BCECE Engineering BCECE Engineering Solved Paper-2010

  • question_answer The value of \[\alpha \in (-\pi ,0)\]satisfying \[\sin \alpha +\int_{\alpha }^{2a}{\cos 2x\,dx=0}\]is

    A)  0                                            

    B)  \[-\frac{\pi }{3}\]            

    C)         \[-\pi \]                              

    D)         All of these

    Correct Answer: B

    Solution :

    \[\sin \alpha +\int_{\alpha }^{2\alpha }{\cos 2xdx=0}\] \[\Rightarrow \]\[\sin \alpha \frac{1}{2}[\sin 2x]_{\alpha }^{2\alpha }=0\] \[\Rightarrow \]\[\sin \alpha +\frac{1}{2}(sin4\alpha -\sin 2\alpha )=0\] \[\Rightarrow \]\[\sin \alpha +\sin \alpha \cos 3\alpha =0\] \[\Rightarrow \]\[\sin \alpha (1+cos3\alpha )=0\] \[\Rightarrow \]\[\sin \alpha =0\]or \[1+\cos 3\alpha =0\] If \[\sin \alpha =0\] \[\Rightarrow \]\[\alpha =0,\,\pi -\pi \] If \[1+\cos 3\alpha =0\] \[\Rightarrow \]               \[\alpha =\frac{-\pi }{3}\] But, only              \[-\frac{\pi }{3}\in (-\pi ,0)\] Hence, value of \[\alpha \] is \[-\frac{\pi }{3}.\]

adversite


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