A) \[2\]
B) \[-2\]
C) \[1\]
D) \[-1\]
Correct Answer: B
Solution :
[b] We have, \[f(x)=\left| \begin{matrix} \cos & x & 1 \\ 2\sin x & {{x}^{2}} & 2x \\ \tan x & x & 1 \\ \end{matrix} \right|=\left| \begin{matrix} \cos x-\tan x & 0 & 0 \\ 2\sin x & {{x}^{2}} & 2x \\ \tan x & x & 1 \\ \end{matrix} \right|\][Applying \[{{R}_{1}}\to {{R}_{1}}-{{R}_{3}}\]] \[=(\cos \,x-\tan \,x)({{x}^{2}}-2{{x}^{2}})\] [Expanding along \[{{R}_{1}}\]] \[=-{{x}^{2}}(\cos \,x-\tan \,x)\] \[\therefore \,f'(x)=-2x(\cos \,x-\tan x)-{{x}^{2}}(-sinx-{{\sec }^{2}}x)\] \[\therefore \underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{f'(x)}{x} \right]=\underset{x\to 0}{\mathop{\lim }}\,-2(\cos \,\,x\,-\tan x)\] \[+x(\sin x+{{\sec }^{2}}x)\] \[=-2\times 1=-2\]
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