A) \[f(2)\,+f'(2)\,=28\]
B) \[f''(2)\,-f'(2)=0\]
C) \[f(2)\,-f'(2)\,+f''(2)=10\]
D) \[f''(2)-f(2)=4\]
Correct Answer: A
Solution :
\[f(x)=a{{x}^{3}}+b{{x}^{2}}+cx+l\] \[f(3x)=27a{{x}^{3}}\,+9b{{x}^{2}}\,+3cx+d\] \[f'(x)=3a{{x}^{2}}+2bx+c\] \[f''(x)=6ax+2b\] \[f(3x)\,=f'(x)f''(x)\] \[27a=18{{a}^{2}}\] \[a=\frac{3}{2},\,b=0,\,c=0,\,d=0\] \[f(x)\,=\frac{3}{2}\,{{x}^{3}}\] \[f'(x)\,=\frac{9}{2}{{x}^{2}},\,f''(x)\,=9x\]
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