Letf(x)=loge(sinx),(0<x<π) and g(x)=sin−1(ex),(x≥0).If α is a positive real number such that a=(fog)′(α) and b=(fog)(α), then :
A
aα2−bα−a=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
aα2+bα+a=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
aα2+bα−a=−2α2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
aα2−bα−a=1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is Daα2−bα−a=1 f(x)=loge(sinx),(0<x<π) & g(x)=sin−1(e−x),(x≥0) f(g(x))=−x⇒(f(g(x)))′=−1 f(g(α)=−α=b⇒(f(g(α)))′=−1=a b=−α a=−1 Now check for options