Question

If $f\left(x\right)=x^4+a{x}^3+b{x}^2+cx+d$, is a polynomial such that $f\left(1\right)=10,f\left(2\right)=20,f\left(3\right)=30$, then the value of $\frac{f\left(12\right)+f(-8)}{10}$ is equal to

• A. $2018$
• B. $1984$
• C. $60$
• D. $600$

Answer 1

The correct option is B $1984$

We have,

$\begin{array}{c}f\left(1\right)=10f\left(1\right)-10=0\\ f\left(2\right)=20\Rightarrow f\left(2\right)-20=0\Rightarrow f\left(2\right)-10\times 2=0\\ f\left(3\right)=30\Rightarrow f\left(3\right)-30=0\Rightarrow f\left(3\right)-10\times 3=0\\ g\left(a\right)=f\left(x\right)-10x=0\\ \therefore g\left(1\right)=f\left(1\right)-10=0\\ g\left(2\right)=f\left(2\right)-20=0\\ g\left(3\right)=f\left(3\right)-30=0\\ \therefore g\left(1\right)=g\left(2\right)=g\left(3\right)=0\\ \Rightarrow \left(x-1\right)\left(x-2\right)\left(x-3\right)dividesg\left(x\right)\\ g\left(x\right)=f\left(x\right)-10x=\left(x-t\right)\left(x-1\right)\left(x-2\right)\left(x-3\right)\\ f\left(x\right)=10x\left(x-t\right)\left(x-1\right)\left(x-2\right)\left(x-3\right)\\ forx=12\left(12-1\right)\left(12-2\right)\left(12-3\right)=11\times 10\times 9=990\\ forx=-8\left(-8-1\right)\left(-8-2\right)\left(-8-3\right)=-9\times -10\times -11=-990\\ \frac{f\left(-12\right)+f\left(-8\right)}{10}=\frac{10\left(x_1+x_2\right)+990\left(x_1-t\right)+\left(-990\right)\left(x^2-t\right)}{10}\\ let,x_1=12x_2=-8\\ \frac{f\left(+12\right)+f\left(-8\right)}{10}=\frac{10\left(12-8\right)+990\left(12-t\right)-990\left(-8-t\right)}{10}\\ =\frac{40+990\left[12-t+8+t\right]}{10}\\ =\frac{40+990\times 20}{10}=1984.\end{array}$

Hence, this is the answer.