Question

Let $A=\left[\begin{array}{ccc}-3& 0& 2\\ 1& x& 5\\ -2& 0& x^2\end{array}\right],B=\left[\begin{array}{c}2\\ b\\ -1\end{array}\right]$ and $C=\left[\begin{array}{ccc}3& 5& 1\end{array}\right].$ If $\text{tr}$ denotes the trace of a matrix, then the number of integral values of $b$ for which $\text{tr}\left(ABC\right)\le -18$ for all $x\in R,$ is

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is C $5$

$ABC=\left[\begin{array}{ccc}-3& 0& 2\\ 1& x& 5\\ -2& 0& x^2\end{array}\right]\left[\begin{array}{c}2\\ b\\ -1\end{array}\right]\left[\begin{array}{ccc}3& 5& 1\end{array}\right]$

$=\left[\begin{array}{c}-8\\ bx-3\\ -x^2-4\end{array}\right]\left[\begin{array}{ccc}3& 5& 1\end{array}\right]$

$=\left[\begin{array}{ccc}-24& -40& -8\\ 3bx-9& 5bx-15& bx-3\\ -3x^2-12& -5x^2-20& -x^2-4\end{array}\right]$

$\text{tr}\left(ABC\right)=-x^2+5bx-43\le -18$
i.e., $-x^2+5bx-25\le 0$
As the coefficient of $x^2$ is negative, so for this inequality
$D\le 0$
$\Rightarrow 25b^2-100\le 0$
$\Rightarrow b^2-4\le 0$
$\Rightarrow b\in [-2,2]$