Answer:
Option d is correct.
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 - (1)/(4)\sin 2x](https://img.qammunity.org/2021/formulas/mathematics/middle-school/crs42w29z4ws04dhavb5j8ukj03n5cpbtg.png)
Differentiating with respect to x on both sides we get,
![k'(x) = \cos x - (1)/(2) \cos 2x](https://img.qammunity.org/2021/formulas/mathematics/middle-school/cls5zejazj0z8ctf38r1ex0ihymwytc0w8.png)
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://img.qammunity.org/2021/formulas/mathematics/middle-school/zdlmdz7hiqh61dle7yczapgf2woie77ldz.png)
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 = (1)/(2)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/v4oubmfyqlldpfncncvuiwno40j2gzgq99.png)
⇒ c = 0 or
Therefore, option d is correct. (Answer)