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CIE 9709 2025 Nov Paper 11 Q3

A Level / CIE / P1

CIE 9709 2025 Nov Paper 11 Paper · Question 3

题目

Problem

In the expansion of

(px+3)5−(x3+px)4,(px+3)^5-\left(x^3+\frac{p}{x}\right)^4,

the coefficient of x4x^4 is 216.

Find the value of the positive constant pp.

[5]
题目中文翻译

在

(px+3)5−(x3+px)4(px+3)^5-\left(x^3+\frac{p}{x}\right)^4

的展开式中,x4x^4 的系数为 216。

求正常数 pp 的值。

解答

解法一

思路

展开

分别找出两个展开式的 x4x^4 项。第二个展开式里,每把一个 x3x^3 换成 p/xp/x,xx 的幂次会减少 44。

答题过程

展开

In (px+3)5(px+3)^5, the x4x^4 term is (54)(px)4⋅3=15p4x4\binom54(px)^4\cdot3=15p^4x^4. In the second expansion, choosing kk copies of p/xp/x gives exponent 3(4−k)−k=12−4k3(4-k)-k=12-4k. This equals 44 when k=2k=2, so its x4x^4 term is (42)(x3)2(p/x)2=6p2x4\binom42(x^3)^2(p/x)^2=6p^2x^4.

Subtracting the coefficients gives 15p4−6p2=21615p^4-6p^2=216. Dividing by 33 and writing u=p2u=p^2 gives

5u2−2u−72=(u−4)(5u+18)=0.5u^2-2u-72=(u-4)(5u+18)=0.

Since p>0p>0, we require u>0u>0. Thus u=4u=4 and p=2p=2.