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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Afternoon)

  • question_answer
    Let \[f\left( x \right)=5\left| x2 \right|\] and \[g(x)=|x+1|,x\in R.\] If f(x) attains maximum value at a and g(x) attains minimum value at \[\beta \] then \[\underset{x\to -\alpha \beta }{\mathop{\lim }}\,\frac{\left( x-1 \right)\left( {{x}^{2}}-5x+6 \right)}{{{x}^{2}}-6x+8}\]is equal to : [JEE Main 12-4-2019 Afternoon]

    A) 1/2                  

    B) -3/2

    C) 3/2                              

    D) -1/2

    Correct Answer: A

    Solution :

    Maxima of f(x) occured at \[x=2\]i.e. \[\alpha =2\] Minima of g(x) occured at \[x=1\] i.e. \[\beta =-1\] \[\therefore \]\[\underset{x\to 2}{\mathop{\lim }}\,\frac{\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)}{\left( x-2 \right)\left( x-4 \right)}=\frac{1}{2}\]


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