Awesome solution. Just to show variety you can do it as well by connecting PC to get a diagonal of the square and then get it’s length by using this AP^2+PC^2=BP^2+PD^2 (can be proven by Pythagoras) and get PC = sqrt(3) then get the side of square and apply cosine rule on triangle APB
Follow up the following link if you want to solve grade 11 or o-level mathematics like a genius ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-YAM-A3vAXhk.htmlsi=p_qOxs7SaHiezejV
Thanks for this video.But I want to ask something.q 's minimum value is not 16.q can be equal to 11.For exactly p=6 and q=11 5/11 < 6/11 < 4/7 If you look to this unequation , it is True. So why we don't take 11? Also I want to explain this solution: Please tell me if it is True or false. Firstly if these numbers are positive , we can multiply everywhere with 7q.So we will get 35<7p<4q , so p can be 6.And q can be 11 for the minimum value. (If these p,q are positive intigers.)
Hi, thanks for the comment. I double checked the scenario when p=6 and q=11, and it does not satisfy 5/9<p/q<4/7. 5/9 is not less than p/q in this case. I think you read the question wrong when u said 5/11<6/11<4/7. It is supposed to be 5/9<6/11<4/7 which is not true.
To find the range of y, we can analyze the behavior of the function y=3^x. When x increases, the value of y gets very large, thus 2/(3^x+1) approaches 0, which makes y get very close to 1. When x decreases, the value of y gets very small, thus 2/(3^x+1) approaches 2, which makes y get very close to -1. So the interval would be (-1,1).
Bro this is Olympiad question I solved it in less than 5 seconds by seeing the thumbnail, wtf dawn easy, if this is Olympiad level than u will cry after seeing JEE advanced maths papers
I am curious that whether the following solution is acceptable or not. Just give a number for the value of the function such as 1 for positive 1 we need x1 = 3 for negative 1 we need x2 = 3/2 then 1/3 + 2/3 = 1 same answer however I am not sure that it is accepted
Point M was not part of the original question. First, we should find the center and radius of this circle. We can find the center by drawing the perpendicular bisectors of WZ and XY and finding their intersection point. This point happens to be the midpoint of AB, the hypotenuse, which I set as M. Hopefully this makes the problem more clear for you.
It may not be obvious for some how you got from the x^2(y^2+1)+(y^2+1) step to (x^2+1)(y^2+1). You should also mention that the (y^2+1) multiplier was also factored out from both terms leaving (x^2+1). Just for clarity’s sake.
This function is periodic over the integers with a period of 2. You can start evaluating f(6) to move forward to see a period of 2, and you can also move backward with a little more effort. The assumption is that this function is defined over the integers.
you can find the inverse of the first one, f(x) = ax + b => f^-1(x) = (x - b)/a. Then equal to the other inverse, (x - b) / a = bx + a <=> x - b = abx + a^2, you find that: x = abx <=> ab = 1, and a^2 = -b. Furthermore, b = -a^2 and replacing in the other one: a(-a^2) = 1 <=> a = -1. Finally, (-1)^2 = - b, b = - 1, so a + b = -1 + (-1) = -2.
I found another way to find the solution. Step 1) Find the inverse f^-1(x) of f(x) while assuming that the given f^-1(x) is different to the actual f^-1(x). Let's call the given f^-1(x) from now on g(x). Well, you'll get f^-1(x) = 1/a * x - b/a as the inverse of f(x) . Step 2) Equate the calculated f^-1(x) and g(x), which means that 1/a * x - b/a has to be equal to bx + a. Step 3) That leads to a system of equations with I) 1/a = b and II) -b/a = a, which is a solvable system of equation. You'll get a = -1 and b = -1. Step 4) Calculate a + b = -1 + (-1) = -2 (sry for my bad english)
b(ax+b)+a=x, abx+b²+a=x. ab=1, b²+a=0. a=1/b, b²+1/b=0. b isn't 0 or ∞, so b³+1=0. Meaning b is one of the 3 cube roots of -1. Meanwhile, a, being its reciprocal, is b's conjugate. Either a=b=-1, or a=(1±√(3)i)/2 and b=(1-±√(3)i)/2. Meaning either a+b=-2, or a+b=1. The two solutions are -2 and 1.
