NDA 2016
- Exam
- NDA
- Year
- 2016
- Subject
- Mathematics
- Chapter
- Continuity and Differentiability
NDA 2016 · Mathematics · Continuity and Differentiability
Paragraph: Let f(x)=\ array ll -2, & -3 ≤ x ≤ 0 x-2, & 0 Question:...
Paragraph:
Let \mathrm{f}(\mathrm{x})=\left\{\right. and
Question: Which of the following statements are correct? 1. is continuous at . 2. is continuous at . 3. is continuous at . Select the correct answer using the code given below:
Question: Which of the following statements are correct? 1. is continuous at . 2. is continuous at . 3. is continuous at . Select the correct answer using the code given below:
Options
- A1 and 2 only
- B2 and 3 only
- C1 and 3 only
- D1,2 and 3
Correct answer
D. 1,2 and 3
Step-by-step solution
At
For LHL:
For RHL: For $(x=0): g(x)=-2+|-2|=0\begin{aligned}
& \mathrm{LHL}=\lim _{x \rightarrow 0^{-}} g(x)=0 \\
& \mathrm{RHL}=\lim _{x \rightarrow 0^{+}} g(x)=0 \\
& g(0)=0
\end{aligned}\Rightarrow g(x)$ is continuous at $x=0$
At $x=2$
For LHL: $g(x)=|x|-2+|x-2|\begin{aligned}
& g(x)=x-2-(x-2) \\
& g(x)=0
\end{aligned}$
For RHL: $g(x)=|x|-2+|x-2|\begin{aligned}
& g(x)=x-2+x-2 \\
& 2(x)=2 x-4
\end{aligned}$
For $(x=2): g(x)=|x|-2+|x-2|\mathrm{LHL}=\lim _{x \rightarrow 2^{-}} g(x)=0\begin{aligned}
& \mathrm{RHL}=\lim _{x \rightarrow 2^{+}} g(x)=\lim _{x \rightarrow 2} 2 x-4=2(2)-4=0 \\
& g(2)=|2|-2+|2-2| \\
& g(2)=2-2+2-2=0
\end{aligned}\Rightarrow g(x)$ is continuous at $x=2$.
At $x=-1$
For LHL: $g(x)=-2+|-2|=0$
For RHL: $g(x)=-2+|-2|=0$
For $(x=-1): g(x)=-2+|-2|=0\begin{aligned}
& \therefore \mathrm{LHL}=\lim _{x \rightarrow-1^{-}} g(x)=0 \\
& \mathrm{RHL}=\lim _{x \rightarrow-1^{+}} g(x)=0 \\
& g(-1)=0
\end{aligned}$ is differentiable at .