2. Solved Papers
  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
    Let\[f\]be twice differentiable function such that\[f(x)=-f(x)\]and\[f(x)=g(x),\]\[h(x)=\{f{{(x)}^{2}}\}+{{\{g(x)\}}^{2}}.\]then\[h(5)=11,\]is equal to:

    A)  22                                         

    B)  11

    C)  0                            

    D)         20

    E)  none of these

    Correct Answer: B

    Solution :

    We have \[h(x)={{\{f(x)\}}^{2}}+{{\{g(x)\}}^{2}}\] \[\Rightarrow \]\[h(x)=2f(x)f(x)+2g(x)g(x)\] 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\[g(x)=-f(x)\] \[\therefore \]  \[h(x)=-2g(x)g(x)+2g(x)g(x)\]                 \[=0,\forall x\] \[\Rightarrow \]\[h(x)=\]constant for all\[x\] But \[h(5)=11\] Hence\[h(x)=11\]for all\[x\].


You need to login to perform this action.
You will be redirected in 3 sec spinner