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A cubic polynomial $P(x)$ is such that $P(1)=1,P(2)=2,P(3)=3,\text{ }and\text{ }P(4)=5$ . The value of $P(6)$ is:
$\begin{align}
  & A)7 \\
 & B)19 \\
 & C)13 \\
 & D)16 \\
\end{align}$

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Last updated date: 20th Jun 2024
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Answer
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Hint: In this question, we have to find the value of a cubic polynomial at 6. Thus, we start solving this problem by letting a function which is equal to the subtraction of cubic polynomial at a variable and the variable. Then, we will find f(x), where x=0, 1, 2. After that, we know that a cubic equation can be expressed as $f(x)=a(x-1)(x-2)(x-3)$ . After that we will find P(4) and then P(6), to get the required result for the problem.

Complete step by step solution:
According to the question, we have to find the value of $P(6)$ of a cubic equation.
Let there be a function, such that it is equal to the subtraction of cubic polynomial at a variable and the variable, that is
$f(x)=P(x)-x$ --------- (1)
Thus, now we will find f(1), f(2), and f(3) from equation (1), we get
$\begin{align}
  & f(1)=P(1)-1 \\
 & f(1)=1-1 \\
 & f(1)=0 \\
\end{align}$
Similarly,
$\begin{align}
  & f(2)=P(2)-2 \\
 & f(2)=2-2 \\
 & f(2)=0 \\
\end{align}$ , and
$\begin{align}
  & f(3)=P(3)-3 \\
 & f(3)=3-3 \\
 & f(3)=0 \\
\end{align}$
Thus, we get that $f(x)=0$ where x=1, 2, and 3.
Now, we will express the above equation in terms of a cubic polynomial as
$f(x)=a(x-1)(x-2)(x-3)$
$P(x)=a(x-1)(x-2)(x-3)+x$
Thus, $P(x)=f(x)+x$
$P(x)=a(x-1)(x-2)(x-3)+x$ -------- (2)
Now, we will put x = 4 and the value of P(4) from the question in equation (2), we get
$P(4)=a(4-1)(4-2)(4-3)+4$
On further simplification, we get
$5=a(3)(2)(1)+4$
On solving the brackets, we get
$5=6a+4$
Now, we will subtract 4 on both sides in the above equation, we get
$5-4=6a+4-4$
As we know, the same terms with opposite signs cancel out each other, thus we get
$1=6a$
So, we will divide 6 on both sides in the above equation, we get
$\dfrac{1}{6}=\dfrac{6}{6}a$
Thus, on further simplification, we get
$a=\dfrac{1}{6}$ ------------ (3)
Therefore, putting the value of equation (3) in equation (2), we get
$P(x)=\dfrac{1}{6}\left( (x-1)(x-2)(x-3) \right)+x$
Now, let x = 6 in the above equation, we get
$P(6)=\dfrac{1}{6}\left( (6-1)(6-2)(6-3) \right)+6$
On solving the brackets of the above equation, we get
$\begin{align}
  & P(6)=\dfrac{1}{6}\left( (5)(4)(3) \right)+6 \\
 & P(6)=\dfrac{1}{6}\left( 60 \right)+6 \\
\end{align}$
On further simplification, we get
\[\begin{align}
  & P(6)=10+6 \\
 & P(6)=16 \\
\end{align}\]

Therefore, the value of P(6) for the cubic polynomial is equal to 16.

Note: While solving this question, make all the calculations properly to avoid error and confusion. In this type of question, the accurate answer is based on the value of a, which we find out through P(4), so make all the steps properly.