Question

Let $f\left(x\right)=\frac{x^2+2}{\left[x\right]},1\le x\le 3$, where $[.]$ represents greatest integer function, then

• A. $f\left(x\right)$ is increasing in $[1,3]$
• B. Least value of $f\left(x\right)$ is $3$
• C. Greatest value of $f\left(x\right)$ is $\frac{11}{2}$
• D. $f\left(x\right)$ has no greatest value

Answer 1

The correct option is B Least value of $f\left(x\right)$ is $3$

The function $f\left(x\right)$ can be defines as follows,

$f\left(x\right)=\frac{x^2+2}{1}$ when $1\le x<2$

$f\left(x\right)=\frac{x^2+2}{2}$ when $2\le x<3$

$f\left(x\right)=\frac{x^2+2}{3}$ when $x=3$

$Case1-$Taking $1\le x<2$,

$\Rightarrow \frac{d\left(f\right(x\left)\right))}{dx}=2x$

This will be always positive for all $x>0$

$\Rightarrow f\left(x\right)$ will be increasing for $1\le x<2$ and will have the least value at $x=1$ for the considered interval

$\Rightarrow f\left(1\right)=\frac{1^2+2}{1}=\frac{3}{1}=3$

$Case2-$Taking $2\le x<3$,

$\Rightarrow \frac{d\left(f\right(x\left)\right))}{dx}=\frac{2x}{2}=x$

This will be always positive for all $x>0$

$\Rightarrow f\left(x\right)$ will be increasing for $2\le x<3$ and will have the least value at $x=2$ for the considered interval

$\Rightarrow f\left(2\right)=\frac{2^2+2}{2}=\frac{6}{2}=3$

$Case3-$Taking $x=3$,

$\Rightarrow f\left(3\right)=\frac{11}{3}(>3)$

Now, we can see that the minimum value of $f\left(x\right)$ in interval $1\le x\le 3$ will be $3$ attained by $f\left(x\right)$ at $x=1,2$

Hence, the correct answer is option $B$