A) 12
B) 10
C) 6
D) 4
Correct Answer: A
Solution :
[a] Given f?? (x) is continuous at x = 0 \[=\underset{x\to 0}{\mathop{\lim }}\,f''(x)=f''(0)=4\] Now, \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2f(x)-3f(2x)+f(4x)}{{{x}^{2}}}\left[ \frac{0}{0}form \right]\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2f'(x)-6f'(2x)+4f'(4x)}{2x}\left[ \frac{0}{0}form \right]\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2f''(x)-12f''(x)+16f''(4x)}{2}\] [Using L? Hospital Rule successively] \[=\frac{2f''(0)-12f''(0)+16f''(0)}{2}=12\]
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