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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 43

  • question_answer
    If \[f(x)\] is a continuous function \[\forall \,x\in R-\{-2\}\] and satisfies \[{{x}^{3}}-{{x}^{2}}(f(x)+2)-x+2(2f(x)+1)=0\forall \,x\in R-\{-2\},\] then \[f(2)\]is equal to

    A) \[1/4\]                  

    B)        \[1/2\]

    C) \[3/4\]   

    D)        \[1\]

    Correct Answer: C

    Solution :

    [c] Here, \[f(x)=\frac{{{x}^{3}}-2{{x}^{2}}-x+2}{{{x}^{2}}-4},\,\,x\ne \pm 2\] Given that \[f(x)\] is continuous at\[x=2\]. \[\therefore \,\,\,\,\,\,f(2)=\underset{x\to 2}{\mathop{\lim }}\,\,\,f(x)\] Now, \[\underset{x\to 2}{\mathop{\lim }}\,\,\,\,\frac{({{x}^{2}}-1)\,\,(x-2)}{(x+2)\,(x-2)}=\frac{3}{4}\]              


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