Question

A positive function $f$ is such that $f(2-h)=4,f(2+h)=3$ as $h\to 0^{+}$ has a minimum at $x=2.$ Then the value of $f\left(2\right)$ can be

• A. $2$
• B. $3$
• C. $5$
• D. $4$

Answer 1

The correct option is A $2$

For minimum at $x=2:$
$f(2-h)>f\left(2\right)<f(2+h)$ as $h\to 0^{+}$
$\Rightarrow 4>f\left(2\right)$ and $f\left(2\right)<3$
As, $f\left(x\right)>0,$
So, possible values of $f\left(2\right)$ is $(0,3)$
Now, from the given options only $2\in (0,3).$
$\therefore$ The value of $f\left(2\right)$ can be $2.$