JEE Main & Advanced Sample Paper JEE Main Sample Paper-45

  • question_answer
    Let \[0\le \theta <\frac{\pi }{2},\]\[\le \phi <\frac{\pi }{2}\] and \[0\le \psi <\frac{\pi }{2}\] and \[\theta +\psi =\frac{\pi }{2}\]. The product \[\tan \,\theta \,\,\tan \,\phi \,\,\tan \,\psi \] attains the greatest value when

    A)  \[\theta =\phi =\psi \]            

    B)  \[\theta =\frac{\pi }{2}\]

    C)  \[\phi =\psi =\pi \]                 

    D)  any one of \[\theta ,\,\,\phi \,\,\psi \] is \[\frac{\pi }{3}\]

    Correct Answer: A

    Solution :

     \[\because \]                \[AM\ge GM\] \[\frac{\tan \,\theta \,\tan \,\phi \,+\tan \,\psi }{3}\ge \,{{(\tan \,\theta \,\tan \,\phi \,\tan \,\psi )}^{1/3}}\] The greatest vlaue of \[\tan \,\theta \,\tan \,\phi \,\tan \,\psi \] is \[{{\left( \frac{\tan \,\theta \,+\tan \,\phi \,+\tan \,\psi }{3} \right)}^{3}}\] In this case \[AM=GM\] and numbers will be equal. Hence, the greatest value \[=\frac{1}{\sqrt{3}}\times \frac{1}{\sqrt{3}}\times \frac{1}{\sqrt{3}}=\frac{1}{3\sqrt{3}}\] at \[\theta =\phi =\psi =\frac{\pi }{6}\]

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