A) \[k\]
B) \[2k\]
C) \[3k\]
D) \[4k\]
E) \[\frac{k}{3}\]
Correct Answer: C
Solution :
\[\because \]\[f\,(0)=k\]and\[\underset{x\to 0}{\mathop{\lim }}\,\frac{2f(x)-3f(2x)+f(4x)}{{{x}^{2}}}\] Using L Hospitals rule \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2f(x)-6f(2x)+4f\,(4x)}{2x}\] Again using L Hospitals rule \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2f\,(x)-12f\,(2x)+16f\,(4x)}{2}\] \[=\frac{1}{2}[2f\,(0)-12f\,(0)+16f\,(0)]\] \[=\frac{1}{2}[2k-12k+16k]=3k\]
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