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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2003

  • question_answer
    Let\[f(x)\]be twice differentiable such that\[f(x)=-f(x),f(x)=g(x),\]where\[f(x)\]and\[f(x)\]represent the first and second derivatives of\[f(x)\]respectively. Also, if\[h(x)={{[f(x)]}^{2}}+{{[g(x)]}^{2}}\]and\[h(5)=5,\]then\[h(10)\]is equal to:

    A)  3                                            

    B)  10

    C)  13                         

    D)         5

    E)  0

    Correct Answer: D

    Solution :

    We have, \[h(x)={{\{f(x)\}}^{2}}+{{\{g(x)\}}^{2}}\] On differentiating w.r.t.\[x,\]we get \[\Rightarrow \] \[h(x)=2f(x)f(x)+2g(x)g(x)\]      ??(i) Now,\[f(x)=g(x)\]and\[f\,(x)=-f(x)\] \[\Rightarrow \]               \[f\,(x)=g(x)\]and\[f\,(x)=-f(x)\] \[\Rightarrow \]               \[-f(x)=g(x)\] Thus,\[f(x)=g(x)\]and\[g(x)=-f(x)\] From (i) \[h(x)=-2g(x)g(x)+2g(x)g(x)\] \[=0\] \[\Rightarrow \]               \[h(x)=5\]                           \[[\because h(5)=5]\] \[\therefore \]  \[h(10)=5\]


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