Question

lf $\alpha$ is the repeated root of quadratic equation $f\left(x\right)=0$ and $A\left(x\right)$ , $B\left(x\right)$ and $C\left(x\right)$ are polynomials of degree 3,4 and 5 respectively, then $\varphi \left(x\right)=$ $\left|\begin{array}{ccc}A\left(x\right)& B\left(x\right)& C\left(x\right)\\ A\left(\alpha \right)& B\left(\alpha \right)& C\left(\alpha \right)\\ A^{{}^{\prime }}\left(\alpha \right)& B^{{}^{\prime }}\left(\alpha \right)& C^{{}^{\prime }}\left(\alpha \right)\end{array}\right|$ is divisible by

• A. $f\left(x\right)$
• B. $A\left(x\right)$
• C. $B\left(x\right)$
• D. $C\left(x\right)$

Answer 1

The correct option is A $f\left(x\right)$

$\varphi \left(x\right)=\left|\begin{array}{ccc}A\left(x\right)& B\left(x\right)& C\left(x\right)\\ A\left(\alpha \right)& B\left(\alpha \right)& C\left(\alpha \right)\\ A^{\prime }\left(\alpha \right)& B^{\prime }\left(\alpha \right)& C^{\prime }\left(\alpha \right)\end{array}\right|$
$\varphi \left(\alpha \right)=\left|\begin{array}{ccc}A\left(\alpha \right)& B\left(\alpha \right)& C\left(\alpha \right)\\ A\left(\alpha \right)& B\left(\alpha \right)& C\left(\alpha \right)\\ A^{\prime }\left(\alpha \right)& B^{\prime }\left(\alpha \right)& C^{\prime }\left(\alpha \right)\end{array}\right|=0$
Now, ${\varphi }^{\prime }\left(x\right)=\left|\begin{array}{ccc}{A}^{\prime }\left(x\right)& B^{\prime }\left(x\right)& C^{\prime }\left(x\right)\\ A\left(\alpha \right)& B\left(\alpha \right)& C\left(\alpha \right)\\ A^{\prime }\left(\alpha \right)& B^{\prime }\left(\alpha \right)& C^{\prime }\left(\alpha \right)\end{array}\right|+\left|\begin{array}{ccc}A\left(x\right)& B\left(x\right)& C\left(x\right)\\ 0& 0& 0\\ A^{\prime }\left(\alpha \right)& B^{\prime }\left(\alpha \right)& C^{\prime }\left(\alpha \right)\end{array}\right|+\left|\begin{array}{ccc}A\left(x\right)& B\left(x\right)& C\left(x\right)\\ A\left(\alpha \right)& B\left(\alpha \right)& C\left(\alpha \right)\\ 0& 0& 0\end{array}\right|$
$\Rightarrow {\varphi }^{\prime }\left(x\right)=\left|\begin{array}{ccc}{A}^{\prime }\left(x\right)& B^{\prime }\left(x\right)& C^{\prime }\left(x\right)\\ A\left(\alpha \right)& B\left(\alpha \right)& C\left(\alpha \right)\\ A^{\prime }\left(\alpha \right)& B^{\prime }\left(\alpha \right)& C^{\prime }\left(\alpha \right)\end{array}\right|$
${\varphi }^{\prime }\left(\alpha \right)=\left|\begin{array}{ccc}{A}^{\prime }\left(\alpha \right)& B^{\prime }\left(\alpha \right)& C^{\prime }\left(\alpha \right)\\ A\left(\alpha \right)& B\left(\alpha \right)& C\left(\alpha \right)\\ A^{\prime }\left(\alpha \right)& B^{\prime }\left(\alpha \right)& C^{\prime }\left(\alpha \right)\end{array}\right|=0$
So, $\varphi \left(\alpha \right)={\varphi }^{\prime }\left(\alpha \right)=0$
So, by multiple root theorem, $\varphi \left(x\right)$ has a repeated root $\alpha$
Also, given $\alpha$ is a repeated root of $f\left(x\right).$
Hence, $\varphi \left(x\right)$ divisible by $f\left(x\right)$.