2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-23

  • question_answer
    For \[x\in (-1,\,\,\,1)\], the number of solutions of the equation \[{{\tan }^{-1}}(x+{{x}^{2}}+{{x}^{3}}+{{x}^{4}}+.....\,\infty )+\] \[{{\cot }^{-1}}(-6+6x-6{{x}^{2}}+......\infty )=\frac{\pi }{2}\] is

    A)  0                                

    B)  1

    C)  2                                

    D)  4

    Correct Answer: A

    Solution :

    Here, \[x+{{x}^{2}}\,+{{x}^{3}}+{{x}^{4}}+.....\infty =\frac{x}{1-x};\,\,|x|<1\] and \[-6+6x-6{{x}^{2}}+....\infty =\frac{-6}{1+x};\,|x|<1\] So, that \[{{\tan }^{-1}}\,\left( \frac{x}{1-x} \right)+{{\cot }^{-1}}\,\left( \frac{-6}{1+x} \right)=\frac{\pi }{2}\] \[\Rightarrow \,\,\frac{x}{1-x}\,=\frac{-6}{1+x}\]\[\Rightarrow \,{{x}^{2}}+x=-6+6x\] \[\Rightarrow \,{{x}^{2}}-5x+6=0\Rightarrow \,(x-2)\,(x-3)\,=0\Rightarrow \,x=2,3\]But \[-1<x<1\] (Given)


You need to login to perform this action.
You will be redirected in 3 sec spinner