Question

Let $P\left(x\right)$ be a four-degree polynomial with leading coefficient equal to $1.$ If $P\left(1\right)=-1,$ $P\left(2\right)=1,$ $P\left(3\right)=3,$ $P\left(4\right)=5,$ then $\underset{x\to 1}{lim}\frac{P\left(x\right)+3-2x}{(x-1)}$ is equal to

• A. $6$
• B. $5$
• C. $2$
• D. $-6$

Answer 1

The correct option is D $-6$

We observe that $P\left(1\right),P\left(2\right),P\left(3\right),P\left(4\right)$ are in A.P. where $a_n=2n-3$
So, $P\left(x\right)=(x-1)(x-2)(x-3)(x-4)+2x-3$

$\underset{x\to 1}{lim}\frac{P\left(x\right)+3-2x}{(x-1)}$
$=\underset{x\to 1}{lim}\frac{(x-1)(x-2)(x-3)(x-4)+2x-3+3-2x}{(x-1)}$
$=\underset{x\to 1}{lim}(x-2)(x-3)(x-4)=-6$