Functions Question 566
Question: If $ f(x)= \begin{cases} & (x^{2}/a)-a,\ \ when\ x<a \\ & \ \ \ \ \ \ \ \ \ \ \ 0,\ \ when\ x=a\text{,} \\ & a-(x^{2}/a),\ \ \text{when }x>a \\ \end{cases} . $ then
Options:
A) $ \underset{x\to a}{\mathop{\lim }},f(x)=a $
B) $ f(x) $ is continuous at $ x=a $
C) $ f(x) $ is discontinuous at $ x=a $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ f(a)=0 $ $ \underset{x\to a-}{\mathop{\lim }}f(x)=\underset{x\to a-}{\mathop{\lim }},( \frac{x^{2}}{a}-a )=\underset{h\to 0}{\mathop{\lim }}{ \frac{{{(a-h)}^{2}}}{a}-a }=0 $ and $ \underset{x\to a+}{\mathop{\lim }},f(x)=\underset{h\to 0}{\mathop{\lim }},{ a-\frac{{{(a+h)}^{2}}}{a} }=0 $ Hence it is continuous at $ x=a $ .
BETA
