A) \[{{A}_{i}}=-\frac{{{h}_{fe}}}{1+{{h}_{oe}}{{R}_{L}}}\]
B) \[{{h}_{fe}}=50,{{h}_{oe}}=25\mu A{{V}^{-1}}\]
C) \[=25\times {{10}^{-6}}A{{V}^{-1}}\]
D) \[{{R}_{L}}=1k\Omega ={{10}^{3}}\Omega \]
Correct Answer: C
Solution :
Given, \[f(x)=\left| \begin{matrix} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \\ \end{matrix} \right|\] Put \[x=0\] \[\Rightarrow \] \[f(0)=\left| \begin{matrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \\ \end{matrix} \right|\] Expanding along \[{{R}_{1}}\], we get \[f(0)=a(0+bc)=b(ac)=abc-abc=0\]
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