A) \[\frac{15}{12}\]
B) \[-9\frac{11}{12}\]
C) \[-\frac{5}{6}\]
D) \[\frac{5}{6}\]
Correct Answer: B
Solution :
(b): \[-3{{x}^{2}}+5x-12\] \[-3\left[ {{x}^{2}}\frac{-5}{3}x+4 \right]=-3\left[ {{x}^{2}}-2.\frac{5}{6}x+{{\left( \frac{5}{6} \right)}^{2}}-{{\left( \frac{5}{6} \right)}^{2}}+4 \right]\]\[=-3\left[ {{\left\{ x-\frac{5}{6} \right\}}^{2}}+4-\frac{25}{36} \right]=-3\left[ {{\left\{ x-\frac{5}{6} \right\}}^{2}}+\frac{119}{36} \right]\] Maximum value is when\[{{\left\{ x-\frac{5}{6} \right\}}^{2}}\], square term = 0 \[\Rightarrow \]Maximum value \[=-3\times \frac{119}{36}\]
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