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 the values of $a$ and $b$ are

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

Answer 1

Answer: (A)

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

Given $f\left(0\right)\ne 0$ , $f^{\prime }\left(0\right)\ne 0$ and $f^n\left(x\right)$ is continuous at $x=0$

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

Let the limit be $L$

At $x=0$, ${\sin }^{2}x=0$

Therefore for limit $L$ to exist $2f\left(0\right)-3af\left(0\right)+bf\left(0\right)=0$

We know that $f\left(0\right)\ne 0$

$\Rightarrow 3a-b=2$

Now differentiate numerator and denominator of $L$ wrt $x$

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

At $x=0$, $2\sin x\cos x=0$, therefore for limit $L$ to exist $2f^{\prime }\left(0\right)-6a{f}^{\prime }\left(0\right)+8b{f}^{\prime }\left(0\right)=0$

We know that $f^{\prime }\left(0\right)\ne 0$

$\Rightarrow 3a-4b=1$

By solving above two equations, we get $a=\frac{7}{9}$ and $b=\frac{1}{3}$

Therefore option $A$ is correct