Differentiation Question 506
Question: The value of $ \underset{x\to \pi /2}{\mathop{\lim }},{{\tan }^{2}}x(\sqrt{2{{\sin }^{2}}x+3\sin x+4} $ $ -\sqrt{{{\sin }^{2}}x+6\sin x+2)} $ is equal to
Options:
A) $ \frac{1}{10} $
B) $ \frac{1}{11} $
C) $ \frac{1}{12} $
D) $ \frac{1}{8} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ \underset{x\to \pi /2}{\mathop{\lim }},{{\tan }^{2}}x(\sqrt{2{{\sin }^{2}}x+3\sin x+4}-\sqrt{{{\sin }^{2}}x+6\sin x+2}) $ $ =\underset{x\to \pi /2}{\mathop{\lim }},{{\tan }^{2}}x\frac{(2sin^{2}x+3sinx+4-sin^{2}x-6\sin x-2)}{\sqrt{2{{\sin }^{2}}x+3\sin x+4}+\sqrt{{{\sin }^{2}}x+6\sin x+2}} $ $ =\underset{x\to \pi /2}{\mathop{\lim }},\frac{{{\tan }^{2}}x(sin^{2}x-3sinx+2)}{\sqrt{2{{\sin }^{2}}x+3\sin x+4}+\sqrt{{{\sin }^{2}}x+6\sin x+2}} $ $ =,\underset{x\to \pi /2}{\mathop{\lim }},\frac{{{\sin }^{2}}x(sin,x-1)(sin,x-2)}{(1-sin^{2}x)(\sqrt{2{{\sin }^{2}}x+3\sin x+4}+\sqrt{{{\sin }^{2}}x+6\sin x+2})} $ $ \underset{x\to \pi /2}{\mathop{\lim }},\frac{-{{\sin }^{2}}x(sinx-2)}{(1+sinx)(\sqrt{2{{\sin }^{2}}x+3\sin x+4}+\sqrt{{{\sin }^{2}}x+6\sin x+2})} $ $ =\frac{1}{2(\sqrt{9}+\sqrt{9})}=\frac{1}{12}. $
BETA
