Fx-min x- x 2 - x- R then fx is

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Intersection

fx=min x, x 2...

Question

$f (x) = min {x, x^{2}}, \forall x ϵ R$ then $f (x)$ is

discontinuous at

0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

discontinuous at

1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

continuous on

R

No worries! We‘ve got your back. Try BYJU‘S free classes today!

continuous on

0, 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct options are
A discontinuous at $0$
B discontinuous at $1$

Suggest Corrections

Similar questions

For $x ϵ [0, \frac{1}{2}], f (x) = tan ({sin}^{- 1} (\frac{x}{\sqrt{2}} + \sqrt{\frac{1 - x^{2}}{2}}) - {sin}^{- 1} x)$ is

Q. Given

f (x) = | x - 1 | + | x + 1 |

. Then f(x) is

Let f(x) = x + 1; where $x ϵ [0, \infty]$ . Choose the right option.

Let $f (x)$ be defined in $[- 2, 2]$ by $f (x) = {\begin{matrix} m a x {\sqrt{4 - x^{2}}, \sqrt{1 + x^{2}}} & - 2 \leq x \leq 0 m i n {\sqrt{4 - x^{2}}, \sqrt{1 + x^{2}}} & 0 < x \leq 2 \end{matrix}$ . Then $f (x)$ is

Q. A function is given by

f (x) = log x . e^{x}

. Then

f (x)

Join BYJU'S Learning Program

f(x)=min{x,x2},∀xϵR then f(x) is

The correct options are A discontinuous at 0 B discontinuous at 1

$f (x) = min {x, x^{2}}, \forall x ϵ R$ then $f (x)$ is

The correct options are
A discontinuous at $0$
B discontinuous at $1$