A) positive
B) negative
C) zero
D) none
Correct Answer: B
Solution :
\[\alpha <-1.\,\,Let \alpha =-1-p\] \[\beta >1.\,Let \beta =1+q, p>0,\,\,q>0\] Now \[1+\frac{c}{a}+\left| \frac{b}{a} \right|=1+\alpha \beta +\left| -\alpha -\beta \right|\] \[=1+\left( 1+q \right)\left( -1-p \right)+\left| -1-p+1+q \right|\] \[=\,\,\,1-\left( 1+p+q+pq \right)+\left| q-p \right|\] \[=\left\{ \begin{matrix} -p-q-pq+q-p=-2p-pq<0\,\,if\,\,q>p \\ -p-q-pq+p-q=-2q-pq<0\,\,if\,\,q<p \\ \end{matrix}\, \right.\] \[\therefore \,\,1+\frac{c}{a}+\left| \frac{b}{a} \right|<0\]You need to login to perform this action.
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