Christmas Streak 54/88: 2018 CSAT (Korean SAT) #21 of Natural Sciences

If we look into the inequality, we observe that it means:

$\cases{g(x)\ge f(x)~(x\ge e) \\\\ f(x)\ge g(x)~(1\le x<e)}$

Then let's graph $y=f(x)$ and $y=g(x)$ first.

They'll look something like this.

And since we want to minimize the slope of $y=g(x),$ $g(1)$ needs to be maximized to $0.$

So we have that $g(1)=0,$ and for $x>e$ we need $g(x)\ge f(x).$

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Now note that the tangent line at $(e,~1-t)$ of $y=f(x)$ has a slope of $\dfrac{1}{e}.$

We can set this as $y=\dfrac{1}{e}(x-e)+1-t=\dfrac{1}{e}x-t.$

If this line is under the $x$ -axis at $x=1,$ namely, if $t\le\dfrac{1}{e},$ then the minimum of the slope of $y=g(x)$ would occur when $y=g(x)$ passes through $(e,~1-t).$

Then since $y=g(x)$ also passes through $(1,~0),$ $h(t)=\dfrac{1-t}{e-1}~\left(t\ge\dfrac{1}{e}\right).$

We have $h'(t)=-\dfrac{1}{e-1}~\left(t<\dfrac{1}{e}\right)$

Plugging in $t=\dfrac{1}{2e},$ we obtain $h'\left(\dfrac{1}{2e}\right)=\dfrac{1}{e-1}.$

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If $t> \dfrac{1}{e},$ then in order for the slope to be minimal, $y=g(x)$ has to a tangent line of $y=f(x)$ at $x>e.$

Let that point of contact be $(p,~-t+\ln p).$

Since $f'(p)=\dfrac{1}{p},$ we obtain $y=g(x)=\dfrac{1}{p}(x-p)-t+\ln p.$

But we know that $g(1)=0.$ So we get $\dfrac{1}{p}-1-t+\ln p=0~\Leftrightarrow~t+1=\dfrac{1}{p}+\ln p.$

Differentiating both sides by $p,$ we obtain $\dfrac{dt}{dp}=-\dfrac{1}{p^2}+\dfrac{1}{p}~\Leftrightarrow~\dfrac{dp}{dt}=\dfrac{p^2}{p-1}.$

The slope of $y=g(x)$ is $\dfrac{1}{p},$ so we have $h(t)=\dfrac{1}{p}.$

$h'(t)=-\dfrac{1}{p^2}\cdot\dfrac{dp}{dt}=-\dfrac{1}{p^2}\cdot\dfrac{p^2}{p-1}=-\dfrac{1}{p-1}.$

Plugging in $h(a)=\dfrac{1}{e+2}=\dfrac{1}{p},$ we get $h'(a)=-\dfrac{1}{p-1}=-\dfrac{1}{e+1}.$

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From above, the value of $h'\left(\dfrac{1}{2e}\right)\times h'(a)$ is $\boxed{\dfrac{1}{(e-1)(e+1)}}.$