A) all\[m\in R\]
B) all\[m\in [-1,\,\,1]\]
C) all\[m\in R-\{0\}\]
D) None of these
Correct Answer: C
Solution :
\[{{y}^{2}}=4a\left( \frac{y+am}{m} \right)i.e.,\,\,m{{y}^{2}}-4ay-4{{a}^{2}}m=0\] \[m\ne 0;\,\,16{{a}^{2}}+16{{a}^{2}}{{m}^{2}}>0\]which is true\[\forall \,m\]. \[\therefore \] \[m\in R-\{0\}\]
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