A) 2
B) \[-10\]
C) 10
D) None of these
Correct Answer: A
Solution :
Given, \[p(0)=4,p'(0)=3,p''(0)=4,p''(0)=6\] Let \[p(x)=a{{x}^{3}}+b{{x}^{2}}+ex+d\] ...(i) \[\therefore \] At \[x=0,p(0)=4\] \[\Rightarrow \] \[d=4\] \[p'(x)=3a{{x}^{2}}+2bx+c\] \[p'(0)=3\] \[\Rightarrow \] \[c=3\] \[p'\,'(x)=6ax+2b\] \[p'\,'(0)=4\] \[2b=4\] \[\Rightarrow \] \[b=2\] \[p'\,'(x)=6a\] \[p'\,'(0)=6\Rightarrow 6a=6\] \[\Rightarrow \] \[a=1\] On putting the values of a, b, c and d in Eq. (i), we get \[p(x)={{x}^{3}}+2{{x}^{2}}+3x+4\] \[\therefore \] \[p'(x)=3{{x}^{2}}+4x+3\] \[P'(-1)=3(-1)2+4(-1)+3\] \[=3-4+3=2\]
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