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If i substitute 0.4 in equation i get 1 as answer.
Any value from 0.4+ when placed in equation will give a value greater than 1 . It might be slightly higher than 1 but will be higher, so why would it be wrong.?
If i substitute 0.4 in equation i get 1 as answer.
Any value from 0.4+ when placed in equation will give a value greater than 1 . It might be slightly higher than 1 but will be higher, so why would it be wrong.?
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Option (D) gives the complete range of x but yes, (A) satisfies the inequality and gives the partial range. The question needs to be worded better.
For what range of values of 'x' will the inequality 15x - (2/x) > 1?
A. x > 0.4 B. x < 1/3 C. -1/3 < x < 0.4, x > 15/2 D. -1/3 < x < 0, x > 2/5 E. x < -1/3 and x > 2/5
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Responding to a pm:
Quote:
Could you please advice me when to consider zero when solving such question of ranges that hold the inequality true. Some question you and Bunuel consider zero on the number line to define the range points becuase it is very important to know your approach of - + - + . if I ignore zero as a point so I will get this approach wrong. I hope you got my issue idea.
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When x is a factor of the inequality, you need to consider 0 as a transition point. The idea is no different from the other transition points. On the right of 0, x is positive and on the left of 0, x is negative.
If the inequality is: x * (x + 3) > 0, the transition points are 0 and -3. If the inequality is: (x - 2)(x + 3) > 0, the transition points are 2 and -3.
Does this help?
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