CUET Preparation Today
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Slope of tangent = $f'(x)$ $=12x(4x^2-1)$ $⇒f'(2)=12×2(4(2)^2-1)$ $=24×15=360$ $\text{Slope of Normal} = \frac{-1}{\text{Slope of tangent}}$ $=\frac{-1}{360}$ To find the equation, $y-y_1=Slope(x-x_1)$ $⇒y-3=360(x-2)$ $⇒y=360x-717$ |
