Question

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

• A. $-\frac{1}{5}$
• B. $-\frac{1}{4}$
• C. $-\frac{1}{3}$
• D. $-\frac{1}{2}$

Answer 1

The correct option is C $-\frac{1}{3}$

$\because y=f\left(x\right)$
$\therefore \frac{dy}{dx}=f^{\prime }\left(x\right)\Rightarrow \frac{dy}{dx}{|}_{x=0}=f^{\prime }\left(0\right)$
$\therefore$ Slope of normal = - $\frac{1}{f^{\prime }\left(0\right)}=3\Rightarrow f^{\prime }\left(0\right)=-\frac{1}{3}$
Now, $\underset{x\to 0}{lim}\frac{x^2}{\left\{f\right(x{)}^2-5f\left(4x^2\right)+4f\left(7x^2\right)\}}$
Replacing $x^2$ by x, then
$\underset{x\to 0}{lim}\frac{x}{f\left(x\right)-5f\left(4x\right)+4f\left(7x\right)}=\underset{x\to 0}{lim}\frac{x}{\left(f\right(x)-f(0\left)\right)-5\left(f\right(4x)-f(0\left)\right)+4\left(f\right(7x)-f(0\left)\right)}=\underset{x\to 0}{lim}\frac{1}{\left(\frac{f\left(x\right)-f\left(0\right)}{x-0}\right)-5\left(\frac{f\left(4x\right)-f\left(0\right)}{4x-0}\right)+4(\frac{f\left(7x\right)-f\left(0\right)}{7x-0}\times 7)}=\frac{1}{f^{\prime }\left(0\right)-20f^{\prime }\left(0\right)+28f^{\prime }\left(0\right)}=\frac{1}{9f^{\prime }\left(0\right)}=\frac{1}{9}\times -3=-\frac{1}{3}$