Question

If $f:\left( {2,3} \right) \to \left( {0,1} \right)$ is defined by $f\left( x \right) = x - \left[ x \right]$, then ${f^{ - 1}}\left( x \right)$ is
1) $x - 2$
2) $x + 1$
3) $x - 1$
4) $x + 2$

Answer 1

Hint: First we need to determine whether the function is increasing or decreasing. After determining that we can determine whether the extreme values in the domain gives the higher or lower value in the range. On determining that we can easily observe the trend of the function and find the value of the inverse function as required.

Complete step-by-step answer:
Given, $f:\left( {2,3} \right) \to \left( {0,1} \right)$ defined by $f\left( x \right) = x - \left[ x \right]$.
We can clearly observe that the function is increasing.
As, as $x$ increases over $\left( {2,3} \right)$, $\left[ x \right] = 2$, over the whole domain.
So, the value of $f\left( x \right)$ increases over the domain.
Hence, it is an increasing function.
Therefore, we can say about the function over the domain that,
$\mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right) = 0$
And, $\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = 1$
Therefore, now, taking the inverse of the function, we can easily say about the function that,
 ${f^{ - 1}}\left( 0 \right) = 2$
And, ${f^{ - 1}}\left( 1 \right) = 3$
Therefore, we can generalise the values by observing the trend of values we get from the extremums of the range as,
${f^{ - 1}}\left( x \right) = x + 2$
Therefore, the correct option is 4.
So, the correct answer is “Option 4”.

Note: The problems of relation and functions can be solved in a variety of ways. Sometimes by just observing the question, the solution can be obtained pretty fast. And sometimes, calculations have to be made. In this question, carrying out any kind of further calculations may have made it more complicated, which is easily solved by just simple observations. Moreover, the term in the function given as, $f\left( x \right) = x - \left[ x \right] = \left\{ x \right\}$, which gives the value in the interval $[0,1)$, where, if $x$ is itself an integer, then, the value of $\left\{ x \right\}$ is $0$.