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JEE Main & Advanced Mathematics Functions Question Bank Critical Thinking

  • question_answer
    If f is an even function defined on the interval (-5, 5), then four real values of x satisfying the equation  \[f(x)=f\left( \frac{x+1}{x+2} \right)\]  are [IIT 1996]

    A) \[\frac{-3-\sqrt{5}}{2},\ \frac{-3+\sqrt{5}}{2},\ \frac{3-\sqrt{5}}{2},\ \frac{3+\sqrt{5}}{2}\]

    B) \[\frac{-5+\sqrt{3}}{2},\ \frac{-3+\sqrt{5}}{2},\ \frac{3+\sqrt{5}}{2},\ \frac{3-\sqrt{5}}{2}\]

    C) \[\frac{3-\sqrt{5}}{2},\ \frac{3+\sqrt{5}}{2},\ \frac{-3-\sqrt{5}}{2},\ \frac{5+\sqrt{3}}{2}\]

    D) \[-3-\sqrt{5},\ -3+\sqrt{5},\ 3-\sqrt{5},\ 3+\sqrt{5}\]

    Correct Answer: A

    Solution :

    • Since f is an even function \[f\,(-x)=f(x)\,,\forall x\in (-5,\,5).\]          
    • We are given that \[f(x)=f\,\left( \frac{x+1}{x+2} \right)\]           
    • \[\Rightarrow \,\,f(-x)=f\left( \frac{-x+1}{-x+2} \right)\,\Rightarrow \,\,f(x)=f\,\left( \frac{-x+1}{-x+2} \right)\]                                                                 \[[\because \,\,\,f(-x)=f(x)]\]           
    • To find the values of x, we set \[x=\frac{-x+1}{-x+2}\,\,\Rightarrow \,\,x=\frac{3\pm \sqrt{9-4}}{2}=\frac{3\pm \sqrt{5}}{2}\]           
    • Also \[f(x)=f\,\left( \frac{x+1}{x+2} \right)=f(-x)\]           
    • To find the values of x, we set \[-x=\frac{x+1}{x+2}\,\,\Rightarrow \,\,x=\frac{-3\pm \sqrt{9-4}}{2}=\frac{-3\pm \sqrt{5}}{2}\]           
    • Thus the four required values of x are \[\frac{-3-\sqrt{5}}{2},\,\frac{-3+\sqrt{5}}{2},\,\frac{3-\sqrt{5}}{2},\,\frac{3+\sqrt{5}}{2}.\]


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