A Distorting Problem

Calculus Level 3

Polynomials P ( x ) P(x) and Q ( x ) Q(x) with degrees n n and 7 7 respectively are such that

P ( x ) e k x d x = Q ( x ) e 4 x + C , \int P(x) e^{kx} \, dx = Q(x) e^{4x} + C,

where C C denotes the arbitrary constant of integration .

Find the value of n + 7 + k + lim x P ( x ) Q ( x ) \displaystyle n + 7 + k + \lim_{x\to\infty} \dfrac{P(x)}{Q(x)} .

18 19 20 22

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Chris Lewis
Sep 8, 2020

Without loss of generality, we can assume the coefficient of x 7 x^7 in Q ( x ) Q(x) is 1 1 . So, let Q ( x ) = x 7 + R ( x ) Q(x)=x^7+R(x) , where R ( x ) R(x) is a polynomial of degree less than 7 7 .

Differentiating,

P ( x ) e k x = d d x [ ( x 7 + R ( x ) ) e 4 x ] = ( 4 x 7 + 4 R ( x ) + 7 x 6 + R ( x ) ) e 4 x \begin{aligned} P(x)\cdot e^{kx}&=\frac{d}{dx} \left[ \left( x^7+R(x) \right)e^{4x}\right] \\ &=\left( 4x^7+4R(x)+7x^6+R'(x) \right) e^{4x}\end{aligned}

Comparing the two sides, we immediately see that n = 7 n=7 and k = 4 k=4 .

The ratio in the limit is just the ratio of the leading coefficients of P P and Q Q (since they have the same degree), which is 4 4 .

Putting this together, the answer is 7 + 7 + 4 + 4 = 22 7+7+4+4=\boxed{22} .

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...