Question

Equation of normal to curve \(y=x+\frac12sin2x\) at \(x=-\frac{\Pi}{2}\) is:

• A. \(x+\Pi=0\)
• B. \(x=2\Pi\)
• C. \(2x+\Pi=0\)
• D. \(x-\Pi=0\)

Answer 1

The Correct Option is C

To find the equation of the normal to the curve \(y = x + \frac{1}{2}\sin 2x\) at \(x=-\frac{\pi}{2}\), follow these steps:

1. Find the derivative of the curve to determine the slope of the tangent at any point \(x\).
   \[\frac{dy}{dx} = 1 + \cos 2x\]
2. Substitute \(x = -\frac{\pi}{2}\) into \(\frac{dy}{dx}\) to find the slope of the tangent.
   \[\frac{dy}{dx}\bigg|_{x = -\frac{\pi}{2}} = 1 + \cos\left(2 \times -\frac{\pi}{2}\right) = 1 + \cos(-\pi) = 1 - 1 = 0\]
3. A slope of \(0\) means the tangent is horizontal, hence its normal will be vertical. A vertical line through \(x = -\frac{\pi}{2}\) has an equation of the form:\[\text{Normal equation: }x = -\frac{\pi}{2}\]
4. Rewriting the equation of normal: The equation \(x = -\frac{\pi}{2}\) implies the normal line is constant at this x-value. To have it terms with 0 on one side, rewrite as:\[2x + \pi = 0\]

Therefore, the equation of the normal to the curve at \(x = -\frac{\pi}{2}\) is \(2x + \pi = 0\).