11th Class Mathematics Other Series Question Bank Critical Thinking

  • question_answer \[{{2}^{\sin \theta }}+{{2}^{\cos \theta }}\] is greater than [AMU 2000]

    A) \[\frac{1}{2}\]

    B) \[\sqrt{2}\]

    C) \[{{2}^{\frac{1}{\sqrt{2}}}}\]

    D) \[{{2}^{\left( 1-\,\frac{1}{\sqrt{2}} \right)}}\]

    Correct Answer: D

    Solution :

    \[\frac{1}{2}\left[ {{2}^{\sin \theta }}+{{2}^{\cos \theta }} \right]\ge \sqrt{{{2}^{\sin \theta }}{{2}^{\cos \theta }}}\]   (\[\because \text{A}\text{.M}\text{.}\ge \text{G}\text{.M}\text{.}\]) Þ \[{{2}^{\sin \theta }}+{{2}^{\cos \theta }}\ge {{2.2}^{(\sin \theta +\cos \theta )/2}}\]             .....(i) Now \[(\sin \theta +\cos \theta )=\sqrt{2}\sin (\theta +\pi /4)\ge -\sqrt{2}\] for all real q \[{{2}^{\sin \theta }}+{{2}^{\cos \theta }}\ge {{2.2}^{(\sin \theta +\cos \theta )/2}}>2\,.\,{{2}^{-\sqrt{2}/2}}\] Þ \[{{2}^{\sin \theta }}+{{2}^{\cos \theta }}\ge {{2}^{1-(1/\sqrt{2})}}\].


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