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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 41

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

    A) \[a\ge \sqrt{2}\]

    B)        \[a<\sqrt{2}\]

    C) \[a\ge 1\]           

    D)        \[a<1\]

    Correct Answer: A

    Solution :

    We have; \[f(x)=\sin x-\cos x-ax+b\] \[\Rightarrow \,\,f'(x)=\cos x+\sin x-a\] \[\Rightarrow \,f'(x)<0\,\,\forall \,\,x\in R\] \[\Rightarrow \,(\cos x+\sin x)<a\,\,\forall \,\,x\in R\] As the max. value of \[(\cos x+\sin x)\]is \[\sqrt{2}\] The above is possible when \[a\ge \sqrt{2}\]


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