There exists a unique polynomial p(x) of degree 2 or less such that p(0)=0, p(1)

Question

There exists a unique polynomial p(x) of degree 2 or less such that p(0)=0, p(1)=1, and p'(a)=2 for any value of a between 0 and 1 (inclusive) except one value of a, say a0. Determine a0, and give this polynomial for a!=a0

Explanation / Answer

Let the polynomial be f(x). Since, the polynomial, is of degree 2 or less. Let it be px2 + qx + r = 0. Since, f(0) = 0. --> 0 + 0 +r = 0 ---> r=0 Since, f(1) = 1 --> p + q = 1. Now, f'(a)=2 --> 2px + q = 2. for x=a. i.e --> 2pa + q = 2 But if a=1/2 --> p + q = 2. Which is not possible, because p + q = 1. Therefore, a0=1/2. Now, if it is a quadratic expression, its not possible to have a constant slope of 2 in an interval beacuse for quadratic, slope changes at every point. Therefore, it should be linear. Therfore, f(x) = ax. since f(1) = 1 --> a=1. But f1(x)= 1 is not equals to 2. Therefore, linear is also not possible. Hence, there's no such solution possible. Feel free to ask any queries... also, post if you think something's wrong. :)

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