Question

Let $f^{\prime \prime }\left(x\right)$ be continous at $x=0$.

If $\underset{x\to 0}{lim}\frac{2f\left(x\right)-3af\left(2x\right)+bf\left(8x\right)}{{\sin }^{2}x}$ exists and $f\left(0\right)\ne 0,f^{\prime }\left(0\right)\ne 0$, then

• A. $a=-\frac{7}{9}$
• B. $b=\frac{1}{3}$
• C. $a=\frac{7}{9}$
• D. $b=-\frac{1}{3}$

Answer 1

Answer: (B), (C)

The correct options are
B $b=\frac{1}{3}$
C $a=\frac{7}{9}$

We have,

$\underset{x\to 0}{lim}\frac{2f\left(x\right)-3af\left(2x\right)+bf\left(8x\right)}{{\sin }^{2}x}$

$=\underset{x\to 0}{lim}\frac{2f\left(x\right)-3af\left(2x\right)+bf\left(8x\right)}{x^2}$

For the limit to exist, we have

$2f\left(0\right)-3af\left(0\right)+bf\left(0\right)=0$ i.e., $3a-b=2$ $[\because f\left(0\right)\ne 0,$ given$]$ ...(1)

$=\underset{x\to 0}{lim}\frac{2f^{\prime }\left(x\right)-6a{f}^{\prime }\left(2x\right)+8b{f}^{\prime }\left(8x\right)}{2x}$

For the limit to exist, we have

$2f^{\prime }\left(0\right)-6a{f}^{\prime }\left(0\right)=8b{f}^{\prime }\left(0\right)=0$

i.e., $3a-4b=1$ $[\because f(0)\ne 0,$ given$]$ ...(2)

Solving equations (1) and (2), we have

$a=\frac{7}{9}$ and $b=\frac{1}{3}$