11th Class Mathematics Sample Paper Mathematics Olympiad - Sample Paper-3

  • question_answer
    If \[y=si{{n}^{-1}}\left( \frac{5x+12\sqrt{1-{{x}^{2}}}}{13} \right)\] then \[\frac{dy}{dx}\]is equa.to

    A)  \[\frac{1}{\sqrt{1-{{x}^{2}}}}\]                      

    B)  \[\frac{-1}{\sqrt{1-{{x}^{2}}}}\]

    C)  \[\frac{2}{\sqrt{1-{{x}^{2}}}}\]                      

    D)  \[\frac{-2}{\sqrt{1-{{x}^{2}}}}\]

    Correct Answer: B

    Solution :

    [b] putting \[x=\cos \theta \] \[\therefore y=soi{{n}^{-1}}\left\{ \frac{5.\cos \theta +12.\sin \theta }{13} \right\}={{\sin }^{-1}}\left[ \left( \frac{5}{13} \right).\cos \theta +\frac{12}{13} \right].\sin \theta \]Let \[\frac{5}{13}=\sin \alpha ,\] then\[\Rightarrow \frac{12}{13}=\cos \alpha ,\] \[\therefore y={{\sin }^{-1}}\{\sin (\alpha +\theta )\}=\alpha +\theta \] Differentiating w.r.t. x, we have \[\frac{dy}{d\theta }=\frac{d\theta }{dx}=\frac{d}{dx}({{\cos }^{-1}}x)=\frac{-1}{\sqrt{1-{{x}^{2}}}}\] Hence, option [b] is correct.

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