MCQ
A function $f(x)\, = \left\{ {\begin{array}{*{20}{c}}{1 + x,}&{x \le 2}\\{5 - x,}&{x > 2}\end{array}} \right.\,$ is
  • A
    Not continuous at $x = 2$
  • B
    Differentiable at $x = 2$
  • Continuous but not differentiable at $x = 2$
  • D
    None of these

Answer

Correct option: C.
Continuous but not differentiable at $x = 2$
c
(c) $\mathop {\lim }\limits_{h \to {0^ - }} 1 + (2 - h) = 3$, 

$\mathop {\lim }\limits_{h \to {0^ + }} 5 - (2 + h) = 3$, $f(2) = 3$

Hence, $f$ is continuous at $x = 2$

Now $Rf'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{5 - (2 + h) - 3}}{h} = - 1$

$Lf'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{1 + (2 - h) - 3}}{{ - h}} = 1$

$\because Rf'(x) \ne Lf'(x)$;  $f$ is not differentiable at $x = 2$.

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

Given that  $ a$ and $  b$ are two unit non-collinear vectors, if $u = a - (a\,.\,b)\,b$ and $v = a \times b$, then find $|v| =$.
Lines $\frac{1-x}{3}=\frac{y-2}{1}=\frac{z-1}{2}$ and $\frac{x-2}{p}=\frac{y-1}{2}=\frac{z-2}{1}$ are mutually perpendicular to each other then, $p=$ ___________ .
If the direction cosines of a line are $\Big(\frac{1}{\text{c}},\frac{1}{\text{c}},\frac{1}{\text{c}}\Big)$ then:
The solution of the differential equation $\frac{{dy}}{{dx}} = \frac{{1 + {y^2}}}{{1 + {x^2}}}$ is
If $A$ and $B$ are the points $(-3,4,-8)$ and $(5,-6,4)$ respectively, then find the ratio in which $y z$-plane divides $\overrightarrow{A B}$.
If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda}$ and $\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}$ is $\frac{1}{\sqrt{3}}$, then the sum of all possible values of $\lambda$ is
Order of  $\Big( \frac{\text{dy}}{\text{dx}}\Big)^{3}+ \Big( \frac{\text{dy}}{\text{dx}}\Big)^{2}+\text{y}^{4}=0$  is:
  1. 4
  2. 3
  3. 1
  4. 2
If $x = x ( y )$ is the solution of the differential equation $y \frac{d x}{d y}=2 x+y^{3}(y+1) e^{y}, x(1)=0$; then $x(e)$ is equal to
Choose the correct answer from the given four options.
The matrix $\begin{bmatrix}0&-5&8\\5&0&12\\-8&-12&0\end{bmatrix}$ is a:
  1. Diagonal matrix.
  2. Symmetric matrix.
  3. Skew-symmetric matrix.
  4. Scalar matrix.
If $f(x) = \left\{ \begin{array}{l}\frac{{{x^4} - 16}}{{x - 2}},\,\,{\rm{when}}\,\,x \ne 2\\\,\,\,\,\,\,\,\,\,\,\,\,\,16,\,{\rm{when}}\,\,x = 2\end{array} \right.$, then