Question

Let $p(x)4$ 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. $-1$
• B. $1$
• C. $-\frac{1}{2}$
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

Let $p(x)=a{x}^2+bx+c...(i)$
Given, constant term ' $c$ '
$=1$ $\therefore p(x)=a{x}^2+bx+\mathrm{1...}(ii)$
Now, by given condition, $p(1)=2$ (remainder)
$\Rightarrow a+b+1=2$
$\Rightarrow a+b=\mathrm{1...}(iii)$
and $p(-1)=4$ (remainder)
$\Rightarrow a-b+1=4$
$\Rightarrow a-b=\mathrm{3...}(iii)$
On adding Eqs. (iii) and (iv), we get
$2a=4$
$\Rightarrow a=2$ form eys (iii) $b=-1$
On putting the values of a and bin Eq (ii), we get
$p(x)=2x^2-x+1=0$
$\therefore$ Sum of the roots $=-\frac{(\text{ Coefficient of }x)}{\left(\text{ Coefficient of }{x}^2\right)}=\frac{-(-1)}{2}$
$=\frac{1}{2}$