Question

Let $$\begin{gathered} f\left(x\right)=\left\{\left|x-1\right|+aifx\le 1 \\ 2x+3ifx>\hspace{0.17em}1\right\} \end{gathered}$$. If $f\left(x\right)$ has local minimum at $x=1$, then:

• A. $7$
• B. $8$
• C. $9$
• D. $a\le 5$

Answer 1

The correct option is D

$a\le 5$

Explanation for the correct option:

Find the value of '$a$':

Given,

If $x\le 1$,$\Rightarrow f\left(x\right)=\left|x-1\right|+a$,

Then

If $x=-1$, $\Rightarrow f\left(x\right)=\left|-1-1\right|+a=2+a$

If $x=0$, $\Rightarrow f\left(x\right)=\left|0-1\right|+a=1+a$

If $x=1$, $\Rightarrow f\left(x\right)=\left|1-1\right|+a=0+a$

So, as the value of $x$ increases, the value of the function will decrease. therefore, $f\left(x\right)$ is a decreasing function. at $x\le 1$

Similarly,

If $x>1$,$\Rightarrow f\left(x\right)=2x+3$,

clearly, $f\left(x\right)=2x+3$ is an increasing function at $x>1$

$f\left(x\right)$ has local minima at $x=1$[Given]

We know that,

$\Rightarrow f\left(1\right)\le li{m}_{x\to 1^{+}}f\left(x\right)$

Therefore,

$$\begin{gathered} f\left(1\right)\le li{m}_{h\to 0}f(1+h) \\ \Rightarrow a\le li{m}_{h\to 0}2(1+h)+3\left[\because f\left(x\right)=2x+3;f\left(1\right)=a\right] \\ a\le li{m}_{h\to 0}5+2h \\ a\le 5 \end{gathered}$$

Hence, the correct answer is Option (D).