Question

$f\left(x\right)=\frac{\left[x\right]+1}{\left\{x\right\}+1}$ for $f:[0,\frac{5}{2})\to [\frac{1}{2},3)$, where $[.]$ represent the integer function and $\{.\}$ represent the fraction part of $x$. then which of the following is true?

• A. $f\left(x\right)$ is injective discontinuous function
• B. $f\left(x\right)$ is surjective non-differntiable function
• C. $min({lim}_{x\to 1^{-}}f\left(x\right),{lim}_{x\to 1^{+}}f\left(x\right))$
• D. $max\left(xvaluesofpointofdiscontinuity\right)=f\left(1\right)$

Answer 1

The correct option is D $max\left(xvaluesofpointofdiscontinuity\right)=f\left(1\right)$

$f\left(x\right)=\frac{\left[x\right]+1}{\left\{x\right\}+1}$ for $f:[0,\frac{5}{2})\to (\frac{1}{2},3]$

$f\left(1\right)=\frac{\left[1\right]+1}{\left\{1\right\}+1}=\frac{1+1}{0+1}=2$

$f\left(x\right)\{\begin{array}{c}\frac{1}{x+1},0\le x<1\\ \frac{2}{x},1\le x<2\\ \frac{3}{x-1},2\le x<\frac{5}{2}\end{array}$

Clearly, $f\left(x\right)$ is discontinuous and bijective function.

Now,

${lim}_{x\to 1^{-}}f\left(x\right)={lim}_{x\to 1^{-}}\left(\frac{1}{x+1}\right)=\frac{1}{2}$

${lim}_{x\to 1^{+}}f\left(x\right)={lim}_{x\to 1^{+}}\left(\frac{2}{x}\right)=2$

Therefore,

$min({lim}_{x\to 1^{-}}f\left(x\right),{lim}_{x\to 1^{+}}f\left(x\right))={lim}_{x\to 1^{-}}f\left(x\right)=\frac{1}{2}\ne f\left(1\right)$

$max(1,2)=2=f\left(1\right)$

Hence the required answer is (D) $max\left(\text{x values of point of discontuuity}\right)=f\left(1\right)$