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JEE Main & Advanced JEE Main Solved Paper-2015

  • question_answer
    Let f (x) be a polynomial of degree four having extreme values at x = 1 and x =2. If\[\underset{x\to 0}{\mathop{\lim }}\,\left( 1+\frac{f\left( x \right)}{{{x}^{2}}} \right)=3,\]then f(2) is equal to : [JEE Main Solved Paper-2015 ]

    A) 0             

    B) 4

    C) -8                                          

    D) -4

    Correct Answer: A

    Solution :

    \[\underset{x\to 0}{\mathop{\lim }}\,\left( 1+\frac{f\left( x \right)}{{{x}^{2}}} \right)=3,\]has repeated root at 0 \[\Rightarrow \]\[f(x)={{x}^{2}}.\left( a{{x}^{2}}+bx+c \right)\] \[\Rightarrow \]\[\underset{x\to 0}{\mathop{\lim }}\,\left( 1+\frac{{{x}^{2}}.\left( a{{x}^{2}}+bx+c \right)}{{{x}^{2}}} \right)=3\] \[\Rightarrow \]\[c+1=3\Rightarrow c=2\] F (x) has extreme at x = 1 and x = 2 \[\Rightarrow \]\[f'(x)=0\]at\[x=1,2\] \[\Rightarrow \]\[f'(1)=0\Rightarrow 4a+3b+4=0\]           ?(1) \[f'(2)=0\Rightarrow 8a+3b+2=0\]                            ?(2) \[f'(2)=0\Rightarrow 4a+3b+2=0\]\[\Rightarrow \] (1)\[\Rightarrow \](2) \[\Rightarrow a={}^{1}/{}_{2},b=-2\] \[\Rightarrow \]\[f(x)={{x}^{2}}\left( \frac{{{x}^{2}}}{2}-2x+2 \right)\] \[f(2)=4.(4-4)=0\]                                


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