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30 Oct 2021, 11:58
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VeritasKarishma
When I tested cases I tired to cover all numbers:
In (1) we have z^3<z relationship, and we can test 5 cases:
1. When z>1. Any number >1 cubed is larger than itself (example: 2^3=8 => 8>2). Conclusion => z is not >1.
2. Positive fractions (when 0<z<1). Any positive fraction cubed is smaller than itself (example: 1/2^3=1/8 => 1/8<1/2). Conclusion => z could be a positive fraction.
3. When z<-1 Any number <-1 cubed is smaller than itself, it does not loose the minus sign (example: -2^3=-8 => -8<-2). Conclusion => z could be a number < -1.
4. Negative fraction (-1<z<0). Any negative fraction cubed is larger than itself, it is closer to 0 and positive numbers (example -1/2^3=-1/8 => -1/8>-1/2). Conclusion => z is not a negative fraction.
5. Zero or 1. 0^3=0, so 0^3 is not less than 0. 1^3= 1, so 1^3 is not less than 1. Conclusion z is not 0 or 1.
In (2) we take an absolute value of z. This means that any value of z becomes positive. Therefore, only fractions work.
From (1) z could be a number <-1 or a positive fraction. From (2) z could be either a negative or positive fraction. The only thing that overlaps is positive fraction.
=> z must be a positive fraction.
02 Apr 2023, 07:26
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(1) (z + 1)(z)(z – 1) < 0
Thus, z<-1 or 0<z<1
Not sufficient
(2) |z| < 1
=> -1<z<1
Not sufficient
(1)+(2)
=> 0<z<1
Sufficient
Thus, z<-1 or 0<z<1
Not sufficient
(2) |z| < 1
=> -1<z<1
Not sufficient
(1)+(2)
=> 0<z<1
Sufficient
Archit3110
Major Poster
02 Apr 2023, 08:07
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for solving such questions check with below range
-2<-3/2<-1<-1/2<0<1/2<1<3/2<2
target is z>0 or say is it +ve
#1
(z + 1)(z)(z – 1) < 0
this will hold true when z is 1/2 or -2
insufficient
#2
|z| < 1
possible for -1<z<0 and 0<z<1
insufficient
from 1 &2
0<z<1
sufficient
option C
-2<-3/2<-1<-1/2<0<1/2<1<3/2<2
target is z>0 or say is it +ve
#1
(z + 1)(z)(z – 1) < 0
this will hold true when z is 1/2 or -2
insufficient
#2
|z| < 1
possible for -1<z<0 and 0<z<1
insufficient
from 1 &2
0<z<1
sufficient
option C
MHBS
