Let f(x) = |x| + |x - 1|, then : - Vidyadip

MCQ

Let $f(x) = |x| + |x - 1|,$ then :

✓
$f(x)$ is continuous at $x = 0,$ as well as at $x = 1$
B
$f(x)$ is continuous at $x = 0,$ but not at $x = 1$
C
$f(x)$ is continuous at $x = 0,$ but not at $x = 0$
D
None of these

✓

Answer

Correct option: A.

$f(x)$ is continuous at $x = 0,$ as well as at $x = 1$

Since modulus function is everywhere continuous $|x|$ and $|x - 1|$ are also everywhere continuous.
Also,
It is known that if $f$ and $g$ are continuous functions, then $f + g$ will also be continuous.
Thus, $|x| + |x - 1|$ is everywhere continuous.
Hence, $f(x)$ is continuous at $x = 0$ and $x = 1, x = 0$ and $x = 1.$

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

Choose the correct answer The line $y = x + 1$ is a tangent to the curve $y^2 = 4x$ at the point :

If O is the origin, OP = 3 with direction ratios proportional to -1, 2, -2 then the coordinates of P are:

Given $A=\left[\begin{array}{ccc}\sqrt{3} & 1 & -1 \\ 2 & 3 & 0\end{array}\right]$ and $B=\left[\begin{array}{ccc}2 & \sqrt{5} & 1 \\ -2 & 3 & \frac{1}{2}\end{array}\right],$ find $A+B$ $=$ ........

Let $f(x)=\int \frac{2 x}{\left(x^2+1\right)\left(x^2+3\right)} d x$ . If $f(3)=\frac{1}{2}\left(\log _e 5-\log _e 6\right)$, then $f(4)$ is equal to

If $f\left( {\frac{{x - 4}}{{x + 2}}} \right) = 2x + 1,(x \in R = \left\{ {1, - 2} \right\}),$ then $\int {f(x)} \,dx$ is equal to (where $C$ is a constant of integration)

If $f\, \&\, g$ are continuous functions in $[0, a]$ satisfying $f (x) = f (a - x)\, \&\, g (x) + g (a - x) = 4$ then $\int\limits_0^a {\,f\,(x)\,\,.\,\,g\,(x)dx} $ $=$

The direction cosines of the normal to the plane $2x - 3y - 6z - 3 = 0$ are:

The system of equations $kx + 2y\,-z = 1$ ; $(k\,-\,1)y\,-2z = 2$ ; $(k + 2)z = 3$ has unique solution, if $k$ is equal to

The probability distribution of a random variable $X$ is:

$X$	$ 0$	$1$	$2$	$3$	$4$
$P(X)$	$ 0.1$	$ k$	$ 2k$	$ k$	$0.1$

where $k$ is some unknown constant.
The probability that the random variable $X$ takes the value $2$ is:

If $\cot^{-1}(\sqrt{\cos\alpha})-\tan^{-1}(\sqrt{\cos\alpha})=\text{x},$ then $\sin\text{x}$ is equal to: