Question

If $f^{\prime \prime }\left(0\right)=k,k\ne 0$, then the value of $\underset{x\to 0}{lim}\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}$ is

• A. $k$
• B. $2k$
• C. $3k$
• D. $4k$

Answer 1

Answer: (C)

$3k$

Given : $f^{\prime \prime }\left(0\right)=k$, where $k\ne 0$

To find :- $\underset{x\to 0}{lim}\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}$

Consider, $L=\underset{x\to 0}{lim}\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}$

Since, $L$ reduces in $\frac{0}{0}$ form after substituting the limit. So, we apply L' Hoospital's Rule and we get

$L=\underset{x\to 0}{lim}\frac{2f^{\prime }\left(x\right)-3f^{\prime }\left(2x\right)\times 2+f^{\prime }\left(4x\right)\times 4}{2x}$ ......... [By L' Hospital's Rule]

$=\underset{x\to 0}{lim}\frac{2f^{\prime }\left(x\right)-6f^{\prime }\left(2x\right)+4f^{\prime }\left(4x\right)}{2x}$ which again reduces in $\frac{0}{0}$ form after substtuting the limit $x\to 0$

$\therefore L=\underset{x\to 0}{lim}\frac{2f^{\prime \prime }\left(x\right)-6f^{\prime \prime }\left(2x\right)\times 2+4f^{\prime \prime }\left(4x\right)\times 4}{2}$ ...... [By L' Hospital's Rule]

$=\underset{x\to 0}{lim}\frac{2f^{\prime \prime }\left(x\right)-12f^{\prime \prime }\left(2x\right)+16f^{\prime \prime }\left(4x\right)}{2}$

$=\frac{2f^{\prime \prime }\left(0\right)-12f^{\prime \prime }\left(0\right)+16f^{\prime \prime }\left(0\right)}{2}$

$=\frac{18f^{\prime \prime }\left(0\right)-12f^{\prime \prime }\left(0\right)}{2}$

$=\frac{6f^{\prime \prime }\left(0\right)}{2}=\frac{6k}{2}=3k$

Hence, $\underset{x\to 0}{lim}\frac{2f\left(x\right)-3f\left(2x\right)+f\left(4x\right)}{x^2}=3k$