Question

If $f\left(x\right)$ is a thrice differentiable function such that, $\underset{x\to 0}{lim}\frac{f\left(4x\right)-3f\left(3x\right)+3f\left(2x\right)-f\left(x\right)}{x^3}=12$ then the value of $f^{\prime \prime \prime }\left(0\right)$ equal to:

• A. $0$
• B. $1$
• C. $12$
• D. None of these

Answer 1

Answer: (C)

$12$

${lim}_{x\to 0}\frac{f4x)-3f(3x)+3f2x)-f\left(x\right)}{x^3}=12$

Since, the function is thrice differentiable, on differentiating with respect to $x$, we get

${lim}_{x\to 0}\frac{4f^{{}^{\prime }}\left(4x\right)-9f^{{}^{\prime }}\left(3x\right)+6f^{{}^{\prime }}\left(2x\right)-f^{{}^{\prime }}\left(x\right)}{3x^2}$

Again, differentiating with respect to $x$, we get

${lim}_{x\to 0}\frac{16f^{\prime \prime }\left(4x\right)-{27}^f"\left(3x\right)+12f^{\prime \prime }\left(2x\right)-f^{\prime \prime }\left(x\right)}{6x}$

Again differentiating with respect to $x$, we get

$\frac{64f^{{}^{\prime \prime \prime }}\left(4x\right)-81f^{{}^{\prime \prime \prime }}\left(3x\right)+24f^{{}^{\prime \prime \prime }}\left(2x\right)-f^{{}^{\prime \prime \prime }}\left(x\right)}{c}$

(where $x=0$)

$=\frac{64f^{{}^{\prime \prime \prime }}\left(0\right)-81f^{{}^{\prime \prime \prime }}\left(0\right)+24f^{{}^{\prime \prime \prime }}\left(0\right)-f^{{}^{\prime \prime \prime }}\left(0\right)}{6}$

$=\frac{6f^{{}^{\prime \prime \prime }}\left(0\right)}{6}=12$ (given)

$\therefore f^{{}^{\prime \prime \prime }}\left(0\right)=12$