1
Question
The inequality sin−1(sin5)>x2−4x holds if
The inequality sin−1(sin5)>x2−4x holds if
Open in App
Solution
The correct option is D xε(2−√9−2π,2+√9−2π)
Since 3π/2<5<2π, we have sin 5 < 0, so sin−1(sin5)=5−2π. Therefore, the given inequality can be written as 5−2π>x2−4x or x2−4x+(2π−5)<0
⇒[x−4−√16−4(2π−5)2][x−√16−4(2π−5)2]<0
⇒[x−(2−√9−2π)][x−(2+√9−2π)]<0
⇒xε(2−√9−2π,2+√9−2π)
Since 3π/2<5<2π, we have sin 5 < 0, so sin−1(sin5)=5−2π. Therefore, the given inequality can be written as 5−2π>x2−4x or x2−4x+(2π−5)<0
⇒[x−4−√16−4(2π−5)2][x−√16−4(2π−5)2]<0
⇒[x−(2−√9−2π)][x−(2+√9−2π)]<0
⇒xε(2−√9−2π,2+√9−2π)
Suggest Corrections
0
Similar questions