A) \[{{x}^{2}}+1\]
B) \[{{x}^{2}}+2x+1\]
C) \[4x+1\]
D) \[{{x}^{2}}-2x+1\]
Correct Answer: A
Solution :
Let \[f(x)=a{{x}^{2}}+bx+c\] ?..(i) Also, given \[f(0)=1\] \[\Rightarrow \] \[1=a\,{{(0)}^{2}}+b(0)+c\] \[\Rightarrow \] \[c=1\] ?.(ii) and \[f'(2)=4\] on differentiating Eq. (i), w. r. t. x, we get \[f'(x)=2ax+b\] ?..(iii) \[\Rightarrow \] \[f'(2)=2a\,(2)+b\] \[\Rightarrow \] \[4=4a+b\] ?..(iv) Again, differentiating Eq. (iii0, we get \[f''(x)=2a\] But \[f''(1)=2\] (given) \[\therefore \] \[f''(1)=2a\] \[\Rightarrow \] \[2=2a\] \[\Rightarrow \] \[a=1\] On putting \[a=1\] in Eq. (iv), we get \[4=4+b\] \[\Rightarrow \] \[b=0\] On putting \[a=1,\,b=0\] and \[c=1\] in Eq. (i), we get \[f(x)={{x}^{2}}+1\]
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