If u and –3v are greater than 0, and u^(1/2)

Question

If u and –3v are greater than 0, and \sqrt{u} < \sqrt{-3v} , which of the following cannot be true ? A. \frac{u}{3} < -v B. \frac{u}{v} > -3 C. \sqrt{\frac{u}{-v}} < \sqrt{3} D. u + 3v > 0 E. u < -3v

PKN · Answer

Given, -3v is greater than zero, implies that v<0.

Given,  \sqrt{u} < \sqrt{-3v} 
we can square both sides of inequality since they are positive, Squaring both sides, we have
u < -3v-------------(equation 1)-----------------(Option E)

Adding -3v on both sides of (1), we have
u+3v<0--------------( Option D wrong )

Dividing 3 on both sides of (1), we have
 \frac{u}{3} < -v ------------(Option A)

Dividing v on both sides of (1), we have
 \frac{u}{v}  >  -3 ------------(Option B) (Flip of sign since v is negative)---(2)

Multiplying -1 on both sides of (2), we have
 \frac{u}{v}  <  -3 , take square root.
Or,  \sqrt{\frac{u}{-v}} < \sqrt{3} ------------(Option C)

Ans. (D)

Enchanting · Answer

u and -3v are greater than 0 \sqrt{u} < \sqrt{-3v} Since u and -3v are positive squaring the above will not change the inequality. u < -3v Option A: u < -3v \frac{u}{v} < -3 , true Option B: u < -3v \frac{u}{v} > -3v , here v is negative, which reversed the inequality true Option C: u < -3v \frac{u}{v} < -3 \sqrt{\frac{u}{-v}} < \sqrt{-3} , true Option D: u < -3v u + 3v < 0 , u + 3v > 0 is not true Answer Option D .

kayarat600 · Answer

Answers don't make sense at all you can't arbitrarily decide when v ismpostive and when it is negative ein options 1 and 2 v will be positive when squared and so will 3.

Bunuel · Answer

If you solve correctly the answer makes perfect sense. u and –3v are greater than 0 imply u > 0 and v < 0. Squaring \sqrt{u} < \sqrt{-3v} gives u < -3v , so E is correct right away. Dividing u < -3v by 3 gives \frac{u}{3} < -v , so A is correct . Dividing u < -3v by v , which is negative and flipping the sign we get \frac{u}{v} > -3 , so B is correct too. Dividing \sqrt{u} < \sqrt{-3v} by \sqrt{-v} , gives \sqrt{\frac{u}{-v}} < \sqrt{3} , so C is correct . However, adding 3v to u < -3v gives u + 3v < 0 , so D is NOT correct . Answer: D.

YUKESH23 · Answer

I got the explanation but when i tried with number the equation in option B doesn't holds true.

For example i am considering U=6 and V=-1, since u>0 & v<0, substituting in equation C gives me -6>-3 which doesn't holds good.

Please explain this issue

Krunaal · Answer

You cannot take u = 6, and v = -1; there's a constraint given the question,  \sqrt{u} < \sqrt{-3v}

Deeya007 · Answer

Aren't options B & D the same?

B. u/v>−3 => u>-3v

D. u+3v>0 => u>-3v

Bunuel

Bunuel · Answer

No. Since v < 0, when we multiply both sides by v, we must flip the inequality sign. So:

u/v > -3
u < -3v
u + 3v < 0

egmat · Answer

Hi Deeya007,

Good catch noticing they look similar, but B and D actually reduce to opposite things. The slip is in how you simplified B.

What happened with B. You took:

- B: u/v > -3 -> multiplied both sides by v -> u > -3v

That multiplication is the trap. Remember from the thread that -3v > 0 forces v < 0. So v is negative - and when you multiply (or divide) an inequality by a negative number, you must flip the sign.

Doing it correctly:

- u/v > -3, multiply both sides by v (negative, so flip): u < -3v

So B really reduces to u < -3v - which is exactly the fact we derived from the given inequality. That's why B can be true.

Now D:

- D: u + 3v > 0 -> u > -3v

No negative multiplication here, so no flip. D reduces to u > -3v, the opposite of what we know. That's why D cannot be true and is the answer.

So the two are not the same - the missing sign flip in B is the whole difference.

Quick way to feel the flip: take a true statement like 6 > 2. Multiply both sides by -1:

- Without flipping: -6 > -2 (false)
- With flipping: -6 < -2 (true)

Multiplying by anything negative reverses the direction - same rule that turns B's > into a <.

Answer: D

Deeya007

Aren't options B & D the same?

B. u/v>−3 => u>-3v

D. u+3v>0 => u>-3v

Bunuel

If u and –3v are greater than 0, and u^(1/2) < (-3y)^(1/2), which of

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