A) \[\frac{3\alpha +6}{\alpha +1}\]
B) \[\frac{\alpha +6}{2\alpha +1}\]
C) \[\frac{3\alpha +6}{\alpha +2}\]
D) \[\frac{3\alpha +6}{2\alpha +1}\]
Correct Answer: D
Solution :
[d] \[{{f}_{\max }}=\frac{\left( n\alpha \right)\left( 3 \right)\left( 3 \right)+\left( n-n\alpha \right)\left( 6 \right)}{\left( n\alpha \right)\left( 3 \right)+\left( n-n\alpha \right)}=\frac{3\alpha +6}{2\alpha +1}\]
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