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(C) Given $f(x) = \begin{cases} \int_{0}^{x}(5-|t-3|) d t, & x>4 \ x^{2}+b x, & x \leq 4 \end{cases}$ Since $f(x)$ is continuous at $x=4$,we have $\l

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

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(C) Given $f(x) = \begin{cases} \int_{0}^{x}(5-|t-3|) d t, & x>4 \ x^{2}+b x, & x \leq 4 \end{cases}$ Since $f(x)$ is continuous at $x=4$,we have $\lim _{x ightarrow 4^{-}} f(x) = \lim _{x ightarrow 4^{+}} f(x) = f(4)$. $16+4b = \int_{0}^{4}(5-|t-3|) d t = \int_{0}^{3}(5-(3-t)) d t + \int_{3}^{4}(5-(t-3)) d t = \int_{0}^{3}(2+t) d t + \int_{3}^{4}(8-t) d t$. Evaluating the integrals: $[2t + \frac{t^2}{2}]_0^3 + [8t - \frac{t^2}{2}]_3^4 = (6 + 4.5) + ((32-8) - (24-4.5)) = 10.5 + (24 - 19.5) = 10.5 + 4.5 = 15$. So,$16+4b = 15 \implies 4b = -1 \implies b = -\frac{1}{4}$. Now,$f'(x)$ for $x  4$ is $5-|x-3| = 5-(x-3) = 8-x$. $LHD = f'(4^-) = 2(4) - 0.25 = 7.75 = \frac{31}{4}$. $RHD = f'(4^+) = 8-4 = 4$. Since $LHD 
eq RHD$,$f$ is not differentiable at $x=4$. Option $(A)$ is true. $f'(3) = 2(3) - 0.25 = 5.75 = \frac{23}{4}$. $f'(5) = 8-5 = 3$. $f'(3)+f'(5) = 5.75 + 3 = 8.75 = \frac{35}{4}$. Option $(B)$ is true. For $x  0$ when $x > \frac{1}{8}$. Thus $f$ is decreasing on $(-\infty, \frac{1}{8})$ and increasing on $(\frac{1}{8}, 4)$. Option $(C)$ is $NOT$ $TRUE$.

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

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