A) 25
B) ? 39
C) ? 25
D) 39
Correct Answer: B
Solution :
\[f(x)=3{{x}^{4}}-8{{x}^{3}}+12{{x}^{2}}-48x+25\] \ \[f'(x)=12{{x}^{3}}-24{{x}^{2}}+24x-48\] \[=12[{{x}^{3}}-2{{x}^{2}}+2x-4]\]\[=12[(x-2)({{x}^{2}}+2)]\] For maximum and minimum value of the function\[f'(x)=0\] Þ \[x=2\]. Now \[{f}''(x)=12[3{{x}^{2}}-4x+2]\] \ \[{f}''(2)=12\,[12-8+2]=72>0\] Hence the function is minimum at \[x=2\] Minimum value of \[f(x)\]on [0, 3] \[=\min \left\{ f(0),\,f(2),\,f(3) \right\}\]\[=\min \left\{ 25,\,-39,\,16 \right\}\]\[=-39\].
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