If M and m are respectively the largest and the smallest integers that satisfying the inequality 6n^2 −5n≤99, find the value of M−m.


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Question Number 119835 by ZiYangLee last updated on 27/Oct/20

$If M and m are respectively the largest and$ $the smallest integers that satisfying the$ $inequality 6 n^{2} - 5 n ⩽ 99, find the value of$ $M - m .$
Commented by talminator2856791 last updated on 27/Oct/20

$is the question correctly stated ?$
	Answered by TANMAY PANACEA last updated on 27/Oct/20

	$6 n^{2} - 5 n - 99 ⩽ 0$ $6 n^{2} - 27 n + 22 n - 99 ⩽ 0$ $3 n (2 n - 9) + 11 (2 n - 9) ⩽ 0$ $(2 n - 9) (3 n + 11) ⩽ 0$ $c r i t i c a l v a l u e o f n = \frac{9}{2} a n d \frac{- 11}{3}$ $f (n) = 6 n^{2} - 5 n - 99$ $w h e n n > \frac{9}{2} f (n) > 0$ $w h e n n < \frac{- 11}{3} f (n) > 0$ $w h e n \frac{9}{2} > n > \frac{- 11}{3} f (n) < 0$ $a t n = \frac{- 11}{3} f (n) = 0$ $a t n = \frac{9}{2} f (n) = 0$ $M = 4 m = - 3$ $M - m$ $4 - (- 3) = 7$
	Commented by ZiYangLee last updated on 27/Oct/20

	$Correct!$
	Commented by TANMAY PANACEA last updated on 27/Oct/20

	$t h a n k y o u s i r$
	Answered by Olaf last updated on 27/Oct/20

	$6 n^{2} - 5 n - 99 = 0$ $Δ = 25 - 4 \times 6 \times (- 99) = 2401 = 49^{2}$ $n_{1} = \frac{5 - 49}{12} = - \frac{11}{3}$ $n_{2} = \frac{5 + 49}{12} = \frac{9}{2}$ $\Rightarrow 6 n^{2} - 5 n ⩽ 99 if n \in [- \frac{11}{3}; + \frac{9}{2}]$ $M = n_{m a x} = 4$ $m = n_{m i n} = - 3 (if m \in Z)$ $M - m = 4 - (- 3) = 7$ $(or 4 if m \in N)$
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