I think it should be E
reasoning in my view for this choice length is fixed 2 to 6 and i ignored D for |x| part.

satish.nagdev wrote:I think it should be E
reasoning in my view for this choice length is fixed 2 to 6 and i ignored D for |x| part.

Hi:

They want a single finite line graph, and I think I'm not too good with inequalities with absolute value, please anyone, confirm c) and d) solutions:

a) x^4<1 --> x>1
b) x^3 <= 27 --> x<=3 (both are discarded)
c) x^2>=16 --> x>=4 and x<=-4, is it correct? or the symbol shouldn't change: x>=4 and x>=-4 .. either way, it's not a finite line
d) 2<=|x|<=5 -->
1- x>=2 and x<=-2 (pls, confirm change of sign: from >= to <= or just maybe < o nochange).
2- x<=5 and x>=-5, with the = sign in both solutions, but signs inversed, I'm not sure about this. Please confirm.
anyway, none of them is an answer
e) 2<=3x+4<=6 --> x>=-2/3 and x<= 2/3 and we have a infinite set of points.

So, my answer is E, what is the OA pls? Are c) and d) correctly solved?

Thank you,
Silvia

Hi Silvia & Satish,
U both r correct, the OA is E.

@ Silvia - Ur approach to solve C is correct. However, I'm not sure bout the approach to solve D.

@ Satish - Can u pls explain how did u get this 1 step answer? Also why do u eliminate the option with |X| sign when u require to find a solution set? Can u pls explain the logic behind this?

Thanks 2 u two

---LA---