MCQ
The function $f(x)=\left|x^{2}-2 x-3\right| \cdot e^{\left|9 x^{2}-12 x+4\right|}$ is not differentiable at exactly :
  • A
    four points
  • B
    three points
  • two points
  • D
    one point

Answer

Correct option: C.
two points
c
$f(x)=|(x-3)(x+1)| \cdot e^{(3 x-2)^{2}}$

$f(x)= (x-3)(x+1) \cdot e^{(3 x-2)^{2}} ; \quad x \in(3, \infty)$

$\quad\quad\quad-(x-3)(x+1) \cdot e^{(3 x-2)^{2}} ; \quad x \in[-1,3]$

$\quad\quad\quad(x-3) \cdot(x+1) \cdot e^{(3 x-2)^{2}} \quad ; x \in(-\infty,-1)$

Clearly, non-differentiable at $x=-1\, \&\, x=3$

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

Similar questions

The Function $\text{f}(\text{x})=\frac{\lambda+\sin\text{x}+2\cos\text{x}}{\sin\text{x}+\cos\text{x}}$ is increasing, if:
  1. $\lambda<1$
  2. $\lambda>1$
  3. $\lambda<2$
  4. $\lambda>2$
The maximum height is reached in $5$ seconds by a stone thrown vertically upwards and moving under the equation  $10s = 10ut -49{t^2}$, where $ s$ is in metre and $t$ is in second. The value of $u$ is ........ $m/\sec $
Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\vec{c}=x \hat{i}+2 \hat{j}+3 \hat{k}, x \in \mathbb{R}$. If $\vec{d}$ is the unit vector in the direction of $\vec{b}+\vec{c}$ such that $\vec{a} \cdot \vec{d}=1$, then $(\vec{a} \times \vec{b}) \cdot \vec{c}$ is equal to
The order of the differential whose general solution is given by ${\text{y}}=\text{C}_1\cos(2\text{x}+\text{C}_{2})+(\text{C}_{3}+\text{C}_{4})\text{a}^{\text{x}+\text{C}_{5}}+\text{C}_{6}\sin(\text{x}-\text{C}_{7}).$ 
  1. 3
  2. 4
  3. 5
  4. 2
Shortest distance between lines $\frac{{x - 6}}{1} = \frac{{y - 2}}{{ - 2}} = \frac{{z - 2}}{2}$ and $\frac{{x + 4}}{3} = \frac{y}{{ - 2}} = \frac{{z + 1}}{{ - 2}}$ is
$\int_{}^{} {\frac{{{e^{m{{\tan }^{ - 1}}x}}}}{{1 + {x^2}}}dx} $ equals to
If $\text{f(x)}=\begin{cases}\frac{\sin(\text{a}+1)}{\text{x}},&\text{x}<0\\\text{c},&\text{x}=0\\\frac{\sqrt{\text{x+bx}^2}-\sqrt{\text{x}}}{\text{bx}\sqrt{\text{x}}},&\text{x}>0&\end{cases}$ is continuouse at x = 0, then:
  1. $\text{a}=-\frac{3}{2},\text{b}=0,\text{c}=\frac{1}{2}$
  2. $\text{a}=-\frac{3}{2},\text{b}=1,\text{c}=-\frac{1}{2}$
  3. $\text{a}=-\frac{3}{2},\text{b}\in\text{R}-\{0\},\text{c}=\frac{1}{2}$
  4. $\text{None of these}.$
The area bounded by the curve $y = x^2 + 4x + 5,$ the axes of co-ordinates & the minimum ordinate is:
${d \over {dx}}{\tan ^{ - 1}}{{4\sqrt x } \over {1 - 4x}} = $
If $\text{A}+\text{B}+\text{C}=\pi,$ then the value of $\begin{vmatrix}\sin(\text{A}+\text{B}+\text{C})&\sin(\text{A}+\text{C})&\cos\text{C}\\-\sin\text{B}&0&\tan\text{A}\\\cos(\text{A}+\text{B})&\tan(\text{B}+\text{C})&0\end{vmatrix}$ is equal to:
  1. 0
  2. 1
  3. $2\sin\text{B}\tan\text{A}\cos\text{C}$
  4. None of these.