NIMCET 2008 Question Paper with solutions

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Question : 40 of 120

Marks: +1, -0

If

y=\sec ^{-1}(\;{x+1}/{x-1})+\sin ^{-1}(\;{x-1}/{x+1}) x ∈[0, ∞)

and

x ≠ 1

,then

{d y}/{d x}

is equal to

1
$\;{x-1}/{x+1}$
0
$\;{x+1}/{x-1}$

Solution:

y=\sec ^{-1}(\;{x+1}/{x-1})+\sin ^{-1}(\;{x-1}/{x+1})( ∵ \sec ^{-1} x=\cos ^{-1} \;{1}/{x})

⇒ \;\; y=\cos ^{-1}(\;{x-1}/{x+1})+\sin ^{-1}(\;{x-1}/{x+1})

⇒( ∵ \sin ^{-1} x+\cos ^{-1} x=\;{π}/{2})

⇒ \;\; y=\;{π}/{2} ⇒ \;\; \;{d y}/{d x}=0

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