SSCE HSC Mathematics Extension 2 Integration by parts: Suppose $0 \leq x \leq \frac{1}{

Question

Suppose $0 \leq x \leq \frac{1}{\sqrt{2}}$.

(i) Show that $0 \leq \frac{2x^2}{1-x^2} \leq 4x^2$.

(ii) Hence show that $0 \leq \frac{1}{1+t} - 2 \leq 4t^2$.

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Answer

(iv) Hence show that for $0 \leq x \leq \frac{1}{\sqrt{2}}$

$$1 \leq \left( \frac{1+x}{1-x} \right)e^{2x} \leq e^{\frac{4}{3}}.$$

For $x > 0$, let $f(x) = x^n e^x$, where $n$ is an integer and $n \geq 2$.

(b) (i) The two points of inflexion of $f(x)$ occur at $x = a$ and $x = b$, where $0 < a < b$. Find $a$ and $b$ in terms of $n$.

(ii) Show that $\frac{f(b)}{f(a)} = \left( \frac{1+\sqrt{n}}{1-\sqrt{n}} \right)^{n/2} e^{2n/\sqrt{n}}.$

(iii) Using the result of part (a)(iv), show that $1 \leq \frac{f(b)}{f(a)} \leq e^{\frac{4}{n}}.$

(iv) What can be said about the ratio $\frac{f(b)}{f(a)}$ as $n \rightarrow \infty$?

Suppose $0 \leq x \leq \frac{1}{\sqrt{2}}$ - HSC - SSCE Mathematics Extension 2 - Question 8 - 2006 - Paper 1