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