has a second order derivatives 
. If its graph passes through the point (2, 1) and at that point the tangent to the graph is 
, then the function is
A)
B)
C)
D)
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 \[\operatorname{y}= 3x - 5\] is tangent to the graph \[\operatorname{y}=f(x)\]. Hence \[{{\left. \frac{dy}{dx} \right|}_{x=2}}=3\,\,or\,\,f'(2)=3\] Then from (i) \[f'(2)=3{{(2-1)}^{2}}+{{c}_{1}}\] \[=\,\,\,3+3+{{c}_{1}}\Rightarrow \,{{c}_{1}}\,=\,0\,\,i.e.,\,\,f'\,(x)=3{{(x-1)}^{2}}\] \[\therefore \,\,\,\,f(x)={{\left( x-1 \right)}^{3}}\,+{{c}_{2}}\] Given \[f(2) =1\] \[\therefore \,\,\,f(2)=1+{{c}_{2}}\] \[1=1+{{c}_{2}}\,\,\,\Rightarrow \,\,{{C}_{2}} =0.\,\,\,Hence\,\,f(x)={{(x-1)}^{3}}\]
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