Functions Question 23
Question: If $ f(x)= \begin{cases} & ,\frac{1-\cos 4x}{x^{2}},\ \ when,x<0 \\ & ,a,,whenx=0 \\ & \frac{\sqrt{x}}{\sqrt{(16+\sqrt{x})}-4},when,x>0 \\ \end{cases} . $ , is continuous at $ x=0 $ , then the value of ‘a’ will be
[IIT 1990; AMU 2000]
Options:
A) 8
B) ?8
C) 4
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \underset{x\to 0-}{\mathop{\lim }},f(x)=\underset{x\to 0-}{\mathop{\lim }},( \frac{2,{{\sin }^{2}}2x}{{{(2x)}^{2}}} ),4=8 $ $ \underset{x\to 0+}{\mathop{\lim }},f(x)=\underset{x\to 0+}{\mathop{\lim }},\sqrt{16+\sqrt{x}+4}=8 $ . Hence $ a=8 $ .
BETA
