A) 3
B) 4
C) 5
D) Nona of these
Correct Answer: C
Solution :
We have the function \[f(x)=\left\{ \begin{matrix} x & for & x<1 \\ 2-x & for & 1\le x\le 2 \\ -2+3x-{{x}^{2}} & for & x>2 \\ \end{matrix} \right.\] Differentiability at \[x=2\] \[LHD={{\left[ \frac{d}{dx}(2-x) \right]}_{x=2}}={{(-1)}_{x=2}}=-1\] \[RHD={{\left[ \frac{d}{dx}(-2+3x-{{x}^{2}}) \right]}_{x=2}}\] \[={{(3-2x)}_{x=2}}=3-4=-1\] \[\therefore \] \[LHD=RHD\]at \[x=2\] Hence, function is differentiable at\[x=2\] Differentiability at \[x=1\] \[LHD={{\left[ \frac{d}{dx}(x) \right]}_{x=1}}=1\] \[RHD={{\left[ \frac{d}{dx}(2-x) \right]}_{x=1}}=-1\] Here, \[LHD\ne RHD\]at\[x=1\] Hence, the function is not differentiable at\[x=1\].
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