If S is a set of P(x) is polynomial of degree ≤2 such that P(0)=0, P(1)=1, P'(x)>0 ∀x∈(0,1), then

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S=0

S=ax+(1-a)x2∀a∈(0,∞)

S=ax+(1-a)x2∀a∈R

S=ax+(1-a)x2∀a∈(0,2)

answer is D.

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Detailed Solution

Let P(x)=bx2+ax+cAs P(0)=0⇒c=0As P(1)=1⇒a+b=1P(x)=ax+(1-a)x2Now P'(x)=a+2(1-a)xas P'(x)>0 for x∈(0,1)Only option (4) satisfies above condition

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