Question

The set of value(s) of $b$ for which function $f\left(x\right)=\{\begin{array}{cc}{x}^2+b^2-5b+6,& x<0\\ x,& x\ge 0\end{array}$ has a point of minima at $x=0,$ is

• A. $(-\infty ,2]$
• B. $[2,3]$
• C. $[\frac{5}{2},\infty )$
• D. $[3,\infty )$

Answer 1

Answer: (A), (D)

$[3,\infty )$

Given : $f\left(x\right)=\{\begin{array}{cc}{x}^2+b^2-5b+6,& x<0\\ x,& x\ge 0\end{array}$
plotting the graph of $f\left(x\right)$ :

Here, $k=b^2-5b+6$
For point of minima at $x=0:f(0-h)>f\left(0\right)<f(0+h),$ as $h\to 0^{+}$
From graph it is clear that $f\left(0\right)<f(0+h),h\to 0^{+}$ and for $f(0-h)>f\left(0\right),k$ should be greater than or equal to $0.$
$\Rightarrow k\ge 0$
$\Rightarrow b^2-5b+6\ge 0$
$\Rightarrow (b-2)(b-3)\ge 0$
$\Rightarrow b\in (-\infty ,2]\cup [3,\infty )$