Question

Let a function $f: R \rightarrow R$ be defined as :
$f(x)= \begin{cases}\int\limits_0^x(5-|t-3|) d t, & x>4 \\ x^2+b x & , x \leq 4\end{cases}$
where $b \in R$. If $f$ is continuous at $x =4$, then which of the following statements is NOT true ?

• A. $f$ is not differentiable at $x =4$
• B. $f^{\prime}(3)+f^{\prime}(5)=\frac{35}{4}$
• C. $f$ is increasing in $\left(-\infty, \frac{1}{8}\right) \cup(8, \infty)$
• D. $f$ has a local minima at $x=\frac{1}{8}$

Answer 1

Answer: (C)

Given $f(x) \begin{cases}\int\limits_0^x(5-|t-3|) d t, & x>4 \\ x^2+b x & x \leq 4\end{cases}$
$f ( x )$ is continuous at $x =4$
So $\displaystyle\lim _{x \rightarrow 4^{-}} f(x)=\displaystyle\lim _{x \rightarrow 4^{+}} f(x)=f(4)$
So $16+4 b =\int_0^3(2-t) d t+\int\limits_3^4(8-t) d t$
$\Rightarrow 16+4 b=15$
So $b=\frac{-1}{4}$
At $x=4$
$ LHD =2 x + b =\frac{31}{4} $
$ RHD =5-|x-3|=4$
$ LHD \neq RHD$
Option (A) is true
and $f ^{\prime}(3)+ f ^{\prime}(5)=\frac{23}{4}+3=\frac{35}{4}$
Option (B) is true
$ \because f(x)=x^2-\frac{x}{4} \text { at } x \leq 4 $
$f^{\prime}(x)=2 x-\frac{1}{4}$
This function is not increasing.
In the interval in $x \in\left(-\infty, \frac{1}{8}\right)$
Option (C) is NOT TRUE.
This function $f(x)$ is also local minima at
$x=\frac{1}{8}$