KVPY Sample Paper KVPY Stream-SX Model Paper-21

  • question_answer
    The number of real solutions of the equation, \[{{\cot }^{-1}}(x-1)+{{\cot }^{-1}}(6-x)={{\cot }^{-1}}(x-2)\] is.

    A) 0

    B) 1

    C) 2

    D) infinite

    Correct Answer: C

    Solution :

    \[\because \]\[{{\cot }^{-1}}(x-1)+co{{t}^{-1}}(6-x)=co{{t}^{-1}}(x-2)\]
    take cot of both sides, we get
    \[\Rightarrow \]\[\left[ \frac{(x-1)\,\,(6-x)-1}{6-x+x-1} \right]=(x-2)\]\[\Rightarrow \]\[-\frac{{{x}^{2}}+7x-7}{5}=x-2\]\[\Rightarrow \]            \[{{x}^{2}}-2x-3=0\]   \[\Rightarrow \]\[(x-3)\,\,(x+1)=0\]\[\Rightarrow \]   \[x=3\];\[x=-1.\]
    both satisfy the given equation.

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