A) P(-1) is the minimum and P(1) is the maximum of P
B) P(-1) is not minimum but P(1) is the maximum of P
C) P(-1) is the minimum but P(1) is not the maximum of P
D) neither P(-1) is the minimum nor P(1) is the maximum of P
Correct Answer: B
Solution :
\[P(x)={{x}^{4}}+a{{x}^{3}}+b{{x}^{2}}+cx+d\] \[P'(x)=4{{x}^{3}}+3a{{x}^{2}}+2bx+c\] As\[P'(x)=0\]has only root\[x=0\] \[\Rightarrow \] \[c=0\] \[P'(x)=x(4{{x}^{2}}+3ax+2b)\] \[\Rightarrow \]\[4{{x}^{3}}+3ax+2b=0\]has non real root. and\[4{{x}^{2}}+3ax+2b>0\forall x\in [-1,1]\]. As P(−1) < P(1)\[\Rightarrow \]P(1) is the max. of\[P(x)\]in [−1, 1]You need to login to perform this action.
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