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JEE Main & Advanced AIEEE Solved Paper-2007

  • 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  Solved  Paper-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 :

    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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