A) 1
B) 2
C) 3
D) 1/3
Correct Answer: B
Solution :
[b] : We have,\[f(x)={{(x+1)}^{1/3}}-{{(x-1)}^{1/3}}\] \[\therefore \]\[f'(x)=\frac{1}{3}\left[ \frac{1}{{{(x+1)}^{2/3}}}-\frac{1}{{{(x-1)}^{2/3}}} \right]\] \[=\frac{{{(x-1)}^{2/3}}-{{(x+1)}^{2/3}}}{3{{({{x}^{2}}-1)}^{2/3}}}\] Clearly, f'(x) does not exist at \[x=\pm 1\] Also,\[f'(x)=0\] \[\Rightarrow \]\[{{(x-1)}^{2/3}}={{(x+1)}^{2/3}}\Rightarrow x=0\] Clearly, \[f'(x)\ne 0\]for any other value of \[x\in [0,1]\]. The value of f(x) at x = 0 is 2. Hence, the greatest value of f(x) is 2.
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