Question

Let $f^{\prime \prime }\left(x\right)$ be continuous 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 the value of $\frac{3a}{b}$ is ?

• A. $7$
• B. $\frac{7}{9}$
• C. $\frac{1}{7}$
• D. $\frac{1}{3}$

Answer 1

Answer: (A)

$7$

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

$=\frac{(2-3a+b)f\left(0\right)}{0}$

but given limit exists$\Rightarrow (2-3a+b)f\left(0\right)=0$

but $f\left(0\right)\ne 0$

$\Rightarrow 2-3a+b=0\to 1$

So now it is in $\frac{0}{0}$ form

${lim}_{x\to 0}\frac{2f^{\prime }\left(x\right)-3a{f}^{\prime }\left(2x\right)\times 2+b{f}^{\prime }\left(8x\right)\times 8}{2\sin x\cos x}$

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

but again limit exists

$\Rightarrow 2-6a+8b=0\to 2$

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

$=\frac{2-12a+64b}{2}\times f^{\prime \prime }\left(0\right)$

From $1$ and $2$

$3a=7b\Rightarrow \frac{3a}{b}=7$