JEE Main 12 Jan 2019 Morning Question 27
Question: $ \underset{x\to \pi /4}{\mathop{\lim }},\frac{{{\cot }^{3}}x-\tan x}{\cos ( x+\frac{\pi }{4} )} $ is [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) $ 4\sqrt{2} $
B) $ 8\sqrt{2} $
C) 4
D) 8
Show Answer
Answer:
Correct Answer: D
Solution:
$ \underset{x\to \frac{\pi }{4}}{\mathop{\lim }},\frac{{{\cot }^{3}}x-\tan x}{\cos ( x+\frac{\pi }{4} )} $
$ =\underset{x\to \frac{\pi }{4}}{\mathop{\lim }},\frac{1-{{\tan }^{4}}x}{\cos ( x+\frac{\pi }{4} )}=2\underset{x\to \frac{\pi }{4}}{\mathop{\lim }},\frac{1-{{\tan }^{2}}x}{\cos ( x+\frac{\pi }{4} )} $
$ =2\underset{x\to \frac{\pi }{4}}{\mathop{\lim }},\frac{{{\cos }^{2}}x-{{\sin }^{2}}x}{\frac{1}{\sqrt{2}}(\cos ,x-\sin x)}.\frac{1}{{{\cos }^{2}}x} $ $ =4\sqrt{2},\underset{x\to \frac{\pi }{4}}{\mathop{\lim }},(\cos ,x+\sin x)=8 $
