Polynomials and with degrees and respectively are such that
where denotes the arbitrary constant of integration .
Find the value of .
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Without loss of generality, we can assume the coefficient of x 7 in Q ( x ) is 1 . So, let Q ( x ) = x 7 + R ( x ) , where R ( x ) is a polynomial of degree less than 7 .
Differentiating,
P ( x ) ⋅ e k x = d x d [ ( x 7 + R ( x ) ) e 4 x ] = ( 4 x 7 + 4 R ( x ) + 7 x 6 + R ′ ( x ) ) e 4 x
Comparing the two sides, we immediately see that n = 7 and k = 4 .
The ratio in the limit is just the ratio of the leading coefficients of P and Q (since they have the same degree), which is 4 .
Putting this together, the answer is 7 + 7 + 4 + 4 = 2 2 .