Question

Equation of normal drawn to the graph of the function defined as $f\left(x\right)=\frac{\sin x^2}{x},x\ne 0$ and $f\left(0\right)=0$ at the origin is

• A. $x+y=0$
• B. $x-y=0$
• C. $y=0$
• D. $x=0$

Answer 1

The correct option is A $x+y=0$

$y=\frac{\sin x^2}{x}$

$\frac{dy}{dx}=\frac{\left(x\right)\left(\cos x^2\right)\left(2x\right)-\left(\sin x^2\right)\left(1\right)}{x^2}$

$\frac{dy}{dx}=2\cos x^2-\frac{\sin x^2}{x^2}$

at $(x,y)=(0,0)\Rightarrow \frac{dy}{dx}=\underset{x\to 0}{lim}(2\cos x^2-\frac{\sin x^2}{x^2})$

$=2-1=1$

$\therefore$ Slope of normal $=-\frac{dx}{dy}=-1$

$\therefore$ equation of normal $\Rightarrow \frac{y-0}{x-0}=-1$

$\Rightarrow x+y=0$