JEE Main & Advanced JEE Main Paper Phase-I (Held on 07-1-2020 Evening)

  • question_answer
    If \[{{\theta }_{1}}\] and \[{{\theta }_{2}}\] be respectively the smallest and the largest values of \[\theta \] in \[(0,2\pi )-\{\pi \}\] which satisfy the equation, \[2\,{{\cot }^{2}}\theta -\frac{5}{\sin \theta }+4=0\], then \[\int\limits_{{{\theta }_{1}}}^{{{\theta }_{2}}}{{{\cos }^{2}}\,3\theta \,d\theta },\] is equal to [JEE MAIN Held on 07-01-2020 Evening]

    A) \[\frac{\pi }{3}+\frac{1}{6}\]          

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

    C) \[\frac{2\pi }{3}\]                   

    D) \[\frac{\pi }{9}\]

    Correct Answer: B

    Solution :

    [b]  \[\because \,\,\,\,2{{\cot }^{2}}\theta -\frac{5}{\sin \,\theta }+4=0\] \[2+2\,\text{cose}{{\text{c}}^{2}}\,\theta -5\,\text{cosec}\,\theta =0\] \[2\,\text{cose}{{\text{c}}^{2}}\,\theta -4\,\text{cosec}\,\theta -\,\text{cosec}\theta +2=0\] \[\therefore \,\,\,(2\,\text{cosec}\,\,\theta -1)\,(\,\text{cosec}\,\theta -2)\,=0\] \[\therefore \,\,\text{cosec}\,\theta =\frac{1}{2}\] or 2. \[\therefore \,\,\sin \,\theta =2\] or \[\frac{1}{2}.\]       \[\because \,\,\theta \,\in \,(0,2\pi )\] \[\therefore \,\,\,\,{{\theta }_{1}}=\frac{\pi }{6}\] and \[{{\theta }_{2}}=\frac{5\pi }{6},\]                      \[\because \,{{\theta }_{1}}<{{\theta }_{2}}\] \[\therefore \,\,I=\int\limits_{\frac{\pi }{6}}^{\frac{5\pi }{6}}{{{\cos }^{2}}\,3\theta \,d\theta =\int\limits_{\frac{\pi }{6}}^{\frac{5\pi }{6}}{\frac{1+\cos \,6\theta }{2}}}\,d\theta \] \[=\frac{1}{2}\left[ \theta +\frac{\sin \,6\theta }{6} \right]_{\frac{\pi }{6}}^{\frac{5\pi }{6}}=\frac{\pi }{3}\]

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