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JEE Main & Advanced Mathematics Functions Question Bank Functions

  • question_answer
    A real valued function \[f(x)\] satisfies the function equation \[f(x-y)=f(x)f(y)-f(a-x)f(a+y)\] where a is a given constant and \[f(0)=1\], \[f(2a-x)\] is equal to                                         [AIEEE 2005]

    A)                    \[f(a)+f(a-x)\]

    B)            \[f(-x)\]

    C)            \[-f(x)\]

    D)            \[f(x)\]

    Correct Answer: C

    Solution :

               \[f(a-(x-a))=f(a)f(x-a)-f(0)f(x)\]     .....(i)                    Put \[x=0,y=0\]; \[f(0)={{(f(0))}^{2}}-{{[f(a)]}^{2}}\Rightarrow f(a)=0\]                    \[[\because f(0)=1].\] From (i), \[f(2a-x)=-f(x)\].


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