A) \[\frac{17}{8}\]
B) \[\frac{15}{8}\]
C) \[\frac{29}{8}\]
D) \[\frac{33}{8}\]
Correct Answer: D
Solution :
[d] \[\because y=3-6x-8{{x}^{2}}\] For maximum and minimum value, \[\frac{dy}{dx}=0\] \[\therefore \frac{dy}{dx}=0-6-16x=0\] \[16x=-6\] \[x=\frac{-3}{8}\] \[\therefore \frac{{{d}^{2}}y}{d{{x}^{2}}}=-16<0\] Hence, y has maximum value at \[x=\frac{-3}{8}\] \[\therefore {{y}_{at\,x=\frac{-3}{8}}}=3-6\times \left( \frac{-3}{8} \right)-8{{\left( \frac{-3}{8} \right)}^{2}}\] \[=3+\frac{18}{8}-\frac{8\times 9}{64}=\frac{192+144-72}{64}=\frac{264}{64}=\frac{33}{8}\] Hence, option [d] is correct.
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