cosh-1x=?
log[x+x2+1]
log[x–x2+1]
log[x–x2–1]
log[x+x2–1]
Explanation for the correct option:
Step 1. Let cosh-1x=y
⇒ x=coshy
=[ey+e-y]2 ; ∵coshθ=eθ+e-θ2
Step 2. Multiply and divide it by ey
x=[e2y+1]2ey
Step 3. Put k=ey
⇒ k2–2xk+1=0
⇒ k=x±x2–1 ; ∵x=-b±b2-4ac2a
⇒ ey=x±x2–1
Step 4.Take log on both side
y=ln[x±x2–1]
⇒ cosh-1x=log[x+x2–1]
Hence, Option ‘D’ is Correct.