Equation of normal to the curve \(y = x + \displaystyle\frac{1}{2}\sin 2x\) at \(x = -\displaystyle\frac{\pi}{2}\) is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(2x + \pi = 0\)
The point on the curve at \(x = -\frac{\pi}{2}\):
\( \displaystyle y = -\frac{\pi}{2} + \frac{1}{2}\sin(-\pi) = -\frac{\pi}{2} \)
The slope of the tangent is dy/dx:
\( \displaystyle \frac{dy}{dx} = 1 + \frac{1}{2}\cdot 2\cos 2x = 1 + \cos 2x \)
At \(x = -\frac{\pi}{2}\):
\( \displaystyle \frac{dy}{dx} = 1 + \cos(-\pi) = 1 - 1 = 0 \)
The tangent is horizontal, so the normal is vertical: it is the line x = constant through the point \(\left(-\frac{\pi}{2}, -\frac{\pi}{2}\right)\).
\( \displaystyle x = -\frac{\pi}{2} \)
\(2x + \pi = 0\)
Note: This topic was not in the CUET 2026 syllabus.