If \sin ^{-1} x+\cos ^{-1}(1-x)=\sin ^{-1}(-x), then x satisfies the equation

Correct Answer is : None of these

NIMCET 2009 Question Paper with solutions

Question : 26 of 120

Marks: +1, -0

\sin ^{-1} x+\cos ^{-1}(1-x)=\sin ^{-1}(-x)

, then

x

satisfies the equation

Solution:

\sin ^{-1} x+\cos ^{-1}(1-x)=\sin ^{-1}(-x)=-\sin ^{-1}(x)

⇒ \;\; 2 \sin ^{-1} x+\cos ^{-1}(1-x)=0

2 \sin ^{-1} x+\cos ^{-1}(1-x)=0

⇒ \;\; \tan ^{-1}(\;{2 x}/{1+x^{2}})+\tan ^{-1}(\;{2 x-x^{2}}/{1-x})=0

\tan ^{-1}\{\;{\;{2 x}/{1+x^{2}}+{x(2-x)}/{(1-x)}}/{1+\;{2 x^{2}(2-x)}/{(1+x^{2})(1-x)}}\}=0

⇒ \tan ^{-1}\{\;{2 x(1-x)+x(2-x)(1+x^{2})}/{(1+x^{2})(1-x)+2 x^{2}(2-x)}\}=0

⇒ \;\; 2 x(1-x)+x(2-x)(1+x^{2})=0

x[2-2 x+2-x+2 x^{2}-x^{3}]=0

⇒ \;\; x(-x^{3}+2 x^{2}-3 x+4)=0

⇒ \;\; x(x^{3}-2 x^{2}+3 x-4)=0

x=0

satisfies the above equation, which is satisfied by option in

b

, but other value of

x

3 \/ 2

which is impossible. Hence, none of these is correct option.

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