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The equation of the normal to the curve $y = sin x $ at (0, 0) is : |
$x-y= 0 $ $x+y= 0 $ $x=0$ $y=0$ |
$x+y= 0 $ |
The correct answer is Option (2) → $x+y= 0$ $y=\sin x$ $y'=\cos x$ $\left.y'\right]_{(0,0)}=1$ So slope of normal = $\frac{-1}{\left.y'\right]_{(0,0)}}=-1$ so eq: $y-0=-1(x-0)$ $⇒x+y= 0 $ |
