CBSE Class 12 Math 2022 Term I Solved Paper

Question : 33 of 50

Marks: +1, -0

y= sin \; (m sin \;^{-1} x)

, then which one of the following equations is true?

Solution:

Given,

y= sin \; (m ( sin \;^{-1} x) ) \;\;\;⋅⋅⋅⋅⋅⋅⋅(i)

Differentiating both sides w.r.t.

x

, we get

\;{d y}/{d x}\;=\cos (m sin \;^{-1} x) × \;{m}/{{√{1-x^2}}}

⇒ \;\; \;{d y}/{d x}\;=\;{m \cos (m sin \;^{-1} x) }/{{√{1-x^2}}} \;\;\;⋅⋅⋅⋅⋅⋅⋅(ii)

⇒ \;\; ({√{1-x^2}}) y^{'}\;=\;{m \cos (m sin \;^{-1} x) }/{{√{1-x^2}}} \;\;\;⋅⋅⋅⋅⋅⋅⋅(iii)

⇒ \;\; ({√{1-x^2}}) y^{'}=m \cos (m sin \;^{-1} x)

Differentiating again w.r.t. '

x

', we get

y^{' '} ({√{1-x^2}}) +y^{'} \;{(-2 x)}/{2 {√{1-x^2}}}\;

=\;-m^2 sin \; (m sin \;^{-1} x) \;{1}/{{√{1-x^2}}}

⇒ \;\; y^{' '} (1-x^2) -x y^{'}=-m^2 y

⇒ y^{' '} (1-x^2) -x y^{'}+m^2 y=0

\;\text " or, "\; \;\; (1-x^2) \;{d^2 y}/{d x^2}-x \;{d y}/{d x}+m^2 y=0

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