Find the number of points at which the line 1 0 0 y = x intersects the curve y = sin ( x ) .
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Did it same way. The line will intersect the curve once in half a period. And limit of x is = 100/3.14 = 31. It will so extend backwards, so 62. And plus one at the origin.
Nice question.
good question... had to think a lot before drawing the graph..
@Omkar Kulkarni What if cos(x) the same?
I did it the same way, but I forgot to include the domain from -1<x<0. It was pretty frustrating, actually.
− 1 ≤ sin x ≤ 1 ⇒ − 1 ≤ 1 0 0 x ≤ 1 ⇒ − 1 0 0 ≤ x ≤ 1 0 0 1 0 0 / 2 π ≈ 1 5 . 9 1 5 5 For each period 1 0 0 x touches to sin x two times. But 0 are counted two times so real roots are 1 6 × 4 − 1 = 6 3
Image credits: @Chew-Seong Cheong
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1 0 0 y = x and y = sin ( x ) → sin ( x ) = 1 0 0 x
As − 1 ≤ sin ( x ) ≤ 1 , − 1 0 0 ≤ 1 0 0 sin ( x ) ≤ 1 0 0 ⇒ − 1 0 0 ≤ x ≤ 1 0 0 .
Now, the period of sine is 2 π . So the number of waves we will consider will be 2 π 2 0 0 ≈ 3 1 . 8 , which we will consider as 3 2 .
The line will intersect each curve twice, and hence we would have 6 4 intersections. But the intersection at ( 0 , 0 ) is common to both waves.
Therefore the answer, 6 3 .