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Find the greatest integer less than or equal to k

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ProfChaos
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Find the greatest integer less than or equal to k [#permalink] New post   25 Nov 2020, 19:45
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It is given that \(34 - |x^2 - 3x + 8| \leq k\)
Find the greatest integer less than or equal to k

A. 27
B. 28
C. 29
D. 30
E. 31

Source : Indian B-School Entrance exam
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B
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Re: Find the greatest integer less than or equal to k [#permalink] New post   25 Nov 2020, 20:16
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ProfChaos wrote:
It is given that {34-|\(x^2-3x+8\)|}<=k
Find the greatest integer less than or equal to k

A. 27
B. 28
C. 29
D. 30
E. 31

Source : Indian B-School Entrance exam



The value of k depends on the value of x.

The larger the value in the modulus, the smaller will be the difference between 34 and this value.

To get the maximum value of k, we need to get the least value for|\(x^2\)−3x+8|

For negative values of x, the value in the modulus will always keep increasing. Negative values are ruled out.

When x = 0, then |\(x^2\)−3x+8| = 8

When x = 1, |\(x^2\)−3x+8| = 6

When x = 2, |\(x^2\)−3x+8| = 6

When x = 3, |\(x^2\)−3x+8| = 8

From x = 4 onwards, the value in the modulus keeps on increasing

The least value of |\(x^2\)−3x+8| is 6.

Therefore the maximum value of k = 34 - 6 = 28


Option B

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Re: Find the greatest integer less than or equal to k [#permalink] New post   25 Nov 2020, 20:14
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ProfChaos wrote:
It is given that {34-|\(x^2-3x+8\)|}<=k
Find the greatest integer less than or equal to k

A. 27
B. 28
C. 29
D. 30
E. 31

Source : Indian B-School Entrance exam



We have to find the minimum value of \(x^2-3x+8\), and for this we can carry out differentiation =>\(\frac{d}{d(x)}(x^2-3x+8)=0......2x-3=0.....x=\frac{3}{2}\)

The value of \(x^2-3x+8=\frac{9}{4}-\frac{9}{2}+8=\frac{23}{4}=5.75\)

So \(34-5.75\leq{k}....28.25\leq{k}\)
Thus, integer value less than or equal to k is 28

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Re: Find the greatest integer less than or equal to k [#permalink] New post   27 Dec 2022, 04:08
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Re: Find the greatest integer less than or equal to k [#permalink]
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