A) 0
B) 1
C) 2
D) infinite
Correct Answer: C
Solution :
[C]\[\because {{\cot }^{-1}}(x-1)+{{\cot }^{-1}}(6-x)={{\cot }^{-1}}(x-2)\] |
Take cot of both sides, we get |
\[\Rightarrow \left[ \frac{(x-1)\,\,(6-x)-1}{6-x+x-1} \right]=(x-2)\] |
\[\Rightarrow -\frac{{{x}^{2}}+7x-7}{5}=x-2\] |
\[{{x}^{2}}-2x-3=0\] |
\[\Rightarrow (x-3)\,\,(x+1)=0\] |
\[x=3;\,\,\,x=-\,1\] |
both satisfy the given equation |
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