Question

Let, $f\left(x\right)=x^5+a{x}^4+b{x}^3+c{x}^2+d$ such that $f\left(1\right)=1,f\left(2\right)=2,f\left(3\right)=3,f\left(4\right)=4,f\left(5\right)=5$ , then find the value of ‘$d$’

• A. $-120$
• B. $-100$
• C. $0$
• D. $100$

Answer 1

The correct option is A

$-120$

Explanation for the correct option:

Finding the value of $d$:

Let ,$f\left(x\right)=x^5+a{x}^4+b{x}^3+c{x}^2+d$ …(1)

Given that,

$f\left(1\right)=1,f\left(2\right)=2,f\left(3\right)=3,f\left(4\right)=4,f\left(5\right)=5$

Then,

$f\left(1\right)-1=0$, $f\left(2\right)-2=0$, $f\left(3\right)-3=0$, $f\left(4\right)-4=0$,$f\left(5\right)-5=0$

From the above expression, we observe that

$f\left(x\right)-x=0$

or,

$f\left(x\right)-x=(x-1)(x-2)(x-3)(x-4)(x-5)$…(2)

This expression satisfies all the given conditions.

Now put $x=0$ in equation (1), we will get

$f\left(0\right)=d$

Again putting $x=0$ in equation (2), we get

$\begin{array}{rcl}f\left(0\right)-0& =& (0-1)(0-2)(0-3)(0-4)(0-5)\\ & \Rightarrow & d=-120\left[\because f\left(0\right)=d\right]\end{array}$

Hence, the correct answer is (A)$-120$.