Which of the following is/are true?

Fell for trap (3)... When thing appear too easy... its a trap...

Hi chetan2u

I solved the given inequalities by considering different cases. Could you please help?
Statement 1: |3x-19|>5-x
case 1: for 3x-19>=0 --> x>=19/3
3x-19>5-x
x>24/4 --> X>6, which seems in valid as x>=19/3--> x>=6.3
Case 2: for 3x-19<0 --> x<19/3
-(3x-19) > 5-x
3x-19 < x-5
2x < 14
x<7 , which again seems invalid as x < 6.3

Where did I go wrong?

Intuitively I can see that I is true for all values of x. But when we calculate the range: it comes x>6 , x<7 which is essentially 6<x<7. Shouldnt this be the range of answers?

What error or flaw am I commiting here?

Can anyone of you explain: Bunuel

Hi
In this question, if we try any value for x for statement 3 then it will be negative only because the negative sign is given outside the mode and any value which will appear after the mode will become negative. what is it that I'm missing also in the above 2 statements I tried different values and then concluded that any case applied to the given inequality will hold the inequality true. so for 1st and 2nd is my approach correct?

Thanks

As x need not be integer and can be anything, surely there will be a value of x for which the expression 5x-23 will be 0, so we will get -|0| = -0 =0, as 0 is neither negative nor positive

Can someone show the algebric approach for option 1 and 2 by opening the mode. Also does < means or and > means and?

Which of the following is/are true?

I. |3x - 19| > 5 - x holds for all values of x.
II. There is no value of x for which |2x - 10| < 4 - x holds.
III. -|5x - 23| is negative for all values of x.

A. None
B. I only
C. I and II only
D. I and III only
E. I, II and III


I. \(|3x - 19| > 5 - x\) holds for all values of x.

When \(x < \frac{19}{3}\) (notice that \(\frac{19}{3} = 6 \frac{1}{3}\)), \(3x - 19 < 0\) thus in this case \(|3x - 19| = -(3x - 19)\), so we'd get \(-(3x - 19) > 5 - x\):

\(-3x + 19 > 5 - x\);
\(14 > 2x\);
\(x < 7\).

Since we consider \(x<6 \frac{1}{3}\) range, then for this case we'd have \(x < 6 \frac{1}{3}\).

When \(x \geq \frac{19}{3}\), \(3x - 19 \geq 0\) thus in this case \(|3x - 19| = 3x - 19\), so we'd get 3x - 19 > 5 - x:

\(3x - 19 > 5 - x\);
\(4x > 24\);
\(x > 6\).

Since we consider \(x \geq 6 \frac{1}{3}\) range, then for this case we'd have \(x \geq 6 \frac{1}{3}\).

\(x < 6 \frac{1}{3}\) and \(x \geq 6 \frac{1}{3}\) give all values of x. So, \(|3x - 19| > 5 - x\) holds for all values of x.


II. There is no value of x for which |2x - 10| < 4 - x holds.

Here notice that x cannot be more than 4, because if it is then the RHS (the right hand side) becomes negative and it cannot be greater than the LHS (the left hand side), which is an absolute value and therefore is always more than or equal to 0.

So, x must be less than or equal to 4. Now, when \(x \leq 4\), \(2x - 10 < 0\) and thus 2x - 10 = -(2x - 10). So, we'd get: -(2x - 10) < 4 - x:

\(-2x + 10 < 4 - x\);
\(x > 6\).

\(x \leq 4\) (the range we consider) and \(x > 6\) have no overlap, therefore there is no value of x for which |2x - 10| < 4 - x holds.


III. -|5x - 23| is negative for all values of x.

This statement says that \(-|5x - 23| < 0\) for all values of x:

\(|5x - 23| > 0\).

This is clearly wrong because when x = 23/5, \(|5x - 23| = 0\), not greater than 0.

Answer: C.

Hope it's clear.

Expert's Global's explanation-

Statement I:
The minimum value for |3x-19| is zero, when x = 19/3 = 6.33.
For x =19/3 = 6.33, |3x-19| > 5-x.
For a greater value of x, e.g. x = 7, |3x-19| > 5-x. For a smaller value of x, e.g. x = 5, |3x-19| > 5-x.
So, for any value of x, |3x-19| > 5-x holds.

Statement II:
The minimum value for |2x-10| is zero, when x = 5.
For x = 5, |2x-10| > 4-x.
For a greater value of x, e.g. x = 6, |2x-10| > 4-x. For a smaller value of x, e.g. x = 4, |2x-10| > 4-x.
So, for no value of x, |2x-10| < 4-x holds.

Statement III:
The least value of -|5x-23] is zero, which is not negative. So -|5x-23] is negative does not hold for all values of x.
Only statements I and II are correct.
Hence, C is the correct answer choice.

Answer video link