A) 0
B) infinite
C) 1
D) 2
Correct Answer: B
Solution :
Given \[F\left( x \right)=x-\left[ x \right]\]so, \[F\left( x \right)+F\left( \frac{1}{x} \right)=1\] |
\[\Rightarrow x-\left[ x \right]+\frac{1}{x}-\left[ \frac{1}{x} \right]=1\] |
\[\Rightarrow x+\frac{1}{x}=\left[ x \right]+\left[ \frac{1}{x} \right]+1..\left( 1 \right)\] |
\[\because \]RHS is an integer. |
\[\therefore \]LHS must be integer. |
Let \[x+\frac{1}{x}=\left[ x \right]+\frac{1}{x}+1=k,k\in \operatorname{I},\]the equation is \[x+\frac{1}{x}=k\Rightarrow {{x}^{2}}-kx+\operatorname{I}=0,\] |
For real \[x,{{k}^{2}}-4\ge 0\Rightarrow k\le -2\,or\,k\ge 2\]and \[k\in \]I, we can get infinitety many values of k and we get solution for each value of\[k\]. |
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