A) \[S=0\]
B) \[S=ax+(1-a){{x}^{2}}\text{ }\forall a\in (0,\infty )\]
C) \[S=ax+(1-a){{x}^{2}}\text{ }\forall a\in R\]
D) \[S=ax+(1-a){{x}^{2}}\text{ }\forall a\in (0,2)\]
Correct Answer: D
Solution :
Let \[P(x)=b{{x}^{2}}+ax+c\] As \[P(0)=0\Rightarrow c=0\] As \[P(1)=1\Rightarrow a+b=1\] \[P(x)=ax+(1-a){{x}^{2}}\] Now \[{P}'(x)=a+2(1-a)x\] as \[{P}'(x)>0\] for \[x\in (0,\,1)\] Only option (d) satisfies above conditionYou need to login to perform this action.
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