Recent content by vcsharp2003

It seems that none of the options given in the challenge problem under post#13 is true. I counted the number of intersections in post#10 to be greater than 16. I also counted the number of intersections as 2 in post#3 . In that case, shouldn't this challenge problem be wrong?

Its a Math book titled Mathematics 11 that is used in Grade 11 in schools of Ontario, Canada. The publisher is McGraw-Hill Ryerson and it has multiple authors from various school boards of Ontario, Canada. You can see a full legal copy of it at the Internet Archive using the following link...

Thank you so much for this example. It helped greatly in understanding the problem. It seems that the number of intersections has no upper limit.

Sorry, you're correct. I didn't paste the original question at the start. The exact problem as it appears in the Math book is as below. From what I have gathered so far from all the replies is that the answer to this exact question is none of the given options since as the curvature of the...

Yes, the problem seems accurate. It's a challenge problem in a high school text book.

Please give me the clue as I have tried your suggestion, but still I can't see the logic.

If it's a straight line there would be many peaks of the sin wave that it would touch. If we take it further down so it is on the x-axis then also there would be many points on the sine curve that it would intersect.

I tried your recommendation on the desmos.com graphing calculator and I was always getting 2 points of intersection for various circles that I tried. Below are three scenarios that I tried, and they all always intersect at two points. But logically, I cannot arrive at this conclusion of two...

The question is asking for the number of different solutions to the following two equations. $$y=\sin{x}$$ $$(x-a)^2 + (y-b)^2 = r^2$$ Solving these is complex for me due to one of the equations being a trigonometric function. If I substitute y from the first equation into the second equation...

Shouldn't ##m^2-24n \gt 0## rather than ##m^2-24n \geq 0##, since the last term after completing the squares is ##\frac {3}{4} B^2## which will always be positive because ##b \approx 212##?

I think the reason its under Calculus is because the first chapter in our Calculus course is Functions where the concept of ##f(x)## is explained. Here we need to use that concept since we need to know what ##f(1)##,##f(2)## and ##f(3)## mean.

After understanding the question based on your post, I tried to go about it, but its far too complex since another requirement we have is that no calculator is allowed. My analysis is as below. Let ##a=f(1)##, ##b=f(2)## and ##c=f(3)##. I can easily see that ##a= 102 + 7\sin{1} \approx 102##...

The angle in these trigonometric ratios of sin 1, sin 2 and sin 3 must be treated as radians. 1 radian is about 57 degrees, 2 radians is about 114 degrees and 3 radians is about 171 degrees. So, we cannot use the assumption that sin x = x since the angle x in these cases are nowhere close to...

Thanks for your answer. I was assuming that y needs to be substituted by f(x) resulting in an equation involving x, which was what was making it impossible to solve. The reason why I assumed this is because in Calculus we generally take it for granted that y=f(x).

When I look at the left hand side of the equation in above question then I can see that the highest degree of x would be 6 after the denominators are eliminated. I know that a polynomial of degree n will have n roots, but this one is not a pure polynomial since there is also a trigonometric...