MCQ
Which of the following functions differentiable at $x = 0$ ?
  • A
    $cos (|x|)+|x|$
  • B
    $cos (|x|)-|x|$
  • C
    $sin (|x|)+|x|$
  • $sin (|x|)-|x|$

Answer

Correct option: D.
$sin (|x|)-|x|$
d
Let $f(x)=\sin (|x|)-|x|$

$\Rightarrow \quad f(x)=\left\{\begin{array}{ll}{\sin x-x:} & {x \geq 0} \\ {-\sin x+x ;} & {x<0}\end{array}\right.$

$\therefore \quad {f^\prime }\left( {{0^ + }} \right) = 0 = {f^\prime }\left( {{0^ - }} \right)$

$\Rightarrow \quad f(x)$ is differentiable at $x=0$

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

Choose the correct answer from the given four options.
Two events E and F are independent. If $\text{P}(\text{E})=0.3,\text{P}(\text{E}\cup\text{F})=0.5,$ then $\text{P}\Big(\frac{\text{E}}{\text{F}}\Big)-\text{P}\Big(\frac{\text{F}}{\text{E}}\Big)$ equal:
A unit vector perpendicular to the plane of $a = 2i - 6j - 3k$, $b = 4i + 3j - k$ is
The vector $\left( {\hat i \times \vec a.\vec b} \right)\hat i + \left( {\hat j \times \vec a.\vec b} \right)\hat j + \left( {\hat k \times \vec a.\vec b} \right)\hat k$ is equal to
$f(x) = {x^2} - 3x$, then the points at which $f(x) = f'(x)$ are
The region represented by the inequation system x, y ≥ 0, y ≤ 6, x + y ≤ 3 is:
Consider the system of equations:
a1x + b1y + c1z = 0
a2x + b2y + c2z = 0
a3x + b3y + c3z = 0,
if $\begin{vmatrix}\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2\\\text{a}_3&\text{b}_3&\text{c}_3\end{vmatrix}=0$, then the system has
  1. More than two solutions.
  2. One trivial and one non-trivial solutions.
  3. No solutions.
  4. Only trivial solution (0, 0, 0).
If the lines $\frac{{x - 1}}{{ - 3}} = \frac{{y - 2}}{{2k}} = \frac{{z - 3}}{2}$, $\frac{{x - 1}}{{3k}} = \frac{{y - 5}}{1} = \frac{{z - 6}}{{ - 5}}$ are at right angles, then $k =$
Choose the correct answer from the given four options:
The area of the region bounded by the curve $\text{y}=\sin\text{x}$ between the ordinates x = 0, $\text{x}=\frac{\pi}{2}$ and the x-axis is:
  1. $2\text{ sq. units}$
  2. $4\text{ sq. units}$
  3. $3\text{ sq. units}$
  4. $1\text{ sq. units}$
The value of $\int\limits_0^4 {\left\{ {\sqrt x } \right\}dx,} $ where $\{ \}$ denotes the fractional part of $x$ is
If $A = \left[ {\begin{array}{*{20}{c}}1&{ - 2}&1\\2&1&3\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}2&1\\3&2\\1&1\end{array}} \right]$, then ${(AB)^T} = $