A) 4
B) \[-2\]
C) \[-4\]
D) 2
Correct Answer: A
Solution :
[a] \[\because f\left( 2 \right)=2\] \[f'\left( 2 \right)=1\] \[\because \underset{x\to 2}{\mathop{lim}}\,\frac{2{{x}^{2}}-4.f(x)}{x-2}\]Putting\[x=2\], then which is\[\frac{0}{0}\]form. \[=\underset{x\to 2}{\mathop{lim}}\,\frac{4x-4.f(x)}{1}\] [By L' Hospital Rule] \[=\frac{4\times 2-4\times 1}{1}=4\]. Hence, option [a] is correct.You need to login to perform this action.
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