Q70 / 155 · WBJEE 2017 · Mathematics

If $f^{\prime \prime}(0)=k, k \neq 0,$ then the value of $\lim _{x \rightarrow 0} \frac{2 f(x)-3 f(2 x)+f(4 x)}{x^{2}}$ is

  • A
    $k$
  • B
    $2 k$
  • Correct answer
    C
    $3 k$
  • D
    $4 k$

Solution

$\lim _{x \rightarrow 0} \frac{2 f(x)-3 f(2 x)+f(4 x)}{x^{2}}$ $=\lim _{x \rightarrow 0} \frac{2 f^{\prime}(x)-3 f^{\prime}(2 x) \cdot 2+f^{\prime}(4 x) \cdot 4}{2 x}$ $=\lim _{x \rightarrow 0} \frac{f^{\prime}(x)-3 f^{\prime}(2 x)+2 f^{\prime}(4 x)}{x}$ $=\lim _{x \rightarrow 0} \frac{f^{\prime \prime}(x)-3 f''{(2 x)} \cdot 2+2 f^{\prime \prime}(4 x) \cdot 4}{1}$ $=\lim _{x \rightarrow 0} f^{\prime \prime}(x)-6 f''{(2 x)}+8 f''(4 x)$ $=f^{\prime \prime}(0)-6 f^{\prime \prime}(0)+8 f''{(0)}$ $=k-6 k+8 k$ $=3 k$

Hence, the correct answer is option: 'C'.

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