Question

Let $f:\left(1,3\right)\to R$ be a function defined by $f\left(x\right)=\frac{x\left[x\right]}{\left(x^2+1\right)}$ , where$\left[x\right]$ denotes the greatest integer $\le x$. Then the range of $f$ is

• A. $(\frac{2}{5},\frac{3}{5}]\cup \left(\frac{3}{4},\frac{4}{5}\right)$
• B. $(\frac{2}{5},\frac{4}{5}]$
• C. $\left(\frac{3}{5},\frac{4}{5}\right)$
• D. $\left(\frac{2}{5},\frac{1}{2}\right)\cup (\frac{3}{5},\frac{4}{5}]$

Answer 1

The correct option is D

$\left(\frac{2}{5},\frac{1}{2}\right)\cup (\frac{3}{5},\frac{4}{5}]$

Explanation for correct option:

Identifying the range $f$:

Consider the given equation as,

$f:\left(1,3\right)\to R$

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

Rewrite the above Equation as,

$f\left(x\right)=\left\{\begin{array}{ll}\frac{x}{\left(x^2+1\right)}& x\in \left(1,2\right)\\ \frac{2x}{\left(x^2+1\right)}& x\in (2,3]\end{array}\right.......\left(1\right)$

$$\begin{gathered} f\text{'}\left(x\right)=\left\{\begin{array}{ll}\frac{\left(x^2+1\right)\times 1-x\times 2x}{{\left(x^2+1\right)}^2}& x\in \left(1,2\right)\\ \frac{\left(x^2+1\right)\times 2-2x\times 2x}{{\left(x^2+1\right)}^2}& x\in (2,3]\end{array}\right. \\ f\text{'}\left(x\right)=\left\{\begin{array}{ll}\frac{x^2+1-x\times 2x}{{\left(x^2+1\right)}^2}& x\in \left(1,2\right)\\ \frac{2+2x^2-4x^2}{{\left(x^2+1\right)}^2}& x\in (2,3]\end{array}\right. \\ f\text{'}\left(x\right)=\left\{\begin{array}{ll}\frac{1-x^2}{{\left(x^2+1\right)}^2}& x\in \left(1,2\right)\\ \frac{2\left(1-x^2\right)}{{\left(x^2+1\right)}^2}& x\in (2,3]\end{array}\right. \end{gathered}$$

From the above Equation we get negative value . since,$f\left(x\right)$ is decreasing function.

substitute the value in the Equation $\left(1\right)$ we get the range of functions belonging to the,

$$\begin{gathered} f\left(x\right)=\frac{x}{\left(x^2+1\right)}x\in \left(1,2\right) \\ f\left(x\right)=\frac{1}{\left(1+1\right)}=\frac{1}{2} \\ f\left(x\right)=\frac{2}{\left(2^2+1\right)}=\frac{2}{5} \\ f\left(x\right)=\frac{2x}{\left(x^2+1\right)}x\in (2,3] \\ f\left(x\right)=\frac{2\times 2}{\left(2^2+1\right)}=\frac{4}{5} \\ f\left(x\right)=\frac{2\times 3}{\left(3^2+1\right)}=\frac{6}{10}=\frac{3}{5} \end{gathered}$$

Hence, $R_f\in \left(\frac{2}{5},\frac{1}{2}\right)\cup (\frac{3}{5},\frac{4}{5}]$

Therefore, the correct answer is Option (D).