Sep 2019 CSAT mock test for Sciences, #15

Calculus Level 3

Point $\rm A$ with positive $x$ -coordinate is on the curve $y=e^x,$ while point $\rm B$ is on the curve $y=-\ln x,$ satisfying the following conditions:

$\overline{\rm OA}=2\overline{\rm OB},$
$\angle \rm AOB = \dfrac{\pi}{2}.$

Find the slope of line $\rm OA.$ ( $\rm O$ is the origin.)

Only 48.3% of the students got this right -- second hardest among the multiple choices problems.

\dfrac{5}{\ln 5}

e

\dfrac{2}{\ln 2}

\dfrac{e^2}{2}

\dfrac{3}{\ln 3}

1 solution

Boi (보이)
Sep 5, 2019

Shown below is a diagram of the given situation. It is immediate that rotating $\dfrac{\pi}{2}$ the graph $y=-\ln x$ results in $y=e^x,$ hence if we do the same for point $\rm B,$ it lands on $y=e^x$ as $\rm B',$ while also being collinear with $\rm O$ and $\rm A.$

Hence, if ${\rm A}(2t,~e^{2t}),$ then ${\rm B'}(t,~e^t)$ while also $e^{2t}=2e^t.$ Since $e^t=2,$ we find $t=\ln 2.$ The slope is easily calculated to be $\dfrac{2}{\ln 2}.$

You wrote OA=2OB. Then how do you get the coordinates of B as $(2t,e^{2t})$ ?. Also how do you get $e^t=2$ ?

A Former Brilliant Member - 1 year, 9 months ago

As stated above, O, A, B' are collinear, and since OB' = OB = 2OA, we have that B' has an x coordinate twice as that of A. The process of obtaining $e^t=2$ is fairly simple: divide both sides of $e^{2t}=2e^t$ by $e^t.$

Boi (보이) - 1 year, 9 months ago

Aah I get what you mean. It was supposed to be B', not B. Whoopsie

Boi (보이) - 1 year, 9 months ago

@Boi (보이) – But you have written it otherwise. Please check. You have written OA=2OB.

A Former Brilliant Member - 1 year, 9 months ago

@A Former Brilliant Member – Sorry, gotcha

Boi (보이) - 1 year, 9 months ago

Sep 2019 CSAT mock test for Sciences, #15

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