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Georgia [21]
3 years ago
12

What are the points of inflection of k(x)=sinx-(1/4)sin2x

Mathematics
1 answer:
zalisa [80]3 years ago
8 0

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

Differentiating with respect to x on both sides we get,

k'(x) = \cos x - \frac{1}{2} \cos 2x

Again, differentiating with respect to x on both sides we get,

k''(x) = - \sin x + \sin 2x = - \sin x + 2 \sin x \cos x

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}

⇒ c = 0 or c = \pm \frac{\pi}{3}

Therefore, option d is correct. (Answer)

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1. If the line that we are searching for is perpendicular to the line y = -4x, this means that the gradient of our line and the gradient of the perpendicular line will multiply to give -1. Thus if we call the gradient of our line m, then:

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