Question

If $f\left(x\right)=|x-a|+|x+b|,xϵR,b>a>0$. Then

• A. $f^{\prime }\left(a^{+}\right)=1$
• B. $f^{\prime }\left(a^{+}\right)=0$
• C. $f^{\prime }(-b^{+})=0$
• D. $f^{\prime }(-b^{+})=1$

Answer 1

The correct option is C $f^{\prime }(-b^{+})=0$

$f(-b)=|-b-a|+|-b+b|=a+b$

$f\left(a\right)=|a-a|+|a+b|=a+b$

Since $0<a<b$

$f(-b)=f\left(a\right)$

Now, when $x>a$

$f\left(x\right)=x-a+x+b=2x+(b-a)$$f^{\prime }\left(x\right)=2$

So, $f^{\prime }\left(a^{+}\right)=2$

when $-b\le x\le a$

$f(-b)=f\left(a\right)$

So, $f\left(x\right)$ is straight line parallel to x-axis

Thus slope for $-b\le x\le a$ is $0$

Thus $f\left({-b}^{+}\right)=0$

So, option C is correct