Question

If $f\left(x\right)$ is a polynomial of degree $n$ such that $f\left(0\right)=0,f\left(1\right)=\frac{1}{2},....,f\left(n\right)=\frac{n}{n+1}$, then the value of $f(n+1)$ is

• A. $1$ when $n$ is odd
• B. $\frac{n}{n+2}$ when $n$ is even
• C. $-\frac{n}{n+1}$ when $n$ is odd
• D. $-1$ when $n$ is even

Answer 1

Answer: (A), (B)

$\frac{n}{n+2}$ when $n$ is even

$(x+1)f\left(x\right)-x$ is a polynomial of degree $n+1$
$\Rightarrow (x+1)f\left(x\right)-x=k\left(x\right)(x-1)(x-2)....(x-n).....\left(i\right)$
Also, $1=k(-1)(-2)...\left(\right(-n-1\left)\right(Puttingx=-1in\left(i\right))$
$\Rightarrow 1=k(-1{)}^{n+1}(n+1)!$
$\Rightarrow k=\frac{1}{(-1{)}^{n+1}(n+1)!}$
$Byreplacingxwithn+1,weget$​​​​
$\Rightarrow (n+2)f(n+1)-(n+1)=k\left[\right(n+1)!]$
$\Rightarrow (n+2)f(n+1)-(n+1)=(-1{)}^{n+1}$
$f(n+1)=1,\text{if n is odd and}\frac{n}{n+2},\text{if n is even.}$