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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Increasing and Decreasing Function

  • question_answer
    The value of ?a? in order that \[f(x)=\sqrt{3}\] \[\sin x-\cos x-2ax+b\] decreases for all real values of x, is given by

    A)            \[a<1\]

    B)            \[a\ge 1\]

    C)            \[a\ge \sqrt{2}\]

    D)            \[a<\sqrt{2}\]

    Correct Answer: B

    Solution :

               Since \[f(x)=\sqrt{3}\sin x-\cos x-2ax+b\] is decreasing for all real values of \[x,\]therefore \[f'(x)<0\]for all x.                    Þ  \[\sqrt{3}\cos x+\sin x-2a<0\]for all x                    \[\Rightarrow \frac{\sqrt{3}}{2}\cos x+\frac{1}{2}\sin x<a\]for all x                    Þ \[\sin \left( x+\frac{\pi }{3} \right)<a\]for all x Þ\[a\ge 1,\,\left[ \because \sin \left( x+\frac{\pi }{3} \right)\le 1 \right]\].


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