A) \[f(x)=\left| x \right|\], in \[-2\le x\le 2\]
B) \[f(x)=\tan x\], in \[0\le x\le \pi \]
C) \[f(x)=l+{{(x-2)}^{\frac{2}{3}}}\], in \[l\le x\le 3\]
D) \[f(x)=x{{(x-2)}^{2}}\] in \[0\le x\le 2\]
Correct Answer: D
Solution :
| [a] \[f(x)=|x|,\] is not derivable at x = 0. |
| [b] \[f(x)=\tan \,x\] is discontinuous at \[x=\pi \]. |
| [c] \[f(x)\,=1+{{(x-2)}^{\frac{2}{3}}}\] is non- derivable at \[x=2\]. |
| [d] Only function which satisfies Rolle's theorem, is \[f(x)\,=x{{(x-2)}^{2}},\,\,0\le x\le 2\] |
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