Limits Continuity And Differentiability Question 10
Question: If $ f(x)= \begin{cases} & ( x^{2}/a )-a,whenx<0 \\ & 0,whenx=a,then \\ & a-( x^{2}/a ),whenx>a \\ \end{cases} . $
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
