Question

Let a function $f:R\to R$ be defined as :
$f(x)=\{\begin{array}{ll}\underset{0}{\overset{x}{\int }}(5-|t-3|)dt,& x>4\\ x^2+bx& ,x\le 4\end{array}$
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 $f(x)\{\begin{array}{ll}\underset{0}{\overset{x}{\int }}(5-|t-3|)dt,& x>4\\ x^2+bx& x\le 4\end{array}$
$f(x)$ is continuous at $x=4$
So $\underset{x\to 4^{-}}{\lim }f(x)=\underset{x\to 4^{+}}{\lim }f(x)=f(4)$
So $16+4b={\int }_0^3(2-t)dt+\underset{3}{\overset{4}{\int }}(8-t)dt$
$\Rightarrow 16+4b=15$
So $b=\frac{-1}{4}$
At $x=4$
$LHD=2x+b=\frac{31}{4}$
$RHD=5-|x-3|=4$
$LHD\ne RHD$
Option (A) is true
and $f^{\prime }(3)+f^{\prime }(5)=\frac{23}{4}+3=\frac{35}{4}$
Option (B) is true
$\because f(x)=x^2-\frac{x}{4}\text{ at }x\le 4$
$f^{\prime }(x)=2x-\frac{1}{4}$
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
$x=\frac{1}{8}$