The function f(x) = e|x| is: - Vidyadip

MCQ

The function $f(x) = e|x|$ is:

✓
Continuous everywhere but not differentiable at $x = 0$
B
Continuous and differentiable everywhere
C
Not continuous at $x = 0$
D
None of these.

✓

Answer

Correct option: A.

Continuous everywhere but not 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

Explore more

More "M.C.Q (1 Marks)" from CONTINUITY AND DIFFERENTIABILITY→CONTINUITY AND DIFFERENTIABILITY — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

The value of the integral $\int_0^{\pi / 2} \frac{3 \sqrt{\cos \theta}}{(\sqrt{\cos \theta}+\sqrt{\sin \theta})^5} d \theta$ equals

The probability that a bulb produced by a factory will fuse after $150$ days of use is $0.05$. What is the probability that out of $5$ such bulbs none will fuse after $150$ days of use

In a triangle $ABC ,$ if $|\overline{ BC }|=8,|\overline{ CA }|=7,|\overline{ AB }|=10,$ then the projection of the vector $\overline{ AB }$ on $\overline{ AC }$ is equal to ....... .

Choose the correct answer in Exercise:
$\int^{\sqrt{3}}_{1}\frac{\text{dx}}{1+\text{x}^{2}}\text{equals}$

If $\vec{\text{a}},\vec{\text{b}}$ represent the diagonals of a rhombus, then:

$\left| {\,\begin{array}{*{20}{c}}{b + c}& a& a\\b& {c + a}& b\\c& c& {a + b}\end{array}\,} \right| = $

While plotting constraints on a graph paper, terminal points on both the axes are connected by a straight line because:

Let $f$ : $R \rightarrow R$ be a differentiable function $ \&$ $ f (1) = 4$ , then the value of $\mathop {Lim}\limits_{x\,\, \to \,\,1} $$\int\limits_4^{f\,(x)} {\,\frac{{2t\,dt}}{{x - 1}}} $; is :

$\int_{0}^{\pi /2}{\frac{dx}{{{a}^{2}}{{\cos }^{2}}x+{{b}^{2}}{{\sin }^{2}}x}}\,=$

A square $ABCD$ of diagonal $2a$ is folded along the diagonal $AC$ so that the planes $DAC$ and $BAC$ are at right angle. The shortest distance between $DC$ and $AB$ is