If the solve the problem
Question:

If $\Delta=\left|\begin{array}{ccc}x-2 & 2 x-3 & 3 x-4 \\ 2 x-3 & 3 x-4 & 4 x-5 \\ 3 x-5 & 5 x-8 & 10 x-17\end{array}\right|=$

$\mathrm{Ax}^{3}+\mathrm{Bx}^{2}+\mathrm{Cx}+\mathrm{D}$, then $\mathrm{B}+\mathrm{C}$ is equal to :

  1. $-1$

  2. 1

  3. $-3$

  4. 9


Correct Option: , 3

Solution:

$\Delta=\left|\begin{array}{ccc}x-2 & 2 x-3 & 3 x-4 \\ 2 x-3 & 3 x-4 & 4 x-5 \\ 3 x-5 & 5 x-8 & 10 x-17\end{array}\right|$

$=\mathrm{Ax}^{3}+\mathrm{Bx}^{2}+\mathrm{Cx}+\mathrm{D}$

$\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1} \quad \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{2}$

$\Delta=\left|\begin{array}{ccc}x-2 & 2 x-3 & 3 x-4 \\ x-1 & x-1 & x-1 \\ x-2 & 2(x-2) & 6(x-2)\end{array}\right|$

$=(x-1)(x-2)\left|\begin{array}{ccc}x-2 & 2 x-3 & 3 x-4 \\ 1 & 1 & 1 \\ 1 & 2 & 6\end{array}\right|$

$=-3(x-1)^{2}(x-2)=-3 x^{3}+12 x^{2}-15 x+6$

$\therefore \quad B+C=12-15=-3$

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