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 $x=\frac{1}{8}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The term independent of $x$ in the expansion of $\left( {\frac{1}{{60}} - \frac{{{x^8}}}{{81}}} \right).{\left( {2{x^2} - \frac{3}{{{x^2}}}} \right)^6}$ is equal to

Let $S =(0,2 \pi)-\left\{\frac{\pi}{2}, \frac{3 \pi}{4}, \frac{3 \pi}{2}, \frac{7 \pi}{4}\right\}$. Let $y =$ $y ( x ), x \in S$, be the solution curve of the differential equation $\frac{ dy }{ dx }=\frac{1}{1+\sin 2 x }, y \left(\frac{\pi}{4}\right)=\frac{1}{2}$. if the sum of abscissas of all the points of intersection of the curve $y=y(x)$ with the curve $y=\sqrt{2} \sin x$ is $\frac{k \pi}{12}$, then $k$ is equal to

The third term of a $G.P.$ is the square of first term. If the second term is $8$, then the ${6^{th}}$ term is

Let $f(x) = {x^2} + 4x + 1$. Then

Orthogonal trajectories of family of the curve ${x^{\frac{2}{3}}} + {y^{\frac{2}{3}}} = {a^{\frac{2}{3}}}$, where $'a'$ is any arbitrary constant, is

The coordinates of the foci of the rectangular hyperbola $xy = {c^2}$ are

If $g(x)=\int_{\sin x}^{\sin (2 x)} \sin ^{-1}(t) d t$, then

The centre and radius of the circle $2{x^2} + 2{y^2} - x = 0$ are

The value of the definite integral $\int\limits_0^{\frac{\pi }{2}} {}$ $\sin\, x\, \sin\, 2x\, \sin\, 3x\, dx$ is equal to :

The set $A = \{ x:x \in R,\,{x^2} = 16$ and $2x = 6\} $ equals