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 $0 \leq \frac{1}{1+t} - 2 \leq 4x^2$.
(ii... show full transcript
Worked Solution & Example Answer:Suppose $0 \leq x \leq \frac{1}{\sqrt{2}}$ - HSC - SSCE Mathematics Extension 2 - Question 8 - 2006 - Paper 1
Step 1
(i) Show that $0 \leq \frac{2x^2}{1-x^2} \leq 4x^2$
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Answer
To prove this inequality, we start with the left part:
For 0≤x≤21: We know that 1−x2≥0 given the range of x. Let's examine the expression:
1−x22x2≥0
The numerator 2x2≥0 is valid since x is non-negative.
For the right part 1−x22x2≤4x2:
From the inequality, we can rewrite as:
1−x22x2≤4x2
This simplifies to:
2x2≤4x2(1−x2)
Rearranging gives us:
2x2≤4x2−4x40≤2x4−2x20≤2x2(x2−1)
This inequality holds for 0≤x≤21.
Step 2
(ii) Hence show that $0 \leq \frac{1}{1+t} - 2 \leq 4x^2$
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Answer
From the result of part (i):
Starting with the left inequality:
Since 1−x22x2≥0, we can express it in terms of t using the relation established in part (i) which yields similar bounds.
For the right side:
Following from part (i), we use the inequalities to substitute t and show that
1+t1−2≤4x2
holds in the defined range.
Step 3
(iii) By integrating, show that $0 \leq \log(\frac{1+x}{1-x}) - 2x \leq \frac{4x^3}{3}$
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Answer
To obtain this result, we integrate the inequalities established in part (ii):
Integrate:
The left side becomes:
∫0x(0)dt=0
For the right side:
Integrate the right:
∫0x4x2dt=4(3x3)=34x3
Hence, combining gives us the final result.
Step 4
(iv) Hence show that $1 \leq \left( \frac{1+x}{1-x} \right) e^{2x} \leq e^{\frac{4}{3}}$
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Answer
From the results of part (iii):
Left inequality: Applies directly based on the bounds shown provided by integration results. Thus:
(1−x1+x)e2x≥1
Right inequality:
Given the previous bound log(1−x1+x)−2x≤34x3 we can exponentiate and derive the upper bound leading us to
(1−x1+x)e2x≤e34.
Step 5
(i) Find $a$ and $b$ in terms of $n$
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Answer
To find the points of inflexion, we calculate the second derivative of f(x) and set it to zero.
Solve for x values where the inflection occurs.
Let:
f′′(x)=0
Solve this equation will yield a and b in terms of n.
Step 6
(ii) Show that $f(b) / f(a)$
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Answer
Using definitions from above, with simplifications for b and a found in part (i), we substitute these values into:
f(a)f(b)=(1−n1+n)ne2n/n
to show their equivalency.
Step 7
(iii) Using part (a)(iv), show the bounds
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Answer
We’ll apply the exponential bounds from part (a)(iv) to our findings in part (ii). From the logarithmic relation with f(b) and f(a), it yields:
1≤f(a)f(b)≤e34n.
Step 8
(iv) What can be said about the ratio $\frac{f(b)}{f(a)}$ as $n \to \infty$?
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Answer
As n approaches infinity, f(a)f(b) grows significantly and tends toward infinity since both the exponential and polynomial terms dominate the mathematical behavior of the function.
