NIMCET 2009 Question Paper with solutions

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Question : 3 of 120

Marks: +1, -0

The number of solutions for

\tan ^{-1} √{x(x+1)}+\sin ^{-1} √{x^{2}+x+1}=\;{π}/{2}

is

zero
one
two
infinite

Solution:

∵ \tan ^{-1} √{x(x+1)}=\cos ^{-1} \;{1}/{√{x^{2}+x+1}}

∴ \tan ^{-1} √{x(x+1)}+\sin ^{-1} √{x^{3}+x+1}=\;{π}/{2}

gives

\cos ^{-1} \;{1}/{√{x^{2}+x+1}}+\sin ^{-1} √{x^{2}+x+1}=\;{π}/{2}

which holds if

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

⇒ \;\; x^{2}+x+1=1

⇒ \;\; x(x+1)=0 ⇒ x=0

or

-1

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