A) \[1\]
B) \[2\]
C) \[3\]
D) \[1/3\]
Correct Answer: B
Solution :
We have, \[f(x)={{(x+1)}^{1/3}}-{{(x-1)}^{1/3}}\] \[\therefore \] \[f'(x)=\frac{1}{2}\left[ \frac{1}{{{(x+1)}^{2/3}}}-\frac{1}{{{(x-1)}^{2/3}}} \right]\] Clearly, \[f'(x)\]does not exist at \[x=\pm 1\] for maxima or minima, put \[f'(x)=0\], 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\[2\]. Note: If any function has no critical point in the given interval \[[a,\,\,b]\], then we check the maximum value of\[x=a,\,\,\,b\].
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