A) Exists and is equal to -2
B) Does not exist
C) Exist and is equal to 0
D) Exists and is equal to 2
Correct Answer: A
Solution :
\[f(x)=\left| \begin{matrix} \cos x & x & 1 \\ 2\sin x & {{x}^{2}} & 2x \\ \tan x & x & 1 \\ \end{matrix} \right|\] \[=\cos x({{x}^{2}}-2{{x}^{2}})-x(2\sin x-2x\tan x)+1(2x\sin x-{{x}^{2}}\tan x)\] \[=-{{x}^{2}}\cos x-2x\sin x+2{{x}^{2}}\tan x+2x\sin x-{{x}^{2}}\tan x\] \[={{x}^{2}}\tan x-{{x}^{2}}\cos x\] \[={{x}^{2}}(\tan x-\cos x)\] \[\int_{{}}^{'}{(x)=2x(\tan x-\cos x)+{{x}^{2}}({{\sec }^{2}}x+\sin x)}\] \[\therefore \underset{x\to 0}{\mathop{\lim }}\,\frac{f'(x)}{x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{2x(\tan x-\cos x)+{{x}^{2}}({{\sec }^{2}}x+\sin x)}{x}\] \[=2(0-1)+0=-2\] \[Hence,\therefore \underset{x\to 0}{\mathop{\lim }}\,\frac{f'(x)}{x}=-2\]
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