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is defined as the least integer greater than x for all odd v
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Updated on: 08 Jan 2020, 15:17
Updated on: 08 Jan 2020, 15:17
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Question Stats:
71% (01:06) correct
28% (01:22) wrong based on 80 sessions
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\(\boxed{x}\)is defined as the least integer greater than x for all odd values of x, and the greatest integer less than x for all even values of x. What is the value of \(\boxed{-2}-\boxed{5}\)?
(A) –12
(B) –9
(C) –8
(D) –7
(E) 3
(A) –12
(B) –9
(C) –8
(D) –7
(E) 3
Re: is defined as the least integer greater than x for all odd v
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02 Aug 2018, 06:35
02 Aug 2018, 06:35
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Expert Reply
Explanation
If \(x\) is odd, \(\alpha (x)\) equals the least integer greater than x (e.g., if x = 3, then the “least integer greater than 3” is equal to 4)
If \(x\) is even, \(\alpha (x)\) equals the greatest integer less than x (e.g., if x = 6, the “greatest integer less than x” is equal to 5.
Since –2 is even, \(\alpha (-2)\) the greatest integer less than –2, or –3.
Since \(5\) is odd, \(\alpha(5)=\) the least integer greater than 5, or 6.
Thus, \(\alpha(-2)-\alpha(5) = -3 - 6 = -9\).
If \(x\) is odd, \(\alpha (x)\) equals the least integer greater than x (e.g., if x = 3, then the “least integer greater than 3” is equal to 4)
If \(x\) is even, \(\alpha (x)\) equals the greatest integer less than x (e.g., if x = 6, the “greatest integer less than x” is equal to 5.
Since –2 is even, \(\alpha (-2)\) the greatest integer less than –2, or –3.
Since \(5\) is odd, \(\alpha(5)=\) the least integer greater than 5, or 6.
Thus, \(\alpha(-2)-\alpha(5) = -3 - 6 = -9\).
Re: is defined as the least integer greater than x for all odd v
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19 Oct 2018, 23:30
19 Oct 2018, 23:30
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when x is odd then f(x)= x+1, f(5)= 5+1=6
when x is even then f(x)= x-1= -2-1= -3
-3-6=-9
when x is even then f(x)= x-1= -2-1= -3
-3-6=-9
