A) \[f''\left( x \right)=2,x\in \left( 1,3 \right)\]
B) \[f''\left( x \right)=f'\left( x \right)=5,\,\] for some \[x\in \left( 2,3 \right)\]
C) \[f''\left( x \right)=3,\]\[x\in \left( 2,3 \right)\]
D) \[f''\left( x \right)=2\] for some \[x\in \left( 1,3 \right)\]
Correct Answer: D
Solution :
| We have, \[f\left( 1 \right)=1,f\left( 2 \right)=4,f\left( 3 \right)=9\] |
| \[\therefore \] \[f\left( x \right)={{\left( x \right)}^{2}}\] |
| Let \[g\left( x \right)=f(x)-{{x}^{2}}\] |
| \[\therefore \,\,g\left( 1 \right)=g\left( 2 \right)=g\left( 3 \right)=0\] |
| From Rolle's theorem on g(x) |
| \[g'\left( x \right)=0x\in (1,2)\] |
| Similarly, \[g'\left( x \right)=0\]for at least \[x\in \left( 2,3 \right)\] |
| \[g''\left( x \right)=f''(x)-2\] |
| \[g''\left( x \right)=0\] if \[g''\left( x \right)=2\]for some \[x\in \left( 1,3 \right)\] |
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