Question

If f(x) be a polynomial such that f(1)=1,f(2)=2,f(3)=3 and f(4)=16.Find the value of f(5).

• A. $53$
• B. $-19$
• C. $-12$
• D. $-6$

Answer 1

The correct option is A $53$

Let us assume,

$\Rightarrow f\left(x\right)=a{x}^3+b{x}^2+cx+d$

Then we get,

$\Rightarrow f\left(1\right)=a+b+c+d=1⟶\left(1\right)$

$\Rightarrow f\left(2\right)=8a+4b+2c+d=2⟶\left(2\right)$

$\Rightarrow f\left(3\right)=27a+9b+3c+d=3⟶\left(3\right)$

$\Rightarrow f\left(4\right)=64a+16b+4c+d=16⟶\left(4\right)$

solving $\left(1\right)$ and $\left(2\right)$ we get

$\Rightarrow 7a+3b+c=1⟶\left(5\right)$

solving $\left(1\right)$ and $\left(3\right)$ we get

$\Rightarrow 2ba+8b+2c=1⟶\left(6\right)$

solving $\left(1\right)$ and $\left(4\right)$ we get

$\Rightarrow 63a+15b+3c=15⟶\left(7\right)$

solving $\left(5\right)$ and $\left(7\right)$ we get

$\Rightarrow 7a+b=2$

This gives $a=2,b=-12$

And then $c=23,d=-12$

$\Rightarrow f\left(x\right)=2x^3-12x^2+23x-12$

$\Rightarrow f\left(x\right)=2\times {\left(5\right)}^3-12\times {\left(5\right)}^2+23\left(5\right)-12$

$=53$

Required answer $f\left(5\right)=53.$

Hence, the answer is $53.$