STEP III 1997 question 2 solution

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STEP III 1997 question 2 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP 1997 Solutions > STEP III 1997 question 2 solution

$\displaystyle \\f(t)=\frac{\ln t}{t}\\f'(t)=\frac{1-\ln t}{t^2}\\ f'(t)=0 \text{ at } t=e\\ \therefore e,e^{-1} \text{ is a maximum point }$

$\displaystyle f(t)\rightarrow0, \text{ as } t\rightarrow\infty$

Also, note that f(t) is negative for 0<t<1 and tends to $\displaystyle -\infty$ as t tends to zero (Althought strictly speaking, f(t) is undefined at t=0). f(1)=0 and f(t) positive for t>1. Use these information to sketch your graph, remember to show distinctly that the x-axis is an asymptote for large t. Then from your graph, it can be seen that 2 values of t correspond to a positive value of f(t).

$\displaystyle \\x^{y}=y^{x}\\ y\ln x=x\ln y\\ \frac{\ln x}{x}=\frac{\ln y}{y}$

i)There's a single (positive) value of y which satisfy $\displaystyle x^{y}=y^{x}$ for a given positive value of x within $\displaystyle 0<x\leq1$ (And x=y)

ii)There're 2 values of y which satisfy $\displaystyle x^{y}=y^{x}$ for a given positive value of x for $\displaystyle 1<x<\infty$ (and one of the 2 values of y always is equivalent to x) iii)At x=e, y=e, so there's only one value of y for x=e

Putting together these information, you should be able to sketch your final graph. Your curve must show:

i)A straight line from (not through) the origin with unit gradient. (This correspond to the set of solutions for which x=y for $\displaystyle x^{y}=y^{x}$

ii)A curve with x=1 as asymptote for large y and y=1 as asymptote for large x. This curve intersects the straight line at x=e. This curve looks similar to the shape of y=1/x

Solution by khaixiang

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