A) 1
B) 2
C) 3
D) ?2
Correct Answer: B
Solution :
Given \[f(2)=4,\ \ f'(2)=1\] \[\therefore \underset{x\to 2}{\mathop{\lim }}\,\frac{xf(2)-2f(x)}{x-2}=\underset{x\to 2}{\mathop{\lim }}\,\frac{xf(2)-2f(2)+2f(2)-2f(x)}{x-2}\] \[=\underset{x\to 2}{\mathop{\lim }}\,\frac{(x-2)f(2)}{x-2}-\underset{x\to 2}{\mathop{\lim }}\,\frac{2f(x)-2f(2)}{x-2}\] \[=f(2)-2\underset{x\to 2}{\mathop{\lim }}\,\frac{f(x)-f(2)}{x-2}=f(2)-2f'(2)=4-2(1)=4-2=2\] Aliter: Applying L-Hospital rule, we get \[\underset{x\to 2}{\mathop{\lim }}\,\frac{f(2)-2f'(2)}{1}=2\]
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