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Let a function \mathrm{ f: \mathbb{R} \rightarrow \mathbb{R} } be defined as :

\mathrm{f(x)= \begin{cases}\int_{0}^{x}(5-|t-3|) d t, & x>4 \\ x^{2}+b x & , x \leq 4\end{cases} }
where \mathrm{ \mathrm{b} \in \mathbb{R}.} If \mathrm{f } is continuous at \mathrm{x=4, } then which of the following statements is NOT true?

Option: 1

\mathrm{f \text { is not differentiable at } x=4}


Option: 2

\mathrm{f^{\prime}(3)+f^{\prime}(5)=\frac{35}{4}}


Option: 3

\mathrm{f \text { is increasing in }\left(-\infty, \frac{1}{8}\right) \cup(8, \infty)}


Option: 4

\mathrm{f \text { has a local minima at } x=\frac{1}{8}}


Answers (1)

best_answer

\text{Given} \;\; f(x) \begin{cases}\int_{0}^{x}(5-|t-3|) d t, & x>4 \\ x^{2}+b x, & x \leq 4\end{cases}\\

f(x) \text{ is continuous at } x=4\\

\text{So } \mathrm{ \lim _{x \rightarrow 4^{-}} f(x)=\lim _{x \rightarrow 4^{+}} f(x)=f(4)}\\

\begin{aligned} \text{So } 16+4 b &=\int_{0}^{3}(2-t) d t+\int_{3}^{4}(8-t) d t \\ \Rightarrow 16+4 b &=15 \end{aligned}

\text{So } \mathrm{b=-\frac{1}{4} \text{ at } x=4}

\mathrm{\begin{aligned} & LHD =2 x+b =\frac{31}{4}\\ &RHD =5-|x-3|=4\\ &LHD \neq RHD \end{aligned}}\\

\text{option (A) is true and } f^{\prime}(3)+f^{\prime}(5)=\frac{23}{4}+3=\frac{35}{4}\\

option (B) is true

\begin{aligned} & \therefore f(x)=x^{2}-\frac{x}{4} \text{ at } x \leq 4\\ & \;\;\; f^{\prime}(x)=2 x-\frac{1}{4} \end{aligned}

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 yocal minina at x=\frac{1}{8}
 

Posted by

avinash.dongre

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