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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2001

  • question_answer
    If a, b are two constant positive integers such that\[f(a+x)=b+[{{b}^{3}}+1-3{{b}^{2}}f(x)+3b{{\{f(x)\}}^{2}}\]\[-\{f{{(x)}^{3}}\}{{]}^{1/3}},\]for every\[x\]. Then,\[f(x)\]is a periodic function with the period

    A)  \[a\]                

    B)  \[2a\]

    C)  \[b\]                

    D)  \[2b\]

    Correct Answer: B

    Solution :

     \[f(a+x)=b+[{{b}^{3}}+1-3{{b}^{2}}f(x)\] \[+3b{{\{f(x)\}}^{2}}-\{f{{(x)}^{3}}\}{{]}^{1/3}}\] \[f(a+x)=b+{{[1+{{\{b-f(x)\}}^{3}}]}^{1/3}}\] \[\Rightarrow \] \[f(a+x)-b={{\{f(x)-b\}}^{3}}{{]}^{1/3}}\] \[\Rightarrow \] \[\phi (a+x)={{[1-\{\phi {{(x)}^{3}}\}]}^{1/3}}\] \[\Rightarrow \] \[\phi (x+2a)={{[1-{{\{\phi (x+a)\}}^{3}}]}^{1/3}}=\phi (x)\] \[\Rightarrow \] \[f(x+2a)-b=f(x)-b\] \[\Rightarrow \] \[f(x+2a)=f(x)\] Hence,\[f(x)\]is a periodic function with period \[2a.\]


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