Question

If $∆=\left|\begin{array}{ccc}x-2& 2x-3& 3x-4\\ 2x-3& 3x-4& 4x-5\\ 3x-5& 5x-8& 10x-17\end{array}\right|=A{x}^3+B{x}^2+Cx+D$, then $B+C$ is equal to

• A. $1$
• B. $-1$
• C. $-3$
• D. $9$

Answer 1

The correct option is C

$-3$

The explanation for the correct option:

It is given that

$\left|\begin{array}{ccc}x-2& 2x-3& 3x-4\\ 2x-3& 3x-4& 4x-5\\ 3x-5& 5x-8& 10x-17\end{array}\right|=A{x}^3+B{x}^2+Cx+D$

Operating $R2\to R2-R1$

Operating $R3\to R3-R2$

$\begin{array}{rcl}\left|\begin{array}{ccc}x-2& 2x-3& 3x-4\\ x-1& x-1& x-1\\ x-2& 2(x-2)& 6(x-2)\end{array}\right|& =& A{x}^3+B{x}^2+Cx+D\end{array}$

$\Rightarrow$ $\begin{array}{rcl}(x-1)(x-2)\left|\begin{array}{ccc}x-2& 2x-3& 3x-4\\ 1& 1& 1\\ 1& 2& 6\end{array}\right|& =& A{x}^3+B{x}^2+Cx+D\end{array}$

$\Rightarrow$ $\begin{array}{rcl}(x-1)(x-2)(x-2)(6-2)-(2x-3)(6-1)+(3x-4)(2-1)& =& A{x}^3+B{x}^2+Cx+D\end{array}$

$\Rightarrow$ $\begin{array}{rcl}(x-1)(x-2)4x-8-10x+15+3x-4& =& A{x}^3+B{x}^2+Cx+D\end{array}$

$\Rightarrow$ $\begin{array}{rcl}(x-1)(x-2)(-3x+3)& =& A{x}^3+B{x}^2+Cx+D\end{array}$

$\Rightarrow$ $\begin{array}{rcl}-3{(x-1)}^2(x-2)& =& A{x}^3+B{x}^2+Cx+D\end{array}$

$\Rightarrow$ $\begin{array}{rcl}-3x^3+12x^2-15x+6& =& A{x}^3+B{x}^2+Cx+D\end{array}$

Comparing both sides

$\begin{array}{rcl}A& =& -3,\\ B& =& 12,\\ C& =& -15\end{array}$

$\begin{array}{rcl}\therefore B+C& =& 12-15\\ & =& \mathbf{-}\mathbf{3}\end{array}$

Hence, The correct answer is option (C).