Question

If normal curve $y=f\left(x\right)$ at $x=0$ be given by the equation $3x-y+3=0$, then the value of $\underset{x\to 0}{lim}{x}^2{\left(f\right(x^2)+5f(4x^2)+4f(7x^2\left)\right)}^{-1}$

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $-\frac{1}{5}$
• D. None of these
  

Answer 1

The correct option is D None of these

given

normal at $x=0$ is $3xy+3=0$

So $f^{\prime }\left(0\right)=\frac{-1}{3}$ and

$f\left(0\right)=3$

$\underset{x\to 0}{lim}\frac{x^2}{\left(f\right(x^2)+5f(4x^2)+4f(7x^2\left)\right)}$

Using L-Hospital Rule, differentiate

$\underset{x\to 0}{lim}\frac{2x}{\left(2x{f}^{\prime }\right(x^2)+40x{f}^{\prime }(4x^2)+56x{f}^{\prime }(7x^2\left)\right)}$

$\underset{x\to 0}{lim}\frac{2}{\left(2f^{\prime }\right(x^2)+40f^{\prime }(4x^2)+56f^{\prime }(7x^2\left)\right)}$

Now subsititute $f^{\prime }\left(0\right)=-1/3$

$=\frac{-6}{49}$