Question

Consider the following statements :
1. The function $f\left(x\right)=\sqrt[3]{3x}$ is continuous at all x except at x = 0.
2. The function f(x) = [x] is continuous at x = 2.99 where (.) is the bracket function.
Which of the above statements is/are correct ?

• A. 1 only
• B. 2 only
• C. Both 1 and 2
• D. Neither 1 nor 2

Answer 1

Answer: (B)

2 only

$1.f\left(x\right)=\sqrt[3]{3x}$

$\underset{x\to 0^{+}}{lim}\sqrt[3]{3x}=\underset{x\to 0^{+}}{lim}f\left(x\right)=0$

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}\sqrt[3]{3x}=0$

$f\left(0\right)=0$

LHL$=f\left(0\right)=$RHL

$\therefore$continuous at $0$ as well

$2.f\left(x\right)=\left[x\right]$

$\underset{x\to {2.99}^{+}}{lim}\left[x\right]=2$ $[x\to {2.99}^{+};x=\mathrm{2.9900.....1};\Rightarrow \left[x\right]=2]$

$x\to {2.99}^{+};x=\mathrm{2.9900.....1};\Rightarrow \left[x\right]=2$

$\underset{x\to {2.99}^{-}}{lim}\left[x\right]=\underset{x\to {2.99}^{-}}{lim}2=2$

$f\left(2.99\right)=\left[2.99\right]=2$

LHL$=f\left(0\right)=$RHL

$\Rightarrow$ It is continuous at $2.99$