Let f(x) be twice differentiable function such that f′′(0)=2. Then, limx→0 2f(x)−3f(2x)+f(4x)x2, is

Let f(x) be twice differentiable function such that f′′(0)=2. Then, limx02f(x)3f(2x)+f(4x)x2, is

  1. A

    6

  2. B

    3

  3. C

    12

  4. D

    none of these 

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    Solution:

    We have,

    limx02f(x)3f(2x)+f(4x)  x2              00 form 

    =limx02f(x)6f(2x)+4f(4x)2x     

                                                       [Using L' Hospital's Rule] 

    =limx02f′′(x)12f′′(2x)+16f′′(4x)2  

                                                       [Using L' Hospital's Rule] 

    = 2f′′(0)12f′′(0)+16f′′(0)2=  3f′′(0)=3×2=6

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