A) \[{{(x-1)}^{2}}\]
B) \[{{(x-1)}^{3}}\]
C) \[{{(x+1)}^{3}}\]
D) \[{{(x+1)}^{2}}\]
Correct Answer: B
Solution :
[b] \[f''(x)=6(x-1)\] \[\therefore f'(x)=3{{(x-1)}^{2}}+c\] ...(i) At the point (2, 1), the tangent to the graph is \[y=3x-5\] The slope of the tangent is 3, \[\therefore f'(2)=3{{(2-1)}^{2}}+c=3.\] \[\Rightarrow 3+c=3\Rightarrow c=0\] Therefore, from (i), we get \[f'(x)=3{{(x-1)}^{2}}\] ...(ii) \[\Rightarrow f(x)={{(x-1)}^{3}}+k\] Since the graph passes through (2, 1), \[1={{(2-1)}^{2}}+k\] \[\Rightarrow k=0\] Hence, the equation of the function is \[f(x)={{(x-1)}^{3}}\]
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