A) 1 and 2
B) 1 and -1/2
C) -1 and 2
D) 1
Correct Answer: A
Solution :
Given that, \[f(x)={{(x-2)}^{2/3}}(2x+1)\] \[\Rightarrow \]\[f'(x)=\frac{2}{3}{{(x-2)}^{-1/3}}(2x+1)+2{{(x-2)}^{2/3}}\] Clearly, f' (x) is not defined at x = 2, So x = 2 is a critical point and for another critical point, Put f'(x) = 0. \[\therefore \]\[\frac{2}{3}{{(x-2)}^{-1/3}}(2x+1)+2{{(x-2)}^{2/3}}=0\] \[\Rightarrow \]\[2x+1=-3(x-2)\]\[\Rightarrow \]\[x=1\] Hence, 1 and 2 are two critical points of f(x).
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