Functions Question 212
Question: Let $ f(x)=\frac{1-\tan x}{4x-\pi },\ x\ne \frac{\pi }{4},\ \ x\in
[ 0,\frac{\pi }{2} ] $ , If $ f(x) $ is continuous in $ [ 0,\frac{\pi }{2} ] $ , then $ f( \frac{\pi }{4} ) $ is [AIEEE 2004]
Options:
A) -1
B) $ \frac{1}{2} $
C) $ -\frac{1}{2} $
D) 1
Show Answer
Answer:
Correct Answer: C
Solution:
$ \underset{x\to \frac{\pi }{4}}{\mathop{\lim }},f(x)=\underset{x\to \frac{\pi }{4}}{\mathop{\lim }},\frac{1-\tan x}{4x-\pi },,[ \frac{0}{0}form ] $
$ \underset{x\to \frac{\pi }{4}}{\mathop{\lim }},\frac{-{{\sec }^{2}}x}{4}=\frac{-2}{4}=\frac{-1}{2} $ .
\ For f(x) to be continuous at $ x=\frac{\pi }{4},\ f( \frac{\pi }{4} )=\frac{-1}{2} $
BETA
