A) 1
B) 2
C) 3
D) 1/3
Correct Answer: B
Solution :
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 exists at\[x=\pm 1\] Now, \[f(x)\ne 0\]then \[{{(x-1)}^{2/3}}={{(x+1)}^{2/3}}\] \[\Rightarrow \] \[x=0\] 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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