Let a function f: R R be defined as : f(x)= ll _ 0 ^ x (5-|t-3|) … - Vidyadip

MCQ

Let a function $f: R \rightarrow R$ be defined as :
$f(x)=\left\{\begin{array}{ll} \int_{0}^{x}(5-|t-3|) d t, & x>4 \\ x^{2}+b x & , x \leq 4
\end{array}\right.$
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
$f^{\prime}(3)+f^{\prime}(5)=\frac{35}{4}$
✓
$f$ is increasing in $\left(-\infty, \frac{1}{8}\right) \cup(8, \infty)$
D
$f$ has a local minima at $x=\frac{1}{8}$

✓

Answer

Correct option: C.

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

c
Given $f(x)\left\{\begin{array}{ll}\int_{0}^{x}(5-|t-3|) d t, & x>4 \\ x^{2}+b x & , x \leq 4\end{array}\right.$

$f(x)$ is continuous at $x=4$

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

So $16+4 b =\int_{0}^{3}(2-t) d t+\int_{3}^{4}(8-t) d t$ $\Rightarrow 16+4 b=15$

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

$\because f(x)=x^{2}-\frac{x}{4}$ 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 $x=\frac{1}{8}$

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