A) \[\frac{1}{2}\]
B) \[\frac{3}{2}\]
C) \[\frac{5}{2}\]
D) \[\frac{9}{2}\]
Correct Answer: D
Solution :
Given it has extremum values at \[x=1\] and \[x=2\] \[\Rightarrow f'(1)=0\] and \[\Rightarrow f'(2)=0\] Given \[f(x)\] is a fourth degree polynomial Let \[f(x)=a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+c=0\] Given \[\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{f(x)}{{{x}^{2}}}+1 \right)=3\] \[\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e}{{{x}^{2}}}+1 \right)=3\] \[\underset{x\to 0}{\mathop{\lim }}\,\left( a{{x}^{2}}+bx+c+\frac{d}{x}+\frac{e}{{{x}^{2}}}+1 \right)=3\] For limit to have finite value (in this case 3) value of \['d'\] and \['e'\] must be 0 \[\Rightarrow d=0\And e=0\] Substituting \[x=0\] in limit: \[\Rightarrow c+1=3\] \[\Rightarrow c=2\] \[f'(x)=4a{{x}^{3}}+3b{{x}^{2}}+2cx+d\] Applying \[f'(1)=0,f'(2)=0\] \[4a(1)+3b(1)+2c(1)+d=0\]\[\Rightarrow 1\] \[4a(8)+3b(4)+2c(2)+d=0\Rightarrow 2\] Substituting \[c=2\] and \[d=0\] \[4a+3b+3=0\] \[32a+12b+8=0\] Solving two equations, we get \[a=\frac{1}{2}\] and \[b=-2\] \[f(x)=\frac{{{x}^{4}}}{2}-2{{x}^{3}}+2{{x}^{2}}\] \[f(-1)=\frac{-{{1}^{4}}}{2}-2{{(-1)}^{3}}+2{{(-1)}^{2}}\] Hence \[f(x)=\frac{9}{2}\] Therefore correct option is ?D?
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