=\dfrac{1}{6}\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)+x \\
\end{align}$
Now, we have to find P(6). So putting x=6 in P(x) we get:
$\begin{align}
  & P\left( x \right)=\dfrac{1}{6}\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)+x \\
 & \Rightarrow P\left( 6 \right)=\dfrac{1}{6}\left( 6-1 \right)\left( 6-2 \right)\left( 6-3 \right)+6 \\
 & \Rightarrow P\left( 6 \right)=\dfrac{1}{6}\left( 5 \right)\left( 4 \right)\left( 3 \right)+6 \\
 & \Rightarrow P\left( 6 \right)=10+6 \\
 & \therefore P\left( 6 \right)=16 \\
\end{align}$
Thus, the value of P(6) is 16.

So, the correct answer is “Option D”.

Note: We could have done this question by assuming P(x) to be $P\left( x \right)=a{{x}^{3}}+b{{x}^{2}}+cx+d$ and then we could have made 4 equations by putting the given 4 values of P(x) at different values of x. Thus, we would have had 4 equations with 4 variables and hence we could have solved it and obtained the value of a, b, c and d and hence obtained P(x). But we didn’t do it because the process of solving 4 equations with 4 variables simultaneously would be a long process and the scope for committing mistakes would have also been greater. Hence, we went for the smaller and easier approach.