Question

Let $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}1-x+a,& x\le 1\end{array}\\ \begin{array}{cc}2x+3,& x>1\end{array}\end{array}$. If $f\left(x\right)$ has local minimum at $x=1,$ then $a\le$

• A. $2$
• B. $3$
• C. $5$
• D. None of these

Answer 1

The correct option is C $5$

We have $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}1-x+a,& x\le 1\end{array}\\ \begin{array}{cc}2x+3,& x>1\end{array}\end{array}$

Since $y=1-x+a$ is decreasing function and $y=2x+3$ is increasing function $f\left(x\right)$ will have a local minimum at $x=1$

If $f\left(1\right)\le f\left(x\right)$ for $x>1$

or if $a\le 2x+3$ for $x>1$

$\Rightarrow a\le 5$