Functions Question 414
Question: The function defined by $ f(x),=, \begin{cases} {{( x^{2}+{e^{\frac{1}{2-x}}} )}^{-1}} & , & x\ne 2 \\ k & , & x=2 \\ \end{cases} . $ , is continuous from right at the point x = 2, then k is equal to
[Orissa JEE 2002]
Options:
A) 0
B) 1/4
C) ?1/4
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ f(x)={{[ x^{2}+{e^{\frac{1}{2-x}}} ]}^{-1}} $ and $ f(2)=k $ If $ f(x) $ is continuous from right at $ x=2 $ then $ \underset{x\to {2^{+}}}{\mathop{\lim }},f(x)=f(2)=k $
Þ $ \underset{x\to {2^{+}}}{\mathop{\lim }},{{[ x^{2}+{e^{\frac{1}{2-x}}} ]}^{-1}}=k $
Þ $ k=\underset{h\to 0}{\mathop{\lim }},f(2+h) $
Þ $ k=\underset{h\to 0}{\mathop{\lim }}{{[ {{(2+h)}^{2}}+{e^{\frac{1}{2-(2+h)}}} ]}^{,-1}} $
Þ $ k=\underset{h\to 0}{\mathop{\lim }}{{[ ,4+h^{2}+4h+{e^{-1/h}}, ]}^{,-1}} $
Þ $ k={{[4+0+0+{e^{-\infty }}]}^{,-1}} $
Þ $ k=\frac{1}{4} $ .
BETA
