Question

If
$\Delta \left(x\right)=\left|\begin{array}{ccc}{x}^2-5x+3& 2x-5& 3\\ 3x^2+x+4& 6x+1& 9\\ 7x^2-6x+9& 14x-6& 21\end{array}\right|=a{x}^3+b{x}^2+cx+d$ then

• A. $a=0$
• B. $b=0$
• C. $c=0$
• D. $d=144$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $a=0$
B $b=0$
C $c=0$
D $d=144$

Differentiate both the sides with respect to $x$
$\Delta {}^{\prime }\left(x\right)=\left|\begin{array}{ccc}2x-5& 2x-5& 3\\ 6x+1& 6x+1& 9\\ 14x-6& 14x-6& 21\end{array}\right|+\left|\begin{array}{ccc}{x}^2-5x+3& 2& 3\\ 3x^2+x+4& 6& 9\\ 7x^2-6x+9& 14& 21\end{array}\right|$
$=0+0=0$
$\Rightarrow \Delta \left(x\right)$ is a constant
Thus, $a=0,b=0$ & $c=0$
For value of $d$ , putting $x=0$
$\Delta \left(0\right)=\left|\begin{array}{ccc}{0}^2-5.0+3& 2.0-5& 3\\ {3.0}^2+0+4& 6.0+1& 9\\ {7.0}^2-6.0+9& 14.0-6& 21\end{array}\right|=a{.0}^3+b{.0}^2+c.0+d\Rightarrow d=144$