hb wrote:
Which of the following describes all values of n for which n^2-1\geq{0}
(A) n\geq{1}
(B) n\leq{1}
(C) 0\leq{n}\leq{1}
(D) n\leq{-1} or n\geq{1}
(E) -1\leq{n}\leq{1}]
Answer:
Equation: n^2-1\geq{0}
(n+1).(n-1)\geq{0}
The above inequality can be broken down into the following two inequalities.
(n+1) \geq{0} and (n-1)\geq{0}
(n+1) \geq{0}
n\geq{-1}
(n-1) \geq{0}
n\geq{1}
The Official Answer is D. Why am i not getting the n\leq{-1} ? What am i doing wrong above in my calculation ?
(A) n\geq{1}
(B) n\leq{1}
(C) 0\leq{n}\leq{1}
(D) n\leq{-1} or n\geq{1}
(E) -1\leq{n}\leq{1}]
Answer:
Equation: n^2-1\geq{0}
(n+1).(n-1)\geq{0}
The above inequality can be broken down into the following two inequalities.
(n+1) \geq{0} and (n-1)\geq{0}
(n+1) \geq{0}
n\geq{-1}
(n-1) \geq{0}
n\geq{1}
The Official Answer is D. Why am i not getting the n\leq{-1} ? What am i doing wrong above in my calculation ?
You are missing the case when both multiples are negative:
(n+1) \leq{0} --> n\leq{-1};
(n-1) \leq{0} --> n\leq{1}.
Common range n\leq{-1}.
This can be solved in another way:
n^2-1\geq{0} --> n^2\geq{1} --> n\leq{-1} or n\geq{1}.
Or:
n^2-1\geq{0} --> n^2\geq{1} --> |n|\geq{1} --> n\leq{-1} or n\geq{1}.
Answer: D.
Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html
inequations-inequalities-part-154664.html
inequations-inequalities-part-154738.html
Hope it helps.