Question

If $f^n\left(x\right)$ be continuous at $x=0,f\left(0\right)\ne 0,f^{\prime }\left(0\right)\ne 0$ and $\underset{x\to 0}{lim}\frac{2f\left(x\right)-3af\left(2x\right)+bf\left(8x\right)}{{\sin }^{2}x}$,exists then values of $a$ and $b$ are

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

Answer 1

The correct option is A $a=\frac{7}{9},b=\frac{1}{3}$

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

Limit exists if ${lim}_{x\to 0}2f\left(x\right)-3af\left(2x\right)+bf\left(8x\right)=0$

$\Rightarrow 2f\left(0\right)-3af\left(0\right)+bf\left(0\right)=0$

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

Given$:f\left(0\right)\ne 0\Rightarrow 2-3a+b=0$

$\Rightarrow 3a-b=2$ ......$\left(1\right)$

Using L-Hospitals's Rule

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

$L={lim}_{x\to 0}\frac{2f^{\prime }\left(x\right)-6a{f}^{\prime }\left(2x\right)+8b{f}^{\prime }\left(8x\right)}{\sin 2x}$ is of $\frac{0}{0}$ form

Limit exists if ${lim}_{x\to 0}2f^{\prime }\left(x\right)-6a{f}^{\prime }\left(2x\right)+8b{f}^{\prime }\left(8x\right)=0$

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

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

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

$\Rightarrow 6a-8b=2$

or $3a-4b=1$ ......$\left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$ we get

$\Rightarrow 3a-b-3a+4b=2-1$

$\Rightarrow 3b=1$ or $b=\frac{1}{3}$

substitute $b=\frac{1}{3}$ in $\left(1\right)$ we get

$3a-b=2$

$\Rightarrow 3a=2+\frac{1}{3}=\frac{6+1}{3}=\frac{7}{3}$

$\Rightarrow a=\frac{7}{9}$

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