Answer:
Option d is correct.
Step-by-step explanation:
The point of inflection of a function y = f(x) at a pointy c is given by f''(c) = 0.
Now, the given function is
![k(x) = \sin x - \frac{1}{4}\sin 2x](https://tex.z-dn.net/?f=k%28x%29%20%3D%20%5Csin%20x%20-%20%5Cfrac%7B1%7D%7B4%7D%5Csin%202x)
Differentiating with respect to x on both sides we get,
![k'(x) = \cos x - \frac{1}{2} \cos 2x](https://tex.z-dn.net/?f=k%27%28x%29%20%3D%20%5Ccos%20x%20-%20%5Cfrac%7B1%7D%7B2%7D%20%5Ccos%202x)
Again, differentiating with respect to x on both sides we get,
![k''(x) = - \sin x + \sin 2x = - \sin x + 2 \sin x \cos x](https://tex.z-dn.net/?f=k%27%27%28x%29%20%3D%20-%20%5Csin%20x%20%2B%20%5Csin%202x%20%3D%20-%20%5Csin%20x%20%2B%202%20%5Csin%20x%20%5Ccos%20x)
So, the condition for point of inflection at point c is
k''(c) = 0 = - sin c + 2 sin c cos c
⇒ sin c(2cos c - 1) = 0
⇒ sin c = 0 or ![\cos c = \frac{1}{2}](https://tex.z-dn.net/?f=%5Ccos%20c%20%3D%20%5Cfrac%7B1%7D%7B2%7D)
⇒ c = 0 or
Therefore, option d is correct. (Answer)