Question

Consider a polynomial as $f\left( x \right)={{x}^{4}}+a{{x}^{3}}+b{{x}^{2}}+cx+d$, such that, $f\left( 1 \right)=10,f\left( 2 \right)=20,f\left( 3 \right)=30$, then the value of $\dfrac{f\left( 12 \right)+f\left( -8 \right)}{10}$ is equal to
A.2018
B.1984
C.60
D.600

Answer 1

Hint:Substitute the values of $x=1,2,3$ in $f\left( x \right)={{x}^{4}}+a{{x}^{3}}+b{{x}^{2}}+cx+d$ then we get three equations in a, b, c & d. Now, substitute the values of $x=12,-8$ in $f\left( x \right)={{x}^{4}}+a{{x}^{3}}+b{{x}^{2}}+cx+d$ to get $f\left( 12 \right)\And f\left( -8 \right)$ and then substitute these values in the relation $\dfrac{f\left( 12 \right)+f\left( -8 \right)}{10}$. Simplify this expression by eliminating a, b, c & d.

Complete step-by-step answer:
It is given that:
$f\left( x \right)={{x}^{4}}+a{{x}^{3}}+b{{x}^{2}}+cx+d$
Substituting $x=1$ in the above equation we get,
$f\left( 1 \right)=1+a+b+c+d$
It is given that $f\left( 1 \right)=10$ so equating the above equation to 10 we get,
$\begin{align}
  & f\left( 1 \right)=1+a+b+c+d=10 \\
 & \Rightarrow a+b+c+d=9.......Eq.(1) \\
\end{align}$
Substituting $x=2$ in the given equation in $f\left( x \right)$ we get,
$f\left( 2 \right)=16+8a+4b+2c+d$
It is given that $f\left( 2 \right)=20$ so equating the above equation to 20 we get,
$\begin{align}
  & f\left( 2 \right)=16+8a+4b+2c+d=20 \\
 & \Rightarrow 8a+4b+2c+d=4........Eq.(2) \\
\end{align}$
Substituting $x=3$ in the given equation in $f\left( x \right)$ we get,
$f\left( 3 \right)=81+27a+9b+3c+d$
It is given that $f\left( 3 \right)=30$ so equating the above equation to 30 we get,
$\begin{align}
  & f\left( 3 \right)=81+27a+9b+3c+d=30 \\
 & \Rightarrow 27a+9b+3c+d=-51........Eq.(3) \\
\end{align}$
From the above we have got three equations,
$\begin{align}
  & a+b+c+d=9........Eq.(1) \\
 & 8a+4b+2c+d=4........Eq.(2) \\
 & 27a+9b+3c+d=-51........Eq.(3) \\
\end{align}$
Rewriting the eq. (1) we get,
$d=9-\left( a+b+c \right)$
Substituting the above value of “d” in eq. (2) and eq. (3) we get,
$\begin{align}
  & 8a+4b+2c+9-\left( a+b+c \right)=4 \\
 & \Rightarrow 7a+3b+c=-5..........Eq.(4) \\
\end{align}$
$\begin{align}
  & 27a+9b+3c+9-\left( a+b+c \right)=-51 \\
 & \Rightarrow 26a+8b+2c=-60 \\
 & \Rightarrow 13a+4b+c=-30........Eq.(5) \\
\end{align}$
Subtracting eq. (5) from eq. (4) we get,
$\begin{align}
  & \text{ }7a+3b+c=-5 \\
 & \dfrac{-\left( 13a+4b+c=-30 \right)}{-6a-b=25} \\
\end{align}$
Taking -1 common from the left hand side of the above equation we get,
$-1\left( 6a+b \right)=25$
$\Rightarrow 6a+b=-25.......Eq.\left( 6 \right)$
Solving this relation $\dfrac{f\left( 12 \right)+f\left( -8 \right)}{10}$ we are required to find the values of $f\left( 12 \right)\And f\left( -8 \right)$.
$f\left( 12 \right)={{\left( 12 \right)}^{4}}+a{{\left( 12 \right)}^{3}}+b{{\left( 12 \right)}^{2}}+c\left( 12 \right)+d$
$f\left( -8 \right)={{\left( -8 \right)}^{4}}+a{{\left( -8 \right)}^{3}}+b{{\left( -8 \right)}^{2}}-8c+d$
Substituting the above values in $\dfrac{f\left( 12 \right)+f\left( -8 \right)}{10}$ we get,
$\begin{align}
  & \dfrac{{{\left( 12 \right)}^{4}}+{{\left( -8 \right)}^{4}}+a\left[ {{\left( 12 \right)}^{3}}-{{\left( 8 \right)}^{3}} \right]+b\left[ {{\left( 12 \right)}^{2}}+{{\left( 8 \right)}^{2}}+4c+2d \right]}{10} \\
 & \Rightarrow \dfrac{20736+4096+a\left( 1728-512 \right)+b\left( 208 \right)+4c+2d}{10} \\
 & \Rightarrow \dfrac{24832+a\left( 1216 \right)+b\left( 208 \right)+4c+2d}{10} \\
 & \Rightarrow \dfrac{24832+a\left( 1216 \right)+b\left( 208 \right)+2\left( 2c+d \right)}{10}......Eq.\left( 7 \right) \\
\end{align}$
From eq. (2) we can find the value of 2c + d.
$8a+4b+2c+d=4$
Rearranging the above equation we get,
$2c+d=4-\left( 8a+4b \right)$
Substituting the above value in eq. (7) we get,
$\begin{align}
  & \dfrac{24832+a\left( 1216 \right)+b\left( 208 \right)+2\left( 2c+d \right)}{10} \\
 & \Rightarrow \dfrac{24832+a\left( 1216 \right)+b\left( 208 \right)+2\left( 4-\left( 8a+4b \right) \right)}{10} \\
 & \Rightarrow \dfrac{24840+a\left( 1216 \right)+b\left( 208 \right)+8-16a-8b}{10} \\
 & \Rightarrow 2484+\dfrac{1200a+200b}{10} \\
 & \Rightarrow 2484+20\left( 6a+b \right) \\
\end{align}$
Substituting the value of $\left( 6a+b \right)$ from eq. (6) in the above equation we get,
$6a+b=-25$
$\begin{align}
  & 2484+20\left( -25 \right) \\
 & =2484-500 \\
 & =1984 \\
\end{align}$
From the above solution, we have solved the value of $\dfrac{f\left( 12 \right)+f\left( -8 \right)}{10}$ as 1984.
Hence, the correct option is (b).

Note: This question demands a good command on rearrangement of a, b, c and d to get the required result of the given relation $\dfrac{f\left( 12 \right)+f\left( -8 \right)}{10}$. The value of 6a+b is known, so students must perform rearrangements so as to be able to use it in the final expression. Be careful about the calculations in this problem, you might make silly mistakes in addition, subtraction, multiplication and division while solving the algebraic expressions. If a student goes wrong at any step, then the final result will be affected.