A) f(x) is derivable but not continuous at x = 2.
B) f(x) is continuous but not derivable at x = 2.
C) f(x) is neither continuous nor derivable at x = 2.
D) f(x) is continuous as well as derivable at x = 2.
Correct Answer: B
Solution :
| [b] First we check continuity at \[x=2\] |
| L.H.L. =\[\underset{h\to 0}{\mathop{\lim }}\,f(2-h)=\underset{h\to 0}{\mathop{\lim }}\,3(2-h)-2\] |
| \[=\underset{h\to 0}{\mathop{\lim }}\,4-3h=4\] |
| R.H.L. \[=\underset{h\to 0}{\mathop{\lim }}\,f(2+h)=\underset{h\to 0}{\mathop{\lim }}\,{{(2+h)}^{2}}=4\] |
| Also, \[f(2)={{(2)}^{2}}=4\] |
| Since, L.H.L = R.H.L. = f(2) |
| \[\therefore \] f(x) is continuous at 2. |
| Now, we check for differentiability |
| L.H.D at x = 2 R.H.D at x = 2 |
| \[f'(x)=3x-2\] \[f'(x)={{x}^{2}}\]. |
| \[f'(x)=3\] \[f'(x)=2x\] |
| \[f'(x)\left| _{x=2} \right.=3\] \[f'(x)\left| _{x=2} \right.=4\] |
| Since L.H.D \[\ne \] R.H.D |
| \[\therefore \,f(x)\] is not derivable at \[x=2\] |
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