题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

(i) Given $y=\ln(2x^2+5)$ find

\frac{dy}{dx}

(2)

(ii) A curve $C$ has equation $y=f(x)$ where

f(x)=\frac{21x}{3x^2+k}\qquad x\in\mathbb{R}

where $k$ is a positive integer, $k>1$ .

Given that

\int_1^k f(x)\,dx<7\ln 8

find the greatest possible value of $k$ .

(5)

题目中文翻译

本题必须写出全部解题步骤。

不接受完全依赖计算器技术的解法。

(i) 已知 $y=\ln(2x^2+5)$ ，求

\frac{dy}{dx}

(ii) 曲线 $C$ 的方程为 $y=f(x)$ ，其中

f(x)=\frac{21x}{3x^2+k}\qquad x\in\mathbb{R}

这里 $k$ 是正整数，且 $k>1$ 。

已知

\int_1^k f(x)\,dx<7\ln 8

求 $k$ 的最大可能值。

解答

(i)

Given

y=\ln(2x^2+5)

Use

\frac{\mathrm{d}}{\mathrm{d}x}\ln u=\frac{1}{u}\frac{\mathrm{d}u}{\mathrm{d}x}

where

u=2x^2+5

Then

\frac{\mathrm{d}u}{\mathrm{d}x}=4x

\boxed{\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{4x}{2x^2+5}}

(ii)

We need

\int_1^k \frac{21x}{3x^2+k}\,\mathrm{d}x

Let

u=3x^2+k

Then

\frac{\mathrm{d}u}{\mathrm{d}x}=6x

21x\,\mathrm{d}x=\frac72\,\mathrm{d}u

Therefore

\int \frac{21x}{3x^2+k}\,\mathrm{d}x =\frac72\ln(3x^2+k)+c

Now apply the limits:

\begin{aligned} \int_1^k \frac{21x}{3x^2+k}\,\mathrm{d}x &=\left[\frac72\ln(3x^2+k)\right]_1^k \\ &=\frac72\ln(3k^2+k)-\frac72\ln(3+k) \\ &=\frac72\ln\left(\frac{3k^2+k}{3+k}\right) \end{aligned}

Given

\int_1^k f(x)\,\mathrm{d}x<7\ln8

we have

\frac72\ln\left(\frac{3k^2+k}{3+k}\right)<7\ln8

Divide by $\frac72$ :

\ln\left(\frac{3k^2+k}{3+k}\right)<2\ln8

Since

2\ln8=\ln64

we get

\ln\left(\frac{3k^2+k}{3+k}\right)<\ln64

Hence

\frac{3k^2+k}{3+k}<64

Since $k>1$ , we have $3+k>0$ , so

3k^2+k<64(3+k)

This gives

\begin{aligned} 3k^2+k&<192+64k \\ 3k^2-63k-192&<0 \end{aligned}

Divide by $3$ :

k^2-21k-64<0

Solve the corresponding equation:

k^2-21k-64=0

Using the quadratic formula,

k=\frac{21\pm\sqrt{21^2+4(64)}}{2} =\frac{21\pm\sqrt{697}}{2}

The positive root is

\frac{21+\sqrt{697}}{2}=23.70\ldots

Therefore the inequality is satisfied for

k<23.70\ldots

Since $k$ is an integer and $k>1$ , the greatest possible value is

\boxed{23}

IAL 2025 Oct Q6

题目

解答

(i)

(ii)