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

  • question_answer
    If \[\mathbf{f}\left( \mathbf{x} \right)=\frac{\mathbf{1}-\mathbf{cosx}}{\mathbf{1}-\mathbf{sinx}}\] then \[f'\left( \frac{\pi }{2} \right)\]is equal to

    A)  1                                

    B)         0              

    C)         \[\infty \]                                  

    D)         does not exist

    Correct Answer: D

    Solution :

    [d]\[f(x)=\frac{1-cosx}{1-\sin x}\] \[\left( 1-sinx \right).\frac{d}{dx}\left( 1-cosx \right)\] \[\therefore f'(x)=\frac{-\left( 1-cosx \right)\times \frac{d}{dx}\left( 1-sinx \right)}{{{\left( 1-sinx \right)}^{2}}}\] \[=\frac{\left( 1-sinx \right).sinx-\left( 1-cosx \right)\left( -cosx \right)}{{{\left( 1-sinx \right)}^{2}}}\] \[\frac{sinx-si{{n}^{2}}x+cosx-co{{s}^{2}}x}{{{\left( 1-sinx \right)}^{2}}}=\frac{\left( sinx+cosx \right)-\left( si{{n}^{2}}x+co{{s}^{2}}x \right)}{{{\left( 1-sinx \right)}^{2}}}\]\[f'\left( \frac{\pi }{2} \right)=\frac{\sin \frac{\pi }{2}+\cos \frac{\pi }{2}-1}{{{\left( 1-\sin \frac{\pi }{2} \right)}^{2}}}=\frac{0}{0}\] It doesn't exist. Hence, the correct option is .

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