The slope of normal to the curve \(y = 3x^2 + 3\sin 3x\), at x = 0 is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(-\displaystyle\frac{1}{9}\)
The slope of the tangent is dy/dx:
\( \displaystyle \frac{dy}{dx} = 6x + 9\cos 3x \)
At x = 0:
\(6(0) + 9\cos 0 = 9\)
The normal is perpendicular to the tangent, so its slope is the negative reciprocal:
\( \displaystyle \text{slope of normal} = -\frac{1}{9} \)