If 1

Question

If −1 10 B. x > \frac{17}{3} C. x^2 > 27 D. 3x + x^2 < −2 E. 2x – x^2 < 0

PyjamaScientist · Answer

If −1 10 x max value = 4.999. x min value = -0.999. None of the values for 2x give > 10. Not possible. B. x > \frac{17}{3} 17/3 is approx 5.xyz. That is greater than 5. Not possible. C. x^2 > 27 This means x is greater than 5.xyz. Not possible. D. 3x + x^2 < −2 This gives -2 < x < -1. Not possible. E. 2x – x^2 < 0 This gives the range: 0 < x < 2. Could be true. Correct choice.

achloes · Answer

Bunuel   PyjamaScientist

The following is the Veritas Prep explanation for choice E:

In the second step, I'm not sure why the inequality sign flips from < to >. Never come across this scenario in my practice so far so quite confused.

Appreciate your help!

PyjamaScientist · Answer

Hi  achloes ,
The sign flips because both the sides have been multiplied by -1. 
 Remember: One of the key axioms of inequalities is that when we multiply or divide both sides of an inequality by a negative number, this reverses the direction of the inequality.

achloes · Answer

Oh I see it now - got caught up on the sign flip and missed the change on the LHS. Thank you!

Anjali_1453 · Answer

Can someone explain option 5

Bunuel · Answer

2x - x^2 < 0 x^2 - 2x > 0 x(x - 2) > 0 The critical points are 0 and 2, which create three ranges: x < 0 0 < x < 2 x > 2 Try extreme values for x. If x is 10, then the whole expression is positive. So, in the third range (x > 2), the expression is positive. The trick here is that in the second range it is negative, and in the first range it is positive again: + - +. Therefore, 2x - x^2 < 0 implies x < 0 or x > 2. So, if -1 < x < 5 is true, 2x - x^2 < 0 could also be true because there is an overlap between -1 < x < 5 and x < 0 or x > 2. For example, if x = 3, which is between -1 and 5, then 2x - x^2 < 0 is true.

luisdicampo · Answer

Deconstructing the Question We need the choice that could be true given -110 gives x>5 , impossible since x<5 . B: x>\frac{17}{3} . Since \frac{17}{3}=5.\overline6 , this is impossible. C: x^2>27 would require |x|>\sqrt{27} . But \sqrt{27}>5 , so this is impossible in -12 . Both are possible inside -1

If 1<x<5, which is the following could be true?

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