STEP I 1990 question 9 solution

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STEP I 1990 question 9 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 1990 question 9 solution

$\displaystyle y = \frac{1}{x}, \; \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{x^{2}}$

Therefore the equation of the tangent at the point $\displaystyle \left(b, \frac{1}{b}\right)$ is $\displaystyle y = -\frac{1}{b^{2}}x + \frac{2}{b}$

The co-ordinates of C are at the intersection of $\displaystyle y = -x + 2$ and $\displaystyle y = -\frac{1}{b^{2}} + \frac{2}{b}$

$\displaystyle -x + 2 = -\frac{1}{b^{2}} + \frac{2}{b} \Rightarrow x\left(1-\frac{1}{b^{2}}\right) = 2\left(1-\frac{1}{b}\right)$

$\displaystyle x = \frac{2}{1 + \frac{1}{b}} = \frac{2b}{b+1}$

$\displaystyle y = \frac{2}{b+1}$

Using the expression for the area of a trapezium:

Area of ACC'A' = $\displaystyle \frac{1}{2}\left(1 + \frac{2}{b+1}\right)\left(\frac{2b}{b+1}\right) = \frac{b(b+3)}{(b+1)^{2}}$

Area of CBB'C' = $\displaystyle \frac{1}{2}\left(\frac{2}{b+1} + \frac{1}{b}\right)\left(b - \frac{2b}{b+1}\right) = \frac{(3b+1)(b-1)}{2(b+1)^{2}}$

Sum of Areas = $\displaystyle \frac{2b(b+3) + (3b+1)(b-1)}{2(b+1)^{2}} = \frac{5b^{2} + 4b - 1}{2(b+1)^{2}} = \frac{5b - 1}{2(b+1)}$

As the gradient is constantly increasing (becoming 'less' negative), the area under the curve between 1 and b is greater than the sum of the areas of ACC'A' and CBB'C:

$\displaystyle \int^{b}_{1} \frac{1}{x} \; \mathrm{d}x = \ln b \geq \frac{5b-1}{2(b+1)}, \; b > 1$

$\displaystyle \Rightarrow \ln (1 + z) \geq \frac{5z + 4}{2z}, \; z > 0$

To prove the left-hand inequality, it is sufficient to show that $\displaystyle \frac{5z + 4}{2z} \geq \frac{2z}{2+z}, \; z > 0$

$\displaystyle 5z^{2} + 14z + 8 \geq 4z^{2} \Leftrightarrow z^{2} + 14z + 8 \geq 0, \; z > 0$

Which is clearly true.

The inequality $\displaystyle \ln(1+z) \leq z$ follows immediately from the fact that $1 + z \leq e^{z}$ by the Maclaurin expansion of $e^{z}$ .

Alternatively, note that the area of the trapezium ABB'A' is greater than the area under the curve between 1 and b.

$\displaystyle \ln b \leq \frac{1}{2}\left(1 + \frac{1}{b}\right)\left(b-1 \right), \; b > 1$

$\displaystyle \ln (1+z) \leq \frac{z^{2} + 2z}{2 + 2z}, \; z>0$

$\displaystyle \frac{z^{2} + 2z}{2 + 2z} \leq z \Leftrightarrow z^{2} + 2z \leq 2z^{2} + 2z \Leftrightarrow 0 \leq z^{2}$

Therefore $\displaystyle \frac{2z}{2+z} \leq \ln (1+z) \leq z, \; z>0$

Solution by Dystopia.

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