need help on this extension1 question (1 Viewer)

CM_Tutor · Jul 16, 2021

Note that the two curves given are inverse functions, and thus reflections in $y = x$ .

The points $A$ and $B$ lie on $y = 2 - x$ , which is perpendicular to $y = x$ .

Consider $P\,\left(\cfrac{1}{2},\,\cfrac{3}{2}\right)$ , which is also on $y=2-x$ but further from $y=x$ than is $y=2^x$ because $2^{\frac{1}{2}} = \sqrt{2} = 1.414... < \cfrac{3}{2}$ .

The reflection of $P$ in $y = x$ is $P'\,\left(\cfrac{3}{2},\,\cfrac{1}{2}\right)$ .

The distance $PP'$ is greater than the distance $AB$ as $A$ and $B$ lie between $P$ and $P'$ on $y=2-x$ .

Further, the distance $PP'$ is $\sqrt{1^2 + 1^2} = \sqrt{2}$ .

Thus the distance $AB < \sqrt{2}$ , as required.

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need help on this extension1 question (1 Viewer)

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