Question Number 100130 by bemath last updated on 25/Jun/20

how many integer number satisfy the equation ((sin x−∣x+2∣)/(x^2 −4x−5)) ≥ 0

Answered by mr W last updated on 25/Jun/20

let f(x)=sin x −∣x+2∣ for x≤−2: f(x)=sin x+x+2 f′(x)=cos x+1≥0 ⇒increasing f(x)≤f(−2)=sin (−2)≈−0.909 for x≥−2: f(x)=sin x−x−2 f′(x)=cos x−1≤0 ⇒decreasing f(x)≤f(−2)=sin (−2)≈−0.909 ⇒f(x)=sin x−∣x+2∣≤sin (−2)≈−0.909<0 ((sin x−∣x+2∣)/(x^2 −4x−5)) ≥ 0 since always sin x−∣x+2∣<0 ⇒x^2 −4x−5<0 i.e. (x+1)(x−5)<0 ⇒−1<x<5 ⇒x=0,1,2,3,4 ⇒5 integers

Commented byRasheed.Sindhi last updated on 25/Jun/20

T_(H_A N) k_(S ) S_I R !

Commented bybobhans last updated on 26/Jun/20

cooll great