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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    Let\[f(x)={{(x-7)}^{2}}{{(x-2)}^{7}}{{(x-2)}^{7}},x\in [2,7]\]. The value of\[\theta \in (2,7)\]such that\[f(\theta )=0\]is equal to

    A)  \[\frac{49}{4}\]                               

    B)  \[\frac{53}{9}\]

    C)  \[\frac{53}{7}\]               

    D)         \[\frac{49}{9}\]

    E)  \[\frac{45}{7}\]

    Correct Answer: B

    Solution :

    Given,   \[f(x)={{(x-7)}^{2}}{{(x-2)}^{7}}\] \[\Rightarrow \]               \[f(\theta )={{(\theta -7)}^{2}}{{(\theta -2)}^{7}}\] \[\Rightarrow \]\[f(\theta )=2(\theta -7){{(\theta -2)}^{7}}+7{{(\theta -2)}^{6}}{{(\theta -7)}^{2}}\] Put             \[f(\theta )=0\] \[\Rightarrow \]\[(\theta -7){{(\theta -2)}^{6}}[2(\theta -2)+7(\theta -7)]=0\] \[\Rightarrow \]\[9\theta =53\Rightarrow \theta =\frac{53}{9}\]              


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