# Functions Set 4

**Example – 6**

Plot the curve \(y = \min \left\{ {|x| - 1,\,\,\,|x - 1| - 1,\,\,\,|x - 2| - 1} \right\}\) .

**Solution:** What does such a statement mean? It means that for each value of * x* , we evaluate all the three quantities on the RHS, select the minimum of the three, and plot that value of *y*.

What we can do is plot the separate graphs for these three quantities on the same axes and select those portions that lie ** lowermost,** out of all the three.

The darker line segments on the left hand side diagram show the minimum value out of all the three, considered at each point *x*. This is therefore the graph of *y*.

**Example – 7**

Solve the equation for *x *: \(\begin{align}\frac{1}{{[x]}} + \frac{1}{{[2x]}} = \{ x\} + \frac{1}{3}\end{align}\)

**Solution:** A solution is not evident by mere observation. But it can be noted that the RHS is always positive: therefore, the LHS must be positive and hence * x* > 0.

Also, for the LHS to be defined, * \(x > 1{\rm{\;\;}}and{\rm{\;\;}}2x > 1{\rm{ }} \Rightarrow {\rm{ }}x > 1\;and\;\;hence[x] > 1.\)*

Let *I* be the integral and *f* the fractional part of *x*.

\[ \Rightarrow \frac{1}{I} + \frac{1}{{2I + [2f]}} = f + \frac{1}{3}\]

We have retained \([2f]\) since 2*f* could be greater than 1 and hence \([2f]\) is not necessarily 0.

We will have to consider two different cases separately:

**(i)** \(\fbox{${f < \frac{1}{2}}$}\)** \( \Rightarrow \)** \(\begin{align}\frac{1}{I} + \frac{1}{{2I}}\end{align}\) = \(f + \frac{1}{3}\)

\({\rm{}} \Rightarrow \begin{align}\frac{3}{{2I}}{\rm{ }} = {\rm{}}f + \frac{1}{3}\end{align}\)

Now we can substitute different values of *I*; if these give valid values for *f* such that \(f < \frac{1}{2}\) , we accept these solutions.

\[I = 1{\rm{}} \Rightarrow {\rm{ }}f = \frac{7}{6}\] | [not acceptable] |

\[I = 2{\rm{ }} \Rightarrow f = \frac{5}{{12}}\] | [acceptable] |

\[I = 3{\rm{ }} \Rightarrow {\rm{ }}f = \frac{1}{6}{\rm{}}\] | [acceptable] |

\[I = 4{\rm{ }} \Rightarrow {\rm{ }}f = \frac{1}{{24}}\] | [acceptable] |

\[I = 5{\rm{ }} \Rightarrow {\rm{ }}f < 0\] | [not acceptable] |

No more solutions will exist since *f * becomes \(< {\rm{ }}0{\rm{ }}forI > 5\)

**(ii)** \(\fbox{${f \ge \frac{1}{2}}$}\) \( \Rightarrow \)\(\begin{align}\frac{1}{I} + \frac{1}{{2I + 1}}\end{align}\) = \(f + \frac{1}{3}\)

\(I = 1{\rm{}} \Rightarrow {\rm{ }}f = 1{\rm{}}\) |
[not acceptable] |

\(I = 2 \Rightarrow f = \begin{align}\frac{{11}}{{30}}{\rm{ }}(f\;does\;not\;satisfy\;f \ge \frac{1}{2}){\rm{}}\end{align}\) |
[not acceptable] |

\(I = 3 \Rightarrow {\rm{ }}f =\begin{align} \frac{1}{7}\end{align}\) |
[not acceptable] |

\(I = 4 \Rightarrow f = 1/36\) |
[not acceptable] |

\(I = 5 \Rightarrow f < 0\) |
[not acceptable] |

Hence the valid solutions are \(\begin{align}x = I + f = \frac{{29}}{{12}},\,\,\,\frac{{19}}{6},\,\,\,\frac{{97}}{{24}}\end{align}\)