Given f(x)=loge(sinx),(0<x<π) and g(x)=sin−1(e−x),(x≥0), then we have
fog(x)=f(g(x))
=f(sin−1(e−x))
=loge(sin(sin−1(e−x)))
=loge(e−x)=−x
⇒fog(x)=−x
⇒b=f(g(α))=−α
Now (f(g(x)))′=−1
⇒a=(f(g(α)))′=−1
∴aα2−bα−a=(−1)α2−(−α)α−(−1)
=−α2+α2+1=1.
JEE Main 2019 — Mathematics Algebra
Let f(x)=loge(sinx),(0<x<π) and g(x)=sin−1(e−x),(x≥0). If α is a positive real number such that a=(fog)′(α) and b=(fog)(α), then
Held on 10 Apr 2019 · Verified 6 Jul 2026.
aα2+bα+a=0
aα2+bα−a=−2α
aα2−bα−a=0
aα2−bα−a=1
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