Let f be the function defined above, where c and d are constants. If f is differentiable at x=2, what is the value of c d?.

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Ответ:

igtguith

04.09.2021

Mathematics

"no terms independent of x" basically means there is no constant term, or the coefficient of x ⁰ is zero.

Recall the binomial theorem:

$\displaystyle (a+b)^n=\sum_{k=0}^n\binom nk a^{n-k}b^k$

So we have

$\displaystyle \left(1+ax^2\right)\left(\frac2x-3x\right)^6 = \left(1+ax^2\right) \sum_{k=0}^6 \binom 6k \left(\frac2x\right)^{6-k}(-3x)^k$

$\displaystyle \left(1+ax^2\right)\left(\frac2x-3x\right)^6 = \left(1+ax^2\right) \sum_{k=0}^6 2^{6-k}(-3)^k\binom 6k x^{2k-6}$

$\displaystyle \left(1+ax^2\right)\left(\frac2x-3x\right)^6 = \sum_{k=0}^6 2^{6-k}(-3)^k\binom 6k x^{2k-6}+a\sum_{k=0}^6 2^{6-k}(-3)^k\binom 6k x^{2k-4}$

The first sum contributes a x ⁰ term for 2k - 6 = 0, or k = 3, while the second sum contributes a x ⁰ term for 2k - 4 = 0, or k = 2. The coefficient of the sum of these terms must be zero:

$2^{6-3}(-3)^3\dbinom 63 + a\times2^{6-2}(-3)^2\dbinom 62 = 0$

which reduces to

2160a - 4320 = 0

2160a = 4320

a = 2

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