A) \[{{(x+1)}^{3}}\]
B) \[{{(x-1)}^{3}}\]
C) \[{{(x+1)}^{2}}\]
D) \[{{(x-1)}^{2}}\]
Correct Answer: B
Solution :
Given \[{{f}'}'(x)=6(x-1)\] \[{f}'(x)=3{{(x-1)}^{2}}+{{c}_{1}}\] ?..(i) But at point (2, 1) the line \[y=3x-5\] is tangent to the graph \[y=f(x)\]. Hence \[{{\left. \frac{dy}{dx} \right|}_{x=2}}=3\] or \[{f}'(2)=3\]. Then from (i) \[{f}'(2)=3{{(2-1)}^{2}}+{{c}_{1}}\] \[3=3+{{c}_{1}}\] Þ \[{{c}_{1}}=0\] i.e., \[{f}'(x)=3{{(x-1)}^{2}}\] Given \[f(2)=1\] \[f(x)={{(x-1)}^{3}}+{{c}_{2}}\] Þ \[f(2)=1+{{c}_{2}}\] Þ \[1=1+{{c}_{2}}\] Þ \[{{c}_{2}}=0\] Hence \[f(x)={{(x-1)}^{3}}\].
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