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Could anyone give me a hint for the most rapid way for solving this question (solving it in exactly 2.5 minutes)? Thanks!

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YuiTo Cheng
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2 Answers2

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I doubt this is really doable in 2'30" (unless you are already in full concentration mode).

You spot at once that D has no solution (knowing that $e^x>x$) and C (first degree) has one. Rewrite E as $\cos x=e^{x^2}\ge1$ and it has exactly one solution ($x=0$).

B is quadratic and potentially has two solutions. By checking the discriminant, $76$, it indeed has them.

A is cubic and potentially has three roots. But the derivative $3x^2+1$ is positive so that there are no extrema and a single root.

  • This is a math subject GRE exam question, this is why it has to be solved in this time. –  Sep 06 '17 at 10:22
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    For (A), think of the intersection between the curve $y=x^3$ and the line $y=10-x$, and it's immediate that there's only one real solution. – Hans Lundmark Sep 06 '17 at 10:23
  • @HansLundmark: how is it immediate ? –  Sep 07 '17 at 11:21
  • $x \mapsto x^3$ is a strictly increasing continuous function which takes on all real values, $x \mapsto 10-x$ is a strictly decreasing continuous function which takes on all real values. – Hans Lundmark Sep 07 '17 at 11:47
  • @HansLundmark: so what's the difference with taking the derivative and showing it positive ? –  Sep 07 '17 at 11:52
  • Not much, except that if you're in a hurry, you might want to use that you already know these facts about (the graphs of) $x^3$ and $10-x$. (It wasn't meant as a criticism of your answer, which I upvoted already yesterday.) – Hans Lundmark Sep 07 '17 at 11:54
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I tried in 2.5. minutes, so more a guess than anything else really, for somebody of my limited skills.

So, $E$ has one real solution for $x=0$, as the Gaussian is bounded above by $1$ and the secant is bounded below by the same value, a graphical approach more than rigorous.

$A$ also has one real solution due to, I hope, a quick derivative check.

Same for $C$ and $D$ (due to linearity and and easy exponential inequality respectively).

My best guess is then $B$, which I verified it has two by checking the discriminant.

An aedonist
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