Multiple Scenarios

The Multiple Scenarios skill is very similar to Case checking and Counter Examples, though it may not be immediately obvious how to proceed. You will have to carefully determine the scenarios, and consider each of them before making the final judgement.

If xx is a real number such that x2>10 x^2 > 10 , which of the following statements must be true?

A) x>10 x > \sqrt{10} .
B) x2+2x>10 x^2 + 2x > 10 .
C) x2+2x>3 x^2 + 2x > 3 .
D) x=4 x = 4 .
E) x2+2x≠5 x^2 + 2x \neq 5 .

Solution: If you tried to do this question quickly, you likely got tricked and chose the wrong answer. Options A and B appear a very tempting choice, but they do not satisfy all possible scenarios.
Let's see how to approach this problem. If x2>10 x^2 > 10 , then we know that either x>10 x > \sqrt{10} or x<−10 x < - \sqrt{10} .

Consider statement A. x=−4 x = -4 satisfies x2>10 x^2 > 10 . However, it does not satisfy x>10 x > \sqrt{10} .
Consider statement B. x=−4 x = -4 satisfies x2>10 x^2 > 10 . However, it does not satisfy x2+2x>10 x^2 + 2x > 10 .
Consider statement D. x=−4 x = -4 satisfies x2>10 x^2 > 10 . However, it does not satisfy x=4 x = 4 .
Consider statement E. x=−1−6 x = -1 - \sqrt{6} satisfies x2>10 x^2 > 10 . However, it does not satisfy x2+2x≠5 x^2 + 2x \neq 5 .
From here, we get that only Statement C could be true, so that is our answer. We now proceed to show it. If x>10 x > \sqrt{10} , then clearly x2+2x>x2=10>3 x^2 + 2x > x^2 = 10 > 3 .
If x<−10 x < - \sqrt{10} , then x+1<−10+1<0 x+1 < - \sqrt{10} + 1 < 0 and so (x+1)2>(−10+1)2 (x+1)^2 > ( -\sqrt{10} + 1)^2 . This gives us x2+2x+1>10−210+1 x^2 + 2x + 1 > 10 - 2 \sqrt{10} + 1 . Hence, we get that x2+2x>10−210>3 x^2 + 2x > 10 - 2 \sqrt{10} > 3 .
Thus, the answer is C.

#Practice #Skill

Note by Arron Kau
6 years, 10 months ago

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