A) 1
B) 2
C) 4
D) 5
Correct Answer: A
Solution :
\[|x|\,=\frac{-(m+6)\,\pm \,(m+2)}{2}\,(reject+sign)\] \[|x|=-(m+4)\] For two distinct solution, m + 4 < 0 \[\Rightarrow \,m<-4\] \[\Rightarrow \] Number of integral values of m in (-10, 10] are {-9, -8, -7, -6, -5} i.e., 5 values Alternating: Since equation has two distinct solution and therefore product of the roots must be less than zero. \[\Rightarrow \,2(m+4)<0\Rightarrow \,m<-4\]. \[f(\lambda )\,=\det \,(A-\lambda I)\,=\left| \begin{matrix} 1-\lambda & 2 \\ -1 & 3-\lambda \\ \end{matrix} \right|\,=(1-\lambda )\,(3-\lambda )\,+2\]\[={{\lambda }^{2}}-4\,\lambda +5={{(\lambda -2)}^{2}}+1\] Clearly, \[{{f}_{\min }}\,(\lambda =2)=1\]You need to login to perform this action.
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