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JEE Main & Advanced Mathematics Functions Question Bank Critical Thinking

  • question_answer
    Let \[f(x)=(1+{{b}^{2}}){{x}^{2}}+2bx+1\] and \[m(b)\] the minimum value of \[f(x)\]for a given b. As b varies, the range of m is [IIT Screening 2001]

    A) [0, 1]

    B) \[\left( 0,\ \frac{1}{2} \right]\]

    C) \[\left[ \frac{1}{2},\ 1 \right]\]

    D) \[(0,\ 1]\]

    Correct Answer: D

    Solution :

    • \[f(x)=(1+{{b}^{2}}){{x}^{2}}+2bx+\frac{{{b}^{2}}}{(1+{{b}^{2}})}-\frac{{{b}^{2}}}{1+{{b}^{2}}}+1\]                  
    • \[=(1+{{b}^{2}})\,{{\left( x+\frac{b}{1+{{b}^{2}}} \right)}^{2}}+\frac{1}{1+{{b}^{2}}}\ge \frac{1}{1+{{b}^{2}}}\]                  
    • \ \[m(b)=\frac{1}{1+{{b}^{2}}}\], so range of \[m(b)=(0,\,1]\].


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