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Question

Let P(x) be a polynomial satisfying limxx3P(x)x6+3x2+7=2. If P(1)=2, P(3)=10 and P(5)=26, then the value of P(2)+|P(0)|10 is

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Solution

Let f(x)=P(x)(x2+1)
Then, x=1,3,5 are roots of f(x).
Therefore, we can write f(x) as A(x1)(x3)(x5)
P(x)=A(x1)(x3)(x5)+(x2+1)
Also, P(x) should be of degree 3.

We have,
limxx3P(x)x6+3x2+7=2
A=2
P(x)=2(x1)(x3)(x5)+x2+1
Now, P(2)=11,P(0)=29
|P(0)|+P(2)10=4

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