JEE Main & Advanced Mathematics Trigonometric Equations Question Bank Solution of trigonometrical equations

  • question_answer
    The only value of x for which \[{{2}^{\sin x}}+{{2}^{\cos x}}>{{2}^{1-(1/\sqrt{2})}}\] holds, is

    A) \[\frac{5\pi }{4}\]

    B) \[\frac{3\pi }{4}\]

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

    D) All values of x

    Correct Answer: A

    Solution :

    Since A.M. \[\ge \] G.M. \[\frac{1}{2}({{2}^{\sin x}}+{{2}^{\cos x}})\ge \sqrt{{{2}^{\sin x}}{{.2}^{\cos x}}}\] \[\Rightarrow \] \[{{2}^{\sin x}}+{{2}^{\cos x}}\ge {{2.2}^{\frac{\sin x+\cos x}{2}}}\] \[\Rightarrow \]\[{{2}^{\sin x}}+{{2}^{\cos x}}\ge {{2}^{1+\frac{\sin x+\cos x}{2}}}\] and we know that \[\sin x+\cos x\ge -\sqrt{2}\] \[\therefore \]  \[{{2}^{\sin x}}+{{2}^{\cos x}}>{{2}^{1-(1/\sqrt{2})}}\], for\[x=\frac{5\pi }{4}\]


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