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 $x+y=0$

Let the equation of line is $y=mx$

$f\left(x\right)=\frac{\sin x^2}{x}$

Let $g\left(x\right)=\sin x^2$

For $\left|x\right|<\delta$:

$g\left(x\right)=\sin x^2=x^2-\frac{(x^2{)}^3}{3!}+\frac{(x^2{)}^5}{5!}-...=\sum _0^{\infty }(-1{)}^n\frac{(x^2{)}^{(2n+1)}}{(2n+1)!}$

$f\left(x\right)=\frac{x^2-\frac{(x^2{)}^3}{3!}+\frac{(x^2{)}^5}{5!}-...}{x}$

$f\left(x\right)=x-\frac{x^5}{3!}+\frac{x^9}{5!}...$

$\frac{df\left(x\right)}{dx}=1-\frac{5x^4}{3!}+\frac{9x^8}{5!}...$

$m_1=1-\frac{5x^4}{3!}+\frac{9x^8}{5!}...$ $(m_1:$ slope of tangent line at $x=0)$

$m_1=1$

$m=\frac{-1}{m_1}$

$\therefore m=-1$

Equation of normal is:

$y+x=0$

Hence, Option A is correct.