Q72 / 200 · WBJEE 2013 · Mathematics
Let $p(x)$ be a quadratic polynomial with constant term 1. Suppose $p(x),$ when divided by $x-1$ leaves remainder 2 and when divided by $x+1$ leaves remainder $4 .$ Then, the sum of the roots of $p(x)=0$ is
- A-1
- B1
- C$-\frac{1}{2}$
- Correct answerD$\frac{1}{2}$
Solution
Let $\quad p(x)=a x^{2}+b x+c$
Given, constant term 'c' $=1$
$\therefore \quad p(x)=a x^{2}+b x+1$
Now, by given condition, $p(1)=2$ (remainder)
$\begin{array}{lr}\Rightarrow & a+b+1=2 \\ \Rightarrow & a+b=1\end{array}$
$\begin{array}{ll}\text { and } & p(-1)=4 \quad \text { (remainder) }\end{array}$
$\begin{array}{lr}\Rightarrow & a-b+1=4 \\ \Rightarrow & a-b=3\end{array}$
On adding Eqs. (iii) and (iv), we get
$2 a=4 \Rightarrow a=2$ form eys
(iii) $b=-1$
On putting the values of a and b in Eq. (ii), we get $\begin{aligned} p(x)=2 x^{2}-x+1=0 \\ \text { Sum of the roots } &=-\frac{\text { (Coefficient of } x)}{\text { (Coefficient of } x^{2} \text { ) }}=\frac{-(-1)}{2} \\ &=\frac{1}{2} \end{aligned}$
Hence, the correct answer is option: 'D'.