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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2007

  • question_answer
    If\[f(x)=2{{x}^{2}}+bx+c\]and\[f(0)=3\]and \[f(2)=1,\]then\[f(1)\]is equal to

    A)  1            

    B)         2

    C)  0            

    D)         1/2

    E) \[-2\]

    Correct Answer: C

    Solution :

    \[\because \]\[f(x)=2{{x}^{2}}+bx+c\] and\[f(0)=3\]and\[f(2)=1\] \[\therefore \]\[c=3\]and\[8+2b+c=1\] \[\Rightarrow \]\[8+2b+3=1\Rightarrow b=-5\] \[\therefore \] \[f(x)=2{{x}^{2}}-5x+3\] \[\Rightarrow \]\[f(1)=2{{(1)}^{2}}-5(1)+3\] \[=2-5+3=0\]


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