Functions Question 265
Question: The value of k so that the function $ f(x)= \begin{cases} & k(2x-x^{2}),\ \ \ when x<0 \\ & \cos x,when x\ge 0 \\ \end{cases} $ is continuous at $ x=0 $ , is
Options:
A) 1
B) 2
C) 4
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
$ f(0-)=\underset{x\to 0-}{\mathop{\lim }}k(2x-x^{2})=0 $ ; $ f(0+)=\underset{x\to 0+}{\mathop{\lim }}\cos x=1 $
$ \therefore f(0)=\cos x=1 $ Hence no value of k can make $ f(0-)=1. $
BETA
