Question

Consider an unknown polynomial which when divided by $(x-3)$ and $(x-4)$ leaves remainders $2$ and $1$ respectively. Let $R\left(x\right)$ be the remainder when this polynomial is divided by $(x-3)(x-4)$. If $R\left(x\right)=p{x}^2+(q-1)x+6$ has no distinct real roots and $p>0$, then the value of $3p+q$ is

• A. $-2$
• B. $\frac{2}{3}$
• C. $-\frac{1}{3}$
• D. none of these

Answer 1

The correct option is C $-\frac{1}{3}$

$P\left(3\right)=9p+(q-1)3+6=2\Rightarrow 9p+3q+1=0P\left(4\right)=16p+(q-1)4+6=1\Rightarrow 16p+4q+1=0$
Solving these two equation
$p=\frac{1}{12},q=\frac{-7}{12}$
$\therefore 3p+q=\frac{-1}{3}$