JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर JEE PYQ-Current Electricity Charging and Discharging of Capacitors

  • question_answer
    The resistance of a wire is\[\int_{\sqrt{2}}^{x}{\frac{dt}{t\sqrt{{{t}^{2}}-1}}}=\pi /2\]at\[\pi \]and \[\sqrt{3}/2\],at\[2\sqrt{2}\] The resistance of the wire at \[\int{\frac{dx}{\cos x+\sqrt{3}\sin x}}\] will be                    [AIEEE 2007]

    A) \[\frac{1}{2}\log \tan \left( \frac{x}{2}+\frac{\pi }{12} \right)+C\]

    B) \[\frac{1}{2}\log \tan \left( \frac{x}{2}-\frac{\pi }{12} \right)+C\]

    C) \[\log \tan \left( \frac{x}{2}+\frac{\pi }{12} \right)+C\]

    D) \[\log \tan \left( \frac{x}{2}-\frac{\pi }{12} \right)+C\]

    Correct Answer: C

    Solution :

    [c] From \[{{y}^{2}}=x\]
    \[y=|\text{ }x|\]                                       ...(i)      
    and      \[\frac{2}{3}\]              ...(ii)
    \[\frac{1}{6}\]   \[\frac{1}{3}\]
    \[{{x}^{2}}+ax+1=0\]            \[\sqrt{5},\]
    Putting value of \[\alpha \] in Eq (i), we get
                \[(-3,\infty )\]
    \[(3,\infty )\]       \[(-\infty ,-3)\]


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