A) \[{{(x+1)}^{3}}\]
B) \[{{(x-1)}^{3}}\]
C) \[{{(x-1)}^{2}}\]
D) \[{{(x+1)}^{2}}\]
Correct Answer: B
Solution :
Given, \[f\,(x)=6(x-1)\] \[\Rightarrow \] \[f(x)=3{{(x-1)}^{2}}+c\] ...(i) But at (2,1),\[y=3x-5\]is tangent to\[y=f(x)\] \[\therefore \] \[f(2)=3\] \[\therefore \]From Eq. (i), \[3=3+c\] \[\Rightarrow \] \[c=0\] \[\therefore \] \[f(x)=3{{(x-1)}^{2}}\] \[\Rightarrow \] \[f(x)={{(x-1)}^{3}}+k\] Since, it passes through (2, 1). \[\therefore \] \[1={{(2-1)}^{3}}+k\] \[\Rightarrow \] \[k=0\]\[\therefore \]\[f(x)={{(x-1)}^{3}}\]You need to login to perform this action.
You will be redirected in
3 sec