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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