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If p<0 and the value of 1−p/q is between 0 and 1, then which of the following must be true?

I. p2+q2>1

II. p+1/q<0

III. q–p<0

A. Only I
B. Only II
C. Only III
D. I and III
E. II and III


Let try to understand what is given first:

0<1−p/q<1 this imply that 1>p/q>0 also
if p and q both positive than q>p
and if p and q both negative than q<p

since p<0 or negative this mean that q is also negative and smaller than p example p= -2 and q= -4
now look at 3 option
1) p^2 + q^2 >1
what if p =-0.2 and q = -0.8 option 1 not a must be true
2)p+1/q<0
since both are negative
(-ve) +1/(-ve) is always less than 0 so must be true
3)q–p<0
since both are negative and p>q this means that this st. must always be true
ex- p=-0.2 and q=-0.4
q-p = -0.4-(-0.2)=-0.2<0
p=-1 q=-3
q-p = -3-(-1) = -2<0
so must be true

IMO E (2 & 3)
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If p<0 and \(0<1-p/q<1\) then q must be negative as well. Otherwise they will add up and sum with 1.also, since they are negative p>q. Otherwise the given inequality would not work

Statement 1.\(p^2+q^2>1\) this is not necessary the case. Consider p=-0.001 and q=-0,1
Statement 2. They are both negative so it must be true
Statement 3. Also True according to the reasoning above.

The answer is E




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p < 0
And,
0 < 1 - p/q < 1
-1 < -p/q < 0

Multiply by -1
=> 1 > p/q > 0
=> 0 < p/q < 1

Since, p < 0
=> q < 0 and p > q

III. p > q (proved above)

I. p^2 + q^2 > 1

If, p = -2, q = -1
=> 4 + 1 > 1 ---- True

If, p = -1/3, q = -1/2
=> 1/9 + 1/4 = 13/36 < 1 --- False

=> Reject I.

II. p + 1/q < 0

p - negative
q - negative => 1/q => negative

=> Negative Number + Negative Number = Negative Number < 0 --- True (always)

=> II. And III. Are always true.

Answer: E

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