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Question

Let f(x) be a polynomial satisfying limit x tends to infinity x2f(x)(2x5+3)=6, also f(1)=3,f(3)=7 and f(5)=11, then find the value of [f(6)+5f(4)]29.


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Solution

Step- 1: Determine f(x):

limxx2f(x)(2x5+3)=6 is finite and non-zero. f(x) will be a polynomial of degree 3.

f(x)=λ(x-1)(x-3)(x-5)+2x+1

Now,

limxx2λ(x-1)(x-3)(x-5)+2x+1(2x5+3)=6λ2=6λ=12

f(x)=12(x-1)(x-3)(x-5)+2x+1

Step- 2:Find the value of [f(6)+5f(4)]29

f(6)=12[(5)(3)(1)]+12+1=193andf(4)=12[(3)(1)(-1)]+9=-27[f(6)+5f(4)]29=193+5(-27)29=5829=2

Hence, the value of [f(6)+5f(4)]29 is 2


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