Differentiation Question 205
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} $
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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(sinx-1)(sinx-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
